# Vector Partition Identities for $2$D, $3$D and $n$D Lattices

Geoffrey B Campbell  Mathematical Sciences Institute, The Australian National
University, Canberra, ACT, 0200, Australia<EMAIL_ADDRESS>
###### Abstract.

We prove identities generating higher dimensional vector partitions. We derive
theorems for integer lattice points in the 2D first quadrant, then generalize
the approach to find 3D and $n$-space lattice point vector region extensions.
We also state combinatorial identities for Visible Point Vectors in 2D up to
5D and $n$D first hyperquadrant and hyperpyramid lattices. 2D and 3D theorems
for vector partitions with binary components are also derived.

###### Key words and phrases:

Exact enumeration problems, generating functions. Partitions of integers.
Elementary theory of partitions. Combinatorial identities, bijective
combinatorics. Lattice points in specified regions.

###### 2010 Mathematics Subject Classification:

Primary: 05A15; Secondary: 05E40, 11Y11, 11P21

Thanks to Professor Dr Henk Koppelaar, whose suggestions helped summarize, for
this paper, some chapters of the author’s draft book.

##  1\. Preview of results

We show the following selected sample results from this paper. These examples
are intended to inform the reader of a few notable points of content along the
way.

Preview 1. The number of first quadrant 2D partitions of vectors into exactly
three parts $\clubsuit_{2}(a,b)$, is for $|y|<1$, $|z|<1$, generated by

$\sum_{a,b\geq
0}^{\infty}\clubsuit_{2}(a,b)y^{a}z^{b}=\frac{1}{3!}\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1&0\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}&-2\\\
\frac{1}{\left(1-y^{3}\right)\left(1-z^{3}\right)}&\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}$
$=\frac{y^{3}z^{3}+y^{2}z^{2}+y^{2}z+yz^{2}+yz+1}{(1-y)^{3}(1+y)(y^{2}+y+1)(1-z)^{3}(1+z)(z^{2}+z+1)}.$

Preview 2. If $|v|<1,|w|<1,|x|<1,|y|<1,|z|<1$ and $q+r+s+t+u=1$, then

$\prod_{\begin{subarray}{c}\gcd(a,b,c,d,e)=1\\\ a,b,c,d,e\geq
1\end{subarray}}\left(\frac{1}{1-v^{a}w^{b}x^{c}y^{d}z^{e}}\right)^{1/(a^{q}b^{r}c^{s}d^{t}e^{u})}$
$=\exp\left\\{\left(\sum_{g=1}^{\infty}\frac{v^{g}}{g^{q}}\right)\left(\sum_{h=1}^{\infty}\frac{w^{h}}{h^{r}}\right)\left(\sum_{i=1}^{\infty}\frac{x^{i}}{i^{s}}\right)\left(\sum_{j=1}^{\infty}\frac{y^{j}}{j^{t}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{k}}{k^{u}}\right)\right\\}.$

Preview 3. For each of $|y|,|z|<1,$

$\prod_{\begin{subarray}{c}(a,b,c,d,e)=1\\\ a,b,c,d<e\\\ a,b,c,d\geq
0,e>0\end{subarray}}\left(\frac{1}{1-y^{a+b+c+d}z^{e}}\right)^{1/e}=\left(\frac{(1-yz)^{4}(1-y^{3}z)^{4}}{(1-z)(1-y^{2}z)^{6}(1-y^{4}z)}\right)^{1/(1-y)^{4}}$
$=1+\frac{z}{1!}+\begin{vmatrix}1&-1\\\
\left(\frac{1-y^{2}}{1-y}\right)^{4}&1\\\
\end{vmatrix}\frac{z^{2}}{2!}+\begin{vmatrix}1&-1&0\\\
\left(\frac{1-y^{2}}{1-y}\right)^{4}&1&-2\\\
\left(\frac{1-y^{3}}{1-y}\right)^{4}&\left(\frac{1-y^{2}}{1-y}\right)^{4}&1\\\
\end{vmatrix}\frac{z^{3}}{3!}$ $+\begin{vmatrix}1&-1&0&0\\\
\left(\frac{1-y^{2}}{1-y}\right)^{4}&1&-2&0\\\
\left(\frac{1-y^{3}}{1-y}\right)^{4}&\left(\frac{1-y^{2}}{1-y}\right)^{4}&1&-3\\\
\left(\frac{1-y^{4}}{1-y}\right)^{4}&\left(\frac{1-y^{3}}{1-y}\right)^{4}&\left(\frac{1-y^{2}}{1-y}\right)^{4}&1\\\
\end{vmatrix}\frac{z^{4}}{4!}+etc.$

Preview 4. For $|y|<1$, $|z|<1$,

$\prod_{\begin{subarray}{c}m,n\geq 1\\\ m\leq
n;\,(m,n)=1\end{subarray}}\left(\frac{1}{1-y^{m}z^{n}}\right)^{m/n^{2}}=\left(1-yz\right)^{\frac{y}{1-y}}\exp\left\\{\frac{y}{(1-y)^{2}}\left(Li_{2}(z)-Li_{2}(yz)\right)\right\\}.$

##  2\. Defining Vector Partitions

With respect to vector partitions, herein we study the additive decomposition
of vectors $v=\langle a_{1},a_{2},\ldots,a_{n}\rangle$ where each of the
components $a_{1},a_{2},\ldots,a_{n}$ are an integer, usually a non-negative
integer, but this may vary by situation. A partition, whether it be for
integer partitions or it be for vector partitions, is a partially ordered set
or poset. For an integer partition of $n$ say
$\lambda_{1}+\lambda_{2}+\ldots+\lambda_{r}=n$ this means we impose that
$\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{r}\geq 0$ for $\lambda$ as
integers. For a vector partition of $v$ say
$\lambda_{1}+\lambda_{2}+\ldots+\lambda_{r}=v$ this means we impose that
$\|\lambda_{1}\|\geq\|\lambda_{2}\|\geq\ldots\geq\|\lambda_{r}\|$ for each
$\lambda$ as an nD ($n$-dimensional, or $n$-component) vector.

######  Definition 2.1.

A partition of an $n$-dimensional vector $v=\langle
a_{1},a_{2},\ldots,a_{n}\rangle$ is a finite partially ordered sequence of
vectors $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}$ such that
$\sum_{i=1}^{r}\lambda_{i}=v$ under vector addition. The $\lambda_{i}$ vectors
are called parts of the vector partition. A suitable partial ordering for $v$
is for example, a non-increasing norm defined by
$\|v\|=\sqrt{\sum_{i=1}^{n}a_{i}^{2}}$ applied to the $\lambda_{i}$.

We will strive to obtain theorems and conjectures with regard to nD vectors
generally with positive integer $n$, but mostly we shall begin with or attempt
to illustrate or justify these theorems and conjectures by first considering
the 2D and 3D situations, since they are most easily visualized in the
Cartesian co-ordinate settings. We say the $p_{n}(\langle
a_{1},a_{2},\ldots,a_{n}\rangle)$ abbreviated to
$p_{n}(a_{1},a_{2},\ldots,a_{n})$ or simply $p_{n}(v)$ is the number of nD
partitions of vector $v$. Here are a few 2D examples taking $a,b>0$ so
excluding the often included $a=0,\,b=0$.

$p_{2}(1,1)=2;\quad\langle 1,1\rangle=(\langle 1,1\rangle^{1});$

$p_{2}(2,1)=1;\quad\langle 2,1\rangle=(\langle 2,1\rangle^{1});$

$p_{2}(1,2)=1;\quad\langle 1,2\rangle=(\langle 1,2\rangle^{1});$

$p_{2}(2,2)=2;\quad\langle 2,2\rangle=(\langle 2,2\rangle^{1});\quad\langle
1,1\rangle+\langle 1,1\rangle=(\langle 1,1\rangle^{2});$

$p_{2}(3,2)=2;\quad\langle 3,2\rangle=(\langle 3,2\rangle^{1});\quad\langle
2,1\rangle+\langle 1,1\rangle=(\langle 2,1\rangle^{1}\langle 1,1\rangle^{1}).$

##  3\. Vector or Multipartite Partitions from George E Andrews

Andrews [2, Chapter 12, page 203] defines
$P(\textbf{n})=P(n_{1},n_{2},...,n_{r})$ as the number of unrestricted
partitions of the ”$r$-partite” or ”multipartite” number
$(n_{1},n_{2},...,n_{r})$ as an ordered $r$-tuple of non-negative integers not
all zero. In our vector partitions theory we mostly would say that this
multipartitie number is the vector $\mathbf{n}=\langle
n_{1},n_{2},...,n_{r}\rangle$, defining a lattice point in Euclidean
$r$-space. Similarly the $Q(\textbf{n})$ function is the number of partitions
of vectors n into distinct parts. Andrews states the two generating functions

(3.1) $\displaystyle\sum_{n_{1},n_{2},...,n_{r}\geq
0}P(\textbf{n})x_{1}^{n_{1}}...x_{r}^{n_{r}}$ $\displaystyle=$
$\displaystyle\prod_{\begin{subarray}{c}n_{1},n_{2},...,n_{r}\geq 0\\\
not\,all\,zero\end{subarray}}\frac{1}{1-x_{1}^{n_{1}}...x_{r}^{n_{r}}}$ (3.2)
$\displaystyle\sum_{n_{1},n_{2},...,n_{r}\geq
0}Q(\textbf{n})x_{1}^{n_{1}}...x_{r}^{n_{r}}$ $\displaystyle=$
$\displaystyle\prod_{n_{1},n_{2},...,n_{r}\geq
0}(1+x_{1}^{n_{1}}...x_{r}^{n_{r}})$

and also gives the result due to Cheema and Motzkin [23] extending the well-
known Euler theorem that the number of integer partitions into ”odd parts” is
same as the number of integer partitions into ”distinct parts”. In our work
however we often replace $n_{i}\geq 0$ by $n_{i}\geq 1$ or even apply mixed
versions of these.

In the theory resulting from our present paper we make use of the fact that
equations (3.1) and (3.2) correspond to $n$D grids of their coefficients for
$x_{1}^{n_{1}}...x_{r}^{n_{r}}$; which here are $P(n_{1},n_{2},...,n_{r})$,
respectively $Q(n_{1},n_{2},...,n_{r})$. Although the lattice points of theses
grids are $n$-space lattice configurations, we find that a straight line from
the origin to the point $\langle n_{1},n_{2},...,n_{r}\rangle$ has an easily
derived generating function for each point along the line.

The above equations (3.1) and (3.2) when represented by their 2D, 3D, 4D and
5D grids in their respective first quadrants and first hyperquadrants,
obviously become unwieldy to visualize after the 2D extended rectangular
lattice, and 3D extended cubic lattice. What human consciousness can say they
see a 4D extended tesseract (4D cube) lattice, or a 5D extended hypercube
lattice? However we understand this dimensional concept, there is a logical
higher space extension also of the 2D first quadrant diagonal of lattice
points upon the line $y=x$. In the 3D cube first hyperquadrant in an
$x$-$y$-$z$ plane of lattice points there is the diagonal line represented by
the equation $x=y=z$ and the coefficients of those lattice point vectors arise
from setting the generating function $f(x,y,z)$ to say $f(z,z,z)$ which then
is the generating function for the 3D diagonal lattice point vectors in that
hyperquadrant. Similarly for the 4D case generating function $f(w,x,y,z)$
calculate $f(z,z,z,z)$ to give the 4D hyperdiagonal generating function for
lattice points along the line with equation $w=x=y=z$ through the 4D extended
tesseract (hypecube). And so on for the 5D function generated by
$f(v,w,x,y,z)$ giving us the function $f(z,z,z,z,z)$ generating the
coefficients (vector partition sums) for the lattice points along the 5D line
$v=w=x=y=z$. This concept can be applied to any of our 2D, 3D, 4D, etc lattice
point vector grids throughout our present volume. It applies to VPV identities
as hyperquadrant lattice functions, square hyperpyramid lattice functions and
skewed hyperpyramid lattice point vector identities. This includes the many
possibilities of applying polylogarithm formulas and Parametric Euler sum
identities; examples given in our oncoming papers being just the starting
place of a large number of possibilities.

##  4\. Hyperspace line generating functions for different nD slopes

In the previous section we saw how to calculate the hyperdiagonal generating
function for an nD generated vector partition grid. Basically, we have the

######  Statement 4.1.

If the nD generating function for the entire nD grid is the $n$ variable
$f_{n}(z_{1},z_{2},z_{3},...,z_{n})$, the equation of the nD hyperdiagonal
line from the origin is $z_{1}=z_{2}=z_{3}=...=z_{n}$ and so the generating
function for the nD vector partitions along that line is given by the single
variable function $f_{n}(z,z,,...,z)$.

That is, even though an nD space grid is impossible for humans to envision, we
can define a straight line through the hyperdiagonal from the nD origin point,
and formulate an exact value of the vector partition function at any point
along that nD hyperspace line.

Furthermore, taking an arbitrary lattice point vector $\langle
a_{1},a_{2},a_{3},...,a_{n}\rangle$ in the nD grid we can show that the vector
partition function (or nD coefficient) for that lattice point is exactly
evaluated as follows.

######  Statement 4.2.

If the nD generating function for the entire nD grid is the $n$ variable
$f_{n}(z_{1},z_{2},z_{3},...,z_{n})$, the equation of the nD hyper-radial-
from-origin line from the origin to the arbitrary point $\langle
a_{1},a_{2},a_{3},...,a_{n}\rangle$ is based on knowing values of

(4.1) $\langle a_{1},a_{2},a_{3},...,a_{n}\rangle=r\langle
c_{1},c_{2},c_{3},...,c_{n}\rangle$

where $\langle c_{1},c_{2},c_{3},...,c_{n}\rangle$ is a VPV with
$\gcd(c_{1},c_{2},c_{3},...,c_{n})=1$ and $r$ is the unique positive integer
that makes this true. So the straight line from the origin to the point
$\langle a_{1},a_{2},a_{3},...,a_{n}\rangle$ has the defining equation
$c_{1}z_{1}=c_{2}z_{2}=c_{3}z_{3}=...=c_{n}z_{n}$ and so the generating
function for the nD vector partitions along that line is given by the single
variable function $f_{n}(c_{1}z,c_{2}z,c_{3}z,...,c_{n}z)$, which has the
coefficient of $z^{r}$ equal to the vector partition function (or nD
coefficient) for that lattice point $\langle
a_{1},a_{2},a_{3},...,a_{n}\rangle$ exactly evaluated.

We will give numerous examples in later papers where the Visible Point Vector
(VPV) identities yield exact results for nD generating functions.

##  5\. Defining Vector Grids and resulting Partition Grids

The set of first quadrant cartesian 2D lattice point vectors is a countable
set. The lattice points radial from the origin in the radial region from $y=x$
to the positive $y$ axis can be replicated in the region between $y=x$ and the
positive $x$ axis. Therefore, the set of all 2D lattice point vectors in the
first quadrant are also countable. They are shown as follows.

######  Definition 5.1.

The first quadrant 2D vector grid, denoted as
$V_{2}(y\in[1,\infty);z\in[1,\infty))$ is the set of vectors $\langle
a,b\rangle$ such that a and b are positive integers. We say that a vector grid
generally is a collection of vectors in a defined region. For example, in the
following picture, first quadrant 2D vector grid is split into an
$\overline{overlinedUpper}$ vector grid, a Bold Diagonal vector grid and an
underlined Lower vector grid.

Table 1. First quadrant 2D vector grid $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$
---|---|---|---|---|---|---|---|---|---
$8$ | $\overline{\langle 1,8\rangle}$ | $\overline{\langle 2,8\rangle}$ | $\overline{\langle 3,8\rangle}$ | $\overline{\langle 4,8\rangle}$ | $\overline{\langle 5,8\rangle}$ | $\overline{\langle 6,8\rangle}$ | $\overline{\langle 7,8\rangle}$ | $\mathbf{\langle 8,8\rangle}$ | $\cdots$
$7$ | $\overline{\langle 1,7\rangle}$ | $\overline{\langle 2,7\rangle}$ | $\overline{\langle 3,7\rangle}$ | $\overline{\langle 4,7\rangle}$ | $\overline{\langle 5,7\rangle}$ | $\overline{\langle 6,7\rangle}$ | $\mathbf{\langle 7,7\rangle}$ | $\underline{\langle 8,7\rangle}$ | $\cdots$
$6$ | $\overline{\langle 1,6\rangle}$ | $\overline{\langle 2,6\rangle}$ | $\overline{\langle 3,6\rangle}$ | $\overline{\langle 4,6\rangle}$ | $\overline{\langle 5,6\rangle}$ | $\mathbf{\langle 6,6\rangle}$ | $\underline{\langle 7,6\rangle}$ | $\underline{\langle 8,6\rangle}$ | $\cdots$
$5$ | $\overline{\langle 1,5\rangle}$ | $\overline{\langle 2,5\rangle}$ | $\overline{\langle 3,5\rangle}$ | $\overline{\langle 4,5\rangle}$ | $\mathbf{\langle 5,5\rangle}$ | $\underline{\langle 6,5\rangle}$ | $\underline{\langle 7,5\rangle}$ | $\underline{\langle 8,5\rangle}$ | $\cdots$
$4$ | $\overline{\langle 1,4\rangle}$ | $\overline{\langle 2,4\rangle}$ | $\overline{\langle 3,4\rangle}$ | $\mathbf{\langle 4,4\rangle}$ | $\underline{\langle 5,4\rangle}$ | $\underline{\langle 6,4\rangle}$ | $\underline{\langle 7,4\rangle}$ | $\underline{\langle 8,4\rangle}$ | $\cdots$
$3$ | $\overline{\langle 1,3\rangle}$ | $\overline{\langle 2,3\rangle}$ | $\mathbf{\langle 3,3\rangle}$ | $\underline{\langle 4,3\rangle}$ | $\underline{\langle 5,3\rangle}$ | $\underline{\langle 6,3\rangle}$ | $\underline{\langle 7,3\rangle}$ | $\underline{\langle 8,3\rangle}$ | $\cdots$
$2$ | $\overline{\langle 1,2\rangle}$ | $\mathbf{\langle 2,2\rangle}$ | $\underline{\langle 3,2\rangle}$ | $\underline{\langle 4,2\rangle}$ | $\underline{\langle 5,2\rangle}$ | $\underline{\langle 6,2\rangle}$ | $\underline{\langle 7,2\rangle}$ | $\underline{\langle 8,2\rangle}$ | $\cdots$
$1$ | $\mathbf{\langle 1,1\rangle}$ | $\underline{\langle 2,1\rangle}$ | $\underline{\langle 3,1\rangle}$ | $\underline{\langle 4,1\rangle}$ | $\underline{\langle 5,1\rangle}$ | $\underline{\langle 6,1\rangle}$ | $\underline{\langle 7,1\rangle}$ | $\underline{\langle 8,1\rangle}$ | $\cdots$
$z/y$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $\cdots$

In certain situations the set of all overlined upper vectors can be mapped to
the set of all underlined lower vectors. The bold diagonal is defined by
$y=z$. We have in fact

######  Statement 5.1.

For any positive integers such that $a<b$,

$\overline{\langle a,b\rangle}\mapsto\underline{\langle b,a\rangle},$

and

$\underline{\langle b,a\rangle}\mapsto\overline{\langle a,b\rangle}.$

Hence, the Upper grid can map onto the Lower grid, and the Lower grid can map
onto the Upper grid.

While the consideration of the upper vectors and the lower vectors may seem
unnecessary in the present discussion, in later papers we shall make good use
of the symmetries and possible mappings, especially when we come to discuss
the Visible Point Vector (VPV) partitions wherein these symmetries can be
applied to practical examples.

######  Definition 5.2.

A finite 2D vector grid is a collection of vectors in a defined finite
rectangular region. $V_{2}(y\in[a,b];z\in[c,d])$ is notation for

$d$ | $\langle a,d\rangle$ | $\langle a+1,d\rangle$ | $\ldots$ | $\langle b-1,d\rangle$ | $\langle b,d\rangle$
---|---|---|---|---|---
$d-1$ | $\langle a,d-1\rangle$ | $\langle a+1,d-1\rangle$ | $\ldots$ | $\langle b-1,d-1\rangle$ | $\langle b,d-1\rangle$
$d-2$ | $\langle a,d-2\rangle$ | $\langle a+1,d-2\rangle$ | $\ldots$ | $\langle b-1,d-2\rangle$ | $\langle b,d-2\rangle$
$\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$
$c+2$ | $\langle a,c+2\rangle$ | $\langle a+1,c+2\rangle$ | $\ldots$ | $\langle b-1,c+2\rangle$ | $\langle b,c+2\rangle$
$c+1$ | $\langle a,c+1\rangle$ | $\langle a+1,c+1\rangle$ | $\ldots$ | $\langle b-1,c+1\rangle$ | $\langle b,c+1\rangle$
$c$ | $\langle a,c\rangle$ | $\langle a+1,c\rangle$ | $\ldots$ | $\langle b-1,c\rangle$ | $\langle b,c\rangle$
$z/y$ | $a$ | $a+1$ | $\ldots$ | $b-1$ | $b$

and we see from this, that the first 2D quadrant finite grid starting at
$\langle a,c\rangle$ can have $b$ and $d$ extended ”to infinity”, which could
then be written $V_{2}(y\in[a,\infty];z\in[c,\infty])$.

We give some examples now.

$V_{2}(y\in[-1,1];z\in[-1,1])=\begin{tabular}[]{c|cccc}$1$&$\langle-1,1\rangle$&$\langle
0,1\rangle$&$\langle 1,1\rangle$\\\ $0$&$\langle-1,0\rangle$&$\langle
0,0\rangle$&$\langle 1,0\rangle$\\\ $-1$&$\langle-1,-1\rangle$&$\langle
0,-1\rangle$&$\langle 1,-1\rangle$\\\
\hline\cr$z/y$&$-1$&$0$&$1$\end{tabular}$

$V_{2}(y\in[-2,2];z\in[-2,2])=\begin{tabular}[]{c|ccccc}$2$&$\langle-2,2\rangle$&$\langle-1,2\rangle$&$\langle
0,2\rangle$&$\langle 1,2\rangle$&$\langle 2,2\rangle$\\\
$1$&$\langle-2,1\rangle$&$\langle-1,1\rangle$&$\langle 0,1\rangle$&$\langle
1,1\rangle$&$\langle 2,1\rangle$\\\
$0$&$\langle-2,0\rangle$&$\langle-1,0\rangle$&$\langle 0,0\rangle$&$\langle
1,0\rangle$&$\langle 2,0\rangle$\\\
$-1$&$\langle-2,-1\rangle$&$\langle-1,-1\rangle$&$\langle
0,-1\rangle$&$\langle 1,-1\rangle$&$\langle 2,-1\rangle$\\\
$-2$&$\langle-2,-2\rangle$&$\langle-1,-2\rangle$&$\langle
0,-2\rangle$&$\langle 1,-2\rangle$&$\langle 2,-2\rangle$\\\
\hline\cr$z/y$&$-2$&$-1$&$0$&$1$&$2$\end{tabular}$

In our work on vector partitions, it will be a standard approach that we
specify the vector grid, and then based on the type of partition function
applied to the grid, specify a corresponding partition grid. Hence our
partition grid related to the vector grid of Definition 5.2, is as follows.

######  Definition 5.3.

The first quadrant 2D partition grid is the set of 2D partitions denoted

$p_{2}(V_{2}(\langle m,n\rangle\subseteq\\{y\in[a,b]\cup z\in[c,d]\\}))$

of $\langle a,b\rangle$ such that a and b are positive integers. We say that a
partition grid is a range of a partition function applied to a vector grid
domain in a defined region. For example, in the following picture, first
quadrant 2D partition grid is split into an ”$\overline{overlined\;Upper}$”
partition grid, a ”Diagonal upslope” grid, and an ”underlined Lower” partition
grid.

${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\vdots}$ | ${\ddots}$
---|---|---|---|---|---|---|---|---|---|---
$8$ | ${\overline{p_{2}(0,8)}}$ | ${\overline{p_{2}(1,8)}}$ | ${\overline{p_{2}(2,8)}}$ | ${\overline{p_{2}(3,8)}}$ | ${\overline{p_{2}(4,8)}}$ | ${\overline{p_{2}(5,8)}}$ | ${\overline{p_{2}(6,8)}}$ | ${\overline{p_{2}(7,8)}}$ | ${p_{2}(8,8)}$ | ${\cdots}$
$7$ | ${\overline{p_{2}(0,7)}}$ | ${\overline{p_{2}(1,7)}}$ | ${\overline{p_{2}(2,7)}}$ | ${\overline{p_{2}(3,7)}}$ | ${\overline{p_{2}(4,7)}}$ | ${\overline{p_{2}(5,7)}}$ | ${\overline{p_{2}(6,7)}}$ | ${{p_{2}(7,7)}}$ | ${\underline{p_{2}(8,7)}}$ | ${\cdots}$
$6$ | ${\overline{p_{2}(0,6)}}$ | ${\overline{p_{2}(1,6)}}$ | ${\overline{p_{2}(2,6)}}$ | ${\overline{p_{2}(3,6)}}$ | ${\overline{p_{2}(4,6)}}$ | ${\overline{p_{2}(5,6)}}$ | ${p_{2}(6,6)}$ | ${\underline{p_{2}(7,6)}}$ | ${\underline{p_{2}(8,6)}}$ | ${\cdots}$
$5$ | ${\overline{p_{2}(0,5)}}$ | ${\overline{p_{2}(1,5)}}$ | ${\overline{p_{2}(2,5)}}$ | ${\overline{p_{2}(3,5)}}$ | ${\overline{p_{2}(4,5)}}$ | ${p_{2}(5,5)}$ | ${\underline{p_{2}(6,5)}}$ | ${\underline{p_{2}(7,5)}}$ | ${\underline{p_{2}(8,5)}}$ | ${\cdots}$
$4$ | ${\overline{p_{2}(0,4)}}$ | ${\overline{p_{2}(1,4)}}$ | ${\overline{p_{2}(2,4)}}$ | ${\overline{p_{2}(3,4)}}$ | ${p_{2}(4,4)}$ | ${\underline{p_{2}(5,4)}}$ | ${\underline{p_{2}(6,4)}}$ | ${\underline{p_{2}(7,4)}}$ | ${\underline{p_{2}(8,4)}}$ | ${\cdots}$
$3$ | ${\overline{p_{2}(0,3)}}$ | ${\overline{p_{2}(1,3)}}$ | ${\overline{p_{2}(2,3)}}$ | ${p_{2}(3,3)}$ | ${\underline{p_{2}(4,3)}}$ | ${\underline{p_{2}(5,3)}}$ | ${\underline{p_{2}(6,3)}}$ | ${\underline{p_{2}(7,3)}}$ | ${\underline{p_{2}(8,3)}}$ | ${\cdots}$
$2$ | ${\overline{p_{2}(0,2)}}$ | ${\overline{p_{2}(1,2)}}$ | ${p_{2}(2,2)}$ | ${\underline{p_{2}(3,2)}}$ | ${\underline{p_{2}(4,2)}}$ | ${\underline{p_{2}(5,2)}}$ | ${\underline{p_{2}(6,2)}}$ | ${\underline{p_{2}(7,2)}}$ | ${\underline{p_{2}(8,2)}}$ | ${\cdots}$
$1$ | ${\overline{p_{2}(0,1)}}$ | ${p_{2}(1,1)}$ | ${\underline{p_{2}(2,1)}}$ | ${\underline{p_{2}(3,1)}}$ | ${\underline{p_{2}(4,1)}}$ | ${\underline{p_{2}(5,1)}}$ | ${\underline{p_{2}(6,1)}}$ | ${\underline{p_{2}(7,1)}}$ | ${\underline{p_{2}(8,1)}}$ | ${\cdots}$
$0$ | ${p_{2}(0,0)}$ | ${\underline{p_{2}(1,0)}}$ | ${\underline{p_{2}(2,0)}}$ | ${\underline{p_{2}(3,0)}}$ | ${\underline{p_{2}(4,0)}}$ | ${\underline{p_{2}(5,0)}}$ | ${\underline{p_{2}(6,0)}}$ | ${\underline{p_{2}(7,0)}}$ | ${\underline{p_{2}(8,0)}}$ | ${\cdots}$
$z/y$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | ${\cdots}$

Depending on how the partition function is defined, it may be that the set of
all Upper partitions can be mapped to the set of all Lower partitions. The
Diagonal upslope is defined here by $y=z$. We may often seek out or observe
these features and apply them to generating functions for vector partitions in
2D, or generalized to 3D, 4D, and so on. As analogy with partitions into
integers having Ferrers graphs, vector partitions can have generating
functions over different grids, yielding theorems on vector partitions that
arise simply from different equivalences of algebraic rational functions of
several variables. What we often call weighted vector partitions are in fact
functions mapped from the vector partition sums.

For example a 2D vector sum may map to a weighted value as in,

$\langle 1,2\rangle+\langle 1,3\rangle+\langle 2,3\rangle=\langle
4,8\rangle\quad\mapsto\quad f_{2}(4,8)=\binom{2}{1}\binom{3}{1}\binom{3}{2},$

so a vector partition sum maps to a function of the sum of that partition
$\langle 4,8\rangle$, the right side showing the function as a product of
binomial coefficients in this instance.

##  6\. Partitions into exactly one part, two parts, or three parts

We now try to lay the foundation for the language of vector partitions needed
to progress our work in this paper.

######  Statement 6.1.

Obviously, the number of partitions (of any kind) into one part, is 1. The
partition grid for 2D vector partitions into one part as defined above is
therefore trivially,

$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$
---|---|---|---|---|---|---|---|---|---|---
$8$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$7$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$6$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$5$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$4$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$3$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$2$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $\cdots$
$z/y$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $\cdots$

and the generating function for this, say $\hat{O}_{2}(y,z)$, for $|y|$ and
$|z|$ both less than $1$, is

(6.1) $\hat{O}_{2}(y,z)=\sum_{a,b\geq
0}\hat{o}_{2}(a,b)y^{a}z^{b}=\sum_{a,b\geqq
0}y^{a}z^{b}=\frac{1}{(1-y)(1-z)},$

where $\hat{o}_{2}(a,b)$ is the number of partitions of 2D vectors $\langle
a,b\rangle$, with each of $a$, and $b$ non-negative integers, into exactly one
part.

So we have described the simplest vector partition scenario, where there is
only one part in each partition in the 2D first quadrant. In the 3D first
hyperquadrant, the extended cubic lattice of points or vectors $\langle
a,b,c\rangle$ with each of $a$, $b$ and $c$ non-negative integers, we have the
generating function, say $\hat{O}_{3}(x,y,z)$, for $|x|$, $|y|$ and $|z|$ all
less than $1$, is

(6.2) $\hat{O}_{3}(x,y,z)=\sum_{a,b,c\geq
0}\hat{o}_{2}(a,b,c)x^{a}y^{b}z^{c}=\sum_{a,b,c\geqq
0}x^{a}y^{b}z^{c}=\frac{1}{(1-x)(1-y)(1-z)},$

where $\hat{o}_{2}(a,b,c)$ is the number of partitions of 3D vectors $\langle
a,b,c\rangle$ with each of $a$, $b$ and $c$ non-negative integers into exactly
one part. Generally speaking of $n$-space, we have: in the $n$D first
hyperquadrant, the extended hypercubic lattice of points or vectors $\langle
a_{1},a_{2},a_{3},\ldots,a_{n}\rangle$ with each of $a_{1}$, $a_{2}$, $a_{3}$
up to $a_{n}$ all non-negative integers, we have the generating function, say
$\hat{O}_{n}(a_{1},a_{2},a_{3},\ldots,a_{n})$, for $|z_{1}|$, $|z_{2}|$ up to
$|z_{n}|$ all less than $1$, is

(6.3) $\sum_{a_{1},a_{2},a_{3},\ldots,a_{n}\geq
0}\hat{o}_{n}(a_{1},a_{2},a_{3},\ldots,a_{n}){z_{1}}^{a_{1}}{z_{2}}^{a_{2}}{z_{3}}^{a_{3}}\cdots{z_{n}}^{a_{n}}$
$=\sum_{a_{1},a_{2},\ldots,a_{n}\geqq
0}{z_{1}}^{a_{1}}{z_{2}}^{a_{2}}\cdots{z_{n}}^{a_{n}}$
$=\frac{1}{(1-z_{1})(1-z_{2})\cdots(1-z_{n})},$

where $\hat{o}_{2}(a_{1},a_{2},\ldots,a_{n})$ is the number of partitions of
$n$D vectors $\langle a_{1},a_{2},\ldots,a_{n}\rangle$, with each of $a_{1}$,
$a_{2}$ up to $a_{n}$ non-negative integers, into exactly one part.

The next simplest case is then the scenario where each partition $\langle
a,b\rangle$ is into exactly two parts, based on the vector grid for $a$ and
$b$ both greater than or equal to 1. An arbitrary example where the vector
partition is into two parts is for $\langle 5,2\rangle$,

$\langle 5,2\rangle=\langle 4,1\rangle+\langle 1,1\rangle=\langle
3,1\rangle+\langle 2,1\rangle.$

A reminder here, that compositions count different orders of summands whereas
partitions do not. We recall that partitions are partially ordered sets, so
these two vector equations are the only two partitions into two parts using
vectors from the 2D first quadrant grid.

##  7\. Partitions into exactly two parts, and exactly three parts

It is easy to find the number of partitions of 2D first quadrant vectors into
exactly two parts, by a process of observation, then by formalizing this.
However, a much neater approach is given from generating functions in Campbell
[18] leading to a determinant form solution given here.

######  Theorem 7.1.

The number of first quadrant 2D partitions of vectors into two parts
$\spadesuit_{2}(a,b)$, and the number of first quadrant 2D partitions of
vectors into three parts $\clubsuit_{2}(a,b)$, are generated by the following
equations valid for $|y|<1$, $|z|<1$,

$\sum_{a,b\geq
0}^{\infty}\spadesuit_{2}(a,b)y^{a}z^{b}=\frac{1}{2!}\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}$ $=\frac{(1+yz)}{2(1+y)(1+z)(1-y^{2})^{2}(1-z^{2})^{2}};$

and

$\sum_{a,b\geq
0}^{\infty}\clubsuit_{2}(a,b)y^{a}z^{b}=\frac{1}{3!}\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1&0\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}&-2\\\
\frac{1}{\left(1-y^{3}\right)\left(1-z^{3}\right)}&\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}$
$=\frac{y^{3}z^{3}+y^{2}z^{2}+y^{2}z+yz^{2}+yz+1}{(1-y)^{3}(1+y)(y^{2}+y+1)(1-z)^{3}(1+z)(z^{2}+z+1)}.$

The two related partition grids for these are:

for $\spadesuit_{2}(a,b)$,

$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$
---|---|---|---|---|---|---|---|---|---|---
$8$ | $5$ | $9$ | $14$ | $18$ | $23$ | $27$ | $32$ | $36$ | $\mathbf{41}$ | $\cdots$
$7$ | $4$ | $8$ | $12$ | $16$ | $20$ | $24$ | $28$ | $\mathbf{32}$ | $36$ | $\cdots$
$6$ | $4$ | $7$ | $11$ | $14$ | $18$ | $21$ | $\mathbf{25}$ | $28$ | $32$ | $\cdots$
$5$ | $3$ | $6$ | $9$ | $12$ | $15$ | $\mathbf{18}$ | $21$ | $24$ | $27$ | $\cdots$
$4$ | $3$ | $5$ | $8$ | $10$ | $\mathbf{13}$ | $15$ | $18$ | $20$ | $23$ | $\cdots$
$3$ | $2$ | $4$ | $6$ | $\mathbf{8}$ | $10$ | $12$ | $14$ | $16$ | $18$ | $\cdots$
$2$ | $2$ | $3$ | $\mathbf{5}$ | $6$ | $8$ | $9$ | $11$ | $12$ | $14$ | $\cdots$
$1$ | $1$ | $\mathbf{2}$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $\cdots$
$0$ | $\mathbf{1}$ | $1$ | $2$ | $2$ | $3$ | $3$ | $4$ | $4$ | $5$ | $\cdots$
$z\uparrow y\rightarrow$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $\cdots$

and for $\clubsuit_{2}(a,b)$,

$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$
---|---|---|---|---|---|---|---|---|---|---
$8$ | $10$ | $25$ | $50$ | $80$ | $120$ | $165$ | $220$ | $280$ | $\mathbf{350}$ | $\cdots$
$7$ | $8$ | $20$ | $40$ | $64$ | $96$ | $132$ | $176$ | $\mathbf{224}$ | $280$ | $\cdots$
$6$ | $7$ | $16$ | $32$ | $51$ | $76$ | $104$ | $\mathbf{139}$ | $176$ | $220$ | $\cdots$
$5$ | $5$ | $12$ | $24$ | $38$ | $57$ | $\mathbf{78}$ | $104$ | $132$ | $165$ | $\cdots$
$4$ | $4$ | $9$ | $18$ | $28$ | $\mathbf{42}$ | $57$ | $76$ | $96$ | $120$ | $\cdots$
$3$ | $3$ | $6$ | $12$ | $\mathbf{19}$ | $28$ | $38$ | $51$ | $64$ | $80$ | $\cdots$
$2$ | $2$ | $4$ | $\mathbf{8}$ | $12$ | $18$ | $24$ | $32$ | $12$ | $40$ | $\cdots$
$1$ | $1$ | $\mathbf{2}$ | $4$ | $6$ | $9$ | $12$ | $16$ | $20$ | $25$ | $\cdots$
$0$ | $\mathbf{1}$ | $1$ | $2$ | $3$ | $4$ | $5$ | $7$ | $8$ | $10$ | $\cdots$
$z\uparrow y\rightarrow$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $\cdots$

Some notes on the partition grid sequences in theorem 7.1

Observations on the diagonal sequences, and symmetries:

Observation 1: The bold diagonal sequence $1,2,5,8,13,18,25,...$ in the
$\spadesuit_{2}(a,b)$ grid is easily found in the OEIS as sequence number
A000982, and is given by the ceiling function, so we have the exact formula
$\spadesuit_{2}(n,n)=\lceil\frac{(n+1)^{2}}{2}\rceil$. In plain words this
means ”The number of 2D vector partitions of $\langle n,n\rangle$ into exactly
two parts where $n$ is a positive integer is equal to
$\lceil\frac{(n+1)^{2}}{2}\rceil$.”

Observation 2: The $\spadesuit_{2}(a,b)$ grid diagonal has generating function
valid for $|q|<1$,

$\frac{1+q^{2}}{(1+q)(1-q)^{3}}=1+2q+5q^{2}+8q^{3}+13q^{4}+18q^{5}+25q^{6}+\ldots.$

Observation 3: The sequence $\spadesuit_{2}(n,n)$ is also given as a ”floor”
function by $\lfloor\frac{n^{2}+1}{2}\rfloor$

Observation 4: If we partition $n$ in two parts, say $r$ and $s$ so that
$r^{2}+s^{2}$ is minimal, then $\spadesuit_{2}(n,n)=r^{2}+s^{2}$. Geometrical
significance: folding a rod with length $n$ units at right angles in such a
way that the end points are at the least distance, which is given by
$\sqrt{\spadesuit_{2}(n,n)}$ as the hypotenuse of a right triangle with the
sum of the base and height $=n$ units.

Observation 5: A Dirichlet summation. With respect to the diagonal sequence
$\spadesuit_{2}(n,n)$ it is known that

$\sum_{n>0}\frac{1}{\lfloor\frac{n^{2}+1}{2}\rfloor}=\sum_{n>0}\frac{1}{2n^{2}}+\sum_{n\geq
0}\frac{1}{2n^{2}+2n+1}$
$1+\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{13}+\frac{1}{18}+\frac{1}{25}+\ldots=\frac{1}{2}\left(\frac{\pi^{2}}{6}+\pi\tanh\left(\frac{\pi}{2}\right)\right)=2.26312655....$

Observation 6: The bold diagonal sequence $1,2,8,19,42,78,139,224,350,...$ in
the $\clubsuit_{2}(a,b)$ grid is found in the OEIS sequence number A101427,
and is given by the following cases,

$\clubsuit_{2}(n,n)=\left\\{\begin{array}[]{ll}((n+2)^{2}(n+1)^{2}+12(\lfloor
n/2\rfloor+1)^{2}+8)/24,&\hbox{for $n$ divisible by 3$$;}\\\
((n+2)^{2}(n+1)^{2}+12(\lfloor
n/2\rfloor+1)^{2})/24,&\hbox{otherwise.}\end{array}\right.$

Therefore we have an exact formula for $\clubsuit_{2}(n,n)$; the number of 2D
vector partitions of $\langle n,n\rangle$ into exactly three parts; for $n$
any positive integer. The description of this sequence A101427 from OEIS is
”Number of different cuboids with volume $(pq)^{n}$, where $p$, $q$ are
distinct prime numbers”.

As an illustrative example of the formula, the first ten terms of
$((n+2)^{2}(n+1)^{2}+12(\lfloor n/2\rfloor+1)^{2}+8)/24$ are
$1,\frac{7}{3},\frac{25}{3},19,\frac{127}{3},\frac{235}{3},139,\frac{673}{3},\frac{1051}{3},517$,
and the first ten terms of $((n+2)^{2}(n+1)^{2}+12(\lfloor
n/2\rfloor+1)^{2})/24$ are
$\frac{2}{3},2,8,\frac{56}{3},42,78,\frac{416}{3},224,350,\frac{1550}{3}$.

Observation 7: The $\clubsuit_{2}(a,b)$ grid diagonal has generating function
valid for $|q|<1$,

$\frac{q^{6}+3q^{4}+4q^{3}+3q^{2}+1}{(1-q^{3})(1-q^{2})^{2}(1-q)^{2}}=1+2q+8q^{2}+19q^{3}+42q^{4}+78q^{5}+139q^{6}+\ldots.$

Observation 8: Horizontal rows and Vertical columns sequences generating
functions.

Row and column sequences for $\spadesuit_{2}(a,b)$ and $\clubsuit_{2}(a,b)$
have certain observable properties worth mentioning:

  1. (a)

The $j$th row sequence is the same as the $j$th column sequence;

  2. (b)

The $k$th term in the $j$th row sequence is the same as $k$ term in the $j$th
column sequence;

  3. (c)

That is, $\spadesuit_{2}(a,b)=\spadesuit_{2}(b,a)$ and
$\clubsuit_{2}(a,b)=\clubsuit_{2}(b,a)$.

  4. (d)

The generating function for the $j$th row sequence is the same as the
generating function for the $j$th column sequence.

Proof of theorem 7.1.

In Campbell [18], we have the following 2D corollary of the main $n$D
extension of Cauchy’s $q$-binomial theorem, which is equation (3.3) in that
paper. The proof of this $n$D theorem used Cramer’s Rule and gave us the
determinants as coefficients. So, setting $a=0$ in identity (3.3) of [18] we
obtain, for $y$ and $z$ both less than unity,

(7.1) $F_{2}(y,z;0,t)=\prod_{j,k\geq 0}\frac{1}{1-y^{j}z^{k}t}\\\ $
$=1+\frac{1}{\left(1-y\right)\left(1-z\right)}\frac{t}{1!}+\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}\frac{t^{2}}{2!}$
$+\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1&0\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}&-2\\\
\frac{1}{\left(1-y^{3}\right)\left(1-z^{3}\right)}&\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}\frac{t^{3}}{3!}$
$+\begin{vmatrix}\frac{1}{\left(1-y\right)\left(1-z\right)}&-1&0&0\\\
\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}&-2&0\\\
\frac{1}{\left(1-y^{3}\right)\left(1-z^{3}\right)}&\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}&-3\\\
\frac{1}{\left(1-y^{4}\right)\left(1-z^{4}\right)}&\frac{1}{\left(1-y^{3}\right)\left(1-z^{3}\right)}&\frac{1}{\left(1-y^{2}\right)\left(1-z^{2}\right)}&\frac{1}{\left(1-y\right)\left(1-z\right)}\\\
\end{vmatrix}\frac{t^{4}}{4!}$ $+etc.$

The proof of the theorem comes from recognizing that $\spadesuit_{2}(b,a)$ and
$\clubsuit_{2}(a,b)$ are the coefficients of $t^{2}$ and $t^{3}$ respectively
in (7.1). The determinants are simplified by applying the below identities,
setting

$a_{1}=\frac{1}{(1-y)(1-z)},\quad a_{2}=\frac{1}{(1-y^{2})(1-z^{2})},\quad
a_{3}=\frac{1}{(1-y^{3})(1-z^{3})},$

substituted into

$\begin{vmatrix}{a_{1}}&-1\\\ {a_{2}}&{a_{1}}\\\
\end{vmatrix}={a_{1}}^{2}+{a_{2}};$

and

$\begin{vmatrix}{a_{1}}&-1&0\\\ {a_{2}}&{a_{1}}&-2\\\
{a_{3}}&{a_{2}}&{a_{1}}\\\ \end{vmatrix}={a_{2}}^{3}+3{a_{1}}{a_{2}}+2{a_{3}}$

respectively. This gives us the appropriate functions generating the
partitions $\spadesuit_{2}(b,a)$ and $\clubsuit_{2}(a,b)$. Expanding these
generating functions allows us to fill in the relevant partition grids.
$\quad\blacksquare$

###  7.1. The $n$-dimensional $q$-binomial theorem

In his classical account of the theory of partitions, Andrews [2, chapter 2]
shows that many of the time honoured partition identities first given by
Euler, Gauss, Heine and Jacobi derive from the $q$-binomial theorem originally
given by Cauchy [21],

######  Theorem 7.2.

The $q$-binomial theorem. If $\left|q\right|<1,\left|t\right|<1$, for all
complex $a$,

(7.2)
$1+\sum_{k=1}^{\infty}\frac{(1-a)(1-aq)(1-aq^{2})...(1-aq^{k-1})t^{k}}{(1-q)(1-q^{2})(1-q^{3})...(1-q^{k})}=\prod_{k=0}^{\infty}\frac{1-atq^{k}}{1-tq^{k}}.$

The following extension to $n$D of the $q$-binomial theorem was proved in
Campbell [18] and 22 years earlier in his thesis [15]. We begin with the

######  Definition 7.1.

Define the function $F_{n}(t)$ for all complex numbers $a,t$ with $|a|,|t|<1$,
and for $n\geq 1$ with $|x_{1}|,|x_{2}|,...,|x_{n}|<1$ by the sequence of
functions $F_{n}(t)$ with

(7.3) $F_{0}(-;a,t)=\frac{1-at}{1-t}$

and for all of
$\left|x_{1}\right|,\left|x_{2}\right|,\left|x_{3}\right|,...,\left|x_{n}\right|<1$,
by

(7.4)
$F_{n}(x_{1},x_{2},x_{3},...,x_{n};a,t)=\prod_{\alpha_{1},\alpha_{2},\alpha_{3},...,\alpha_{n}\geq
0}\frac{1-x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}at}{1-x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}t}\equiv\sum_{k=0}^{\infty}{{}_{n}}{A_{k}}t^{k}$

where $\alpha_{1},\alpha_{2},\alpha_{3},...,\alpha_{n}$ may be all zero.

By the way, note that from this definition the right side of (7.2) is
$F_{1}(q;a,t)$, so it is evident we are working on a generalised $q$-binomial
expression. Next, based on this definition we can assert the $n$-space
$q$-binomial theorem, extended making $q$ redundant replaced by $n$ separate
$x_{i}$ variables,

######  Theorem 7.3.

For each $|x_{i}|<1$, with $1\leq i\leq n$,

(7.5) ${{}_{n}}{A_{k}}(x_{1},x_{2},x_{3},...,x_{n};a)=\det(a_{ij})/k!$

where the determinant is of order $k$, and

$a_{ij}=\begin{cases}\frac{1-a^{i-j+1}}{(1-{x_{1}}^{i-j+1})(1-{x_{2}}^{i-j+1})(1-{x_{3}}^{i-j+1})...(1-{x_{n}}^{i-j+1})},&\text{$i\geq
j$;}\\\ -i,&\text{$i=j-1$;}\\\ 0,&\text{$otherwise$.}\end{cases}$

We write theorem 7.3, restating it as a single equation identity as follows,
and rate it as a distinct theorem for our purposes.

######  Theorem 7.4.

For each $|x_{i}|<1$, with $1\leq i\leq n$,

(7.6)
$F_{n}(x_{1},x_{2},x_{3},...,x_{n};a,t)=\prod_{\alpha_{1},\alpha_{2},\alpha_{3},...,\alpha_{n}\geq
0}\frac{1-x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}at}{1-x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}t}\\\
$
$=1+\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}\frac{t}{1!}$
$+\begin{vmatrix}\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-1\\\
\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{n}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}\\\
\end{vmatrix}\frac{t^{2}}{2!}$
$+\begin{vmatrix}\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-1&0\\\
\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{n}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-2\\\
\frac{1-a^{3}}{\left(1-{x_{1}}^{3}\right)\left(1-{x_{2}}^{3}\right)\cdots\left(1-{x_{n}}^{3}\right)}&\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{n}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}\\\
\end{vmatrix}\frac{t^{3}}{3!}$
$+\begin{vmatrix}\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-1&0&0\\\
\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{n}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-2&0\\\
\frac{1-a^{3}}{\left(1-{x_{1}}^{3}\right)\left(1-{x_{2}}^{3}\right)\cdots\left(1-{x_{n}}^{3}\right)}&\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{r}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}&-3\\\
\frac{1-a^{4}}{\left(1-{x_{1}}^{4}\right)\left(1-{x_{2}}^{4}\right)\cdots\left(1-{x_{n}}^{4}\right)}&\frac{1-a^{3}}{\left(1-{x_{1}}^{3}\right)\left(1-{x_{2}}^{3}\right)\cdots\left(1-{x_{n}}^{3}\right)}&\frac{1-a^{2}}{\left(1-{x_{1}}^{2}\right)\left(1-{x_{2}}^{2}\right)\cdots\left(1-{x_{n}}^{2}\right)}&\frac{1-a}{\left(1-{x_{1}}\right)\left(1-{x_{2}}\right)\cdots\left(1-{x_{n}}\right)}\\\
\end{vmatrix}\frac{t^{4}}{4!}$ $+etc.$

We are now in a position to state the $n$-dimensional vector partition
generalization of theorem 7.1.

######  Theorem 7.5.

The number of first hyperquadrant $n$D partitions of vectors into two parts
$\spadesuit_{n}(a_{1},a_{2},\ldots,a_{n})$, and the number of first
hyperquadrant $n$D partitions of vectors into three parts
$\clubsuit_{n}(a_{1},a_{2},\ldots,a_{n})$, are generated by the following
equations valid for $|z_{i}|<1$, with $1\leq i\leq n$,

$\sum_{a_{1},a_{2},\ldots,a_{n}\geq
0}^{\infty}\spadesuit_{n}(a_{1},a_{2},\ldots,a_{n}){z_{1}}^{a_{1}}{z_{2}}^{a_{2}}\cdots{z_{n}}^{a_{n}}=\frac{1}{2!}\begin{vmatrix}\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}&-1\\\
\frac{1}{\left(1-{z_{1}}^{2}\right)\left(1-{z_{2}}^{2}\right)\cdots\left(1-{z_{n}}^{2}\right)}&\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}\\\
\end{vmatrix}$ $={c_{1}}^{2}+{c_{2}};$
$where\;c_{1}=\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}\;and\;c_{2}=\frac{1}{\left(1-{z_{1}^{2}}\right)\left(1-{z_{2}^{2}}\right)\cdots\left(1-{z_{n}^{2}}\right)}.$

Also

$\sum_{a_{1},a_{2},\ldots,a_{n}}^{\infty}\clubsuit_{n}(a_{1},a_{2},\ldots,a_{n}){z_{1}}^{a_{1}}{z_{2}}^{a_{2}}\cdots{z_{n}}^{a_{n}}$
$=\frac{1}{3!}\begin{vmatrix}\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}&-1&0\\\
\frac{1}{\left(1-{z_{1}}^{2}\right)\left(1-{z_{2}}^{2}\right)\cdots\left(1-{z_{n}}^{2}\right)}&\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}&-2\\\
\frac{1}{\left(1-{z_{1}}^{3}\right)\left(1-{z_{2}}^{3}\right)\cdots\left(1-{z_{n}}^{3}\right)}&\frac{1}{\left(1-{z_{1}}^{2}\right)\left(1-{z_{2}}^{2}\right)\cdots\left(1-{z_{n}}^{2}\right)}&\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)}\\\
\end{vmatrix}$
$={c_{2}}^{3}+3{c_{1}}{c_{2}}+2{c_{3}};\;where\;c_{1}=\frac{1}{\left(1-{z_{1}}\right)\left(1-{z_{2}}\right)\cdots\left(1-{z_{n}}\right)},$
$c_{2}=\frac{1}{\left(1-{z_{1}^{2}}\right)\left(1-{z_{2}^{2}}\right)\cdots\left(1-{z_{n}^{2}}\right)}\;and\;c_{3}=\frac{1}{\left(1-{z_{1}^{3}}\right)\left(1-{z_{2}^{3}}\right)\cdots\left(1-{z_{n}^{3}}\right)}.$

PROOF: From the coefficients of $t^{2}$ respectively $t^{3}$ and mapping
$x_{1}\mapsto z_{i}$ in theorem 7.4 and setting $a=0$ we obtain the number of
relevant $n$D partitions into exactly two parts and exactly three parts.
$\quad\quad\blacksquare$

Hence, in this section we have demonstrated that in $n$D first hyperquadrants,
the number of vector partitions into exactly integer $m$ number of parts is
derivable to a closed form of a determinant.

##  8\. First 3D hyperquadrant tableau reduced to a 2D tableau

The set of all lattice points in 3D space first hyperquadrant (extended cube)
are similarly a countable set. The lattice point vectors in nD space first
hyperquadrant comprised of points $\left\langle
a_{1},a_{2},a_{3},...,a_{n}\right\rangle$ with each $a_{i}$ a positive integer
is a countable set.

In 3D the set of first hyperquadrant points can be portrayed as shown below,
as 2D horizontal layers or lamina, each one as follows.

(8.1) $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $.\cdot$
---|---|---|---|---|---|---|---
$6$ | $\langle 1,6,a\rangle$ | $\langle 2,6,a\rangle$ | $\langle 3,6,a\rangle$ | $\langle 4,6,a\rangle$ | $\langle 5,6,a\rangle$ | $\langle 6,6,a\rangle$ | $\cdots$
$5$ | $\langle 1,5,a\rangle$ | $\langle 2,5,a\rangle$ | $\langle 3,5,a\rangle$ | $\langle 4,5,a\rangle$ | $\langle 5,5,a\rangle$ | $\langle 6,5,a\rangle$ | $\cdots$
$4$ | $\langle 1,4,a\rangle$ | $\langle 2,4,a\rangle$ | $\langle 3,4,a\rangle$ | $\langle 4,4,a\rangle$ | $\langle 5,4,a\rangle$ | $\langle 6,4,a\rangle$ | $\cdots$
$3$ | $\langle 1,3,a\rangle$ | $\langle 2,3,a\rangle$ | $\langle 3,3,a\rangle$ | $\langle 4,3,a\rangle$ | $\langle 5,3,a\rangle$ | $\langle 6,3,a\rangle$ | $\cdots$
$2$ | $\langle 1,2,a\rangle$ | $\langle 2,2,a\rangle$ | $\langle 3,2,a\rangle$ | $\langle 4,2,a\rangle$ | $\langle 5,2,a\rangle$ | $\langle 6,2,a\rangle$ | $\cdots$
$1$ | $\langle 1,1,a\rangle$ | $\langle 2,1,a\rangle$ | $\langle 3,1,a\rangle$ | $\langle 4,1,a\rangle$ | $\langle 5,1,a\rangle$ | $\langle 6,1,a\rangle$ | $\cdots$
$y\uparrow,x\rightarrow,z=a$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $\cdots$

where $a\in\\{1,2,3,\ldots\\}$.

In order to count each lattice point vector in the entire first 3D
hyperquadrant, it suffices to count the following vectors $\left\langle
a,b,c\right\rangle$ such that $0<a<b<c$ and apply symmetries. We can start
with an arrangement like

(8.2) $\begin{array}[]{ccccccccc}\textmd{etc.}&&&&&&&&\\\ \langle
1,3,6\rangle&\langle 2,3,7\rangle&\langle 3,6,7\rangle&\langle
4,6,8\rangle&\langle 5,6,9\rangle&\langle 6,8,9\rangle&\langle
7,8,10\rangle&\langle 8,9,10\rangle&\\\ \langle 1,2,6\rangle&\langle
2,5,6\rangle&\langle 3,5,7\rangle&\langle 4,5,8\rangle&\langle
5,7,8\rangle&\langle 6,7,9\rangle&\langle 7,8,9\rangle&&\\\ \langle
1,4,5\rangle&\langle 2,4,6\rangle&\langle 3,4,7\rangle&\langle
4,6,7\rangle&\langle 5,6,8\rangle&\langle 6,7,8\rangle&&&\\\ \langle
1,3,5\rangle&\langle 2,3,6\rangle&\langle 3,5,6\rangle&\langle
4,5,7\rangle&\langle 5,6,7\rangle&&&&\\\ \langle 1,2,5\rangle&\langle
2,4,5\rangle&\langle 3,4,6\rangle&\langle 4,5,6\rangle&&&&&\\\ \langle
1,3,4\rangle&\langle 2,3,5\rangle&\langle 3,4,5\rangle&&&&&&\\\ \langle
1,2,4\rangle&\langle 2,3,4\rangle&&&&&&&\\\ \langle
1,2,3\rangle&&&&&&&&\end{array}$

where the bottom entry for ”column $a$” is $\langle a,a+1,a+2\rangle$ and each
column has terms $\bigcup_{a<j<k}\langle a,j,k\rangle$ in an order that
exhausts successive $k$ integer terms with respect to the $j$ terms between
$a$ and $k$. The above arrangement of 3D vectors enables us to write the
generating function

######  Theorem 8.1.

(8.3)
$\prod_{0<a<b<c}(1+x^{a}y^{b}z^{c})=1+\sum_{0<i<j<k}p_{3}(\mathfrak{D};i,j,k)x^{i}y^{j}z^{k}$
$\displaystyle=$ $\displaystyle(1+x^{1}y^{2}z^{3})$
$\displaystyle(1+x^{1}y^{2}z^{4})(1+x^{2}y^{3}z^{4})$
$\displaystyle(1+x^{1}y^{3}z^{4})(1+x^{2}y^{3}z^{5})(1+x^{3}y^{4}z^{5})$
$\displaystyle(1+x^{1}y^{2}z^{5})(1+x^{2}y^{4}z^{5})(1+x^{3}y^{4}z^{6})(1+x^{4}y^{5}z^{6})$
$\displaystyle(1+x^{1}y^{3}z^{5})(1+x^{2}y^{3}z^{6})(1+x^{3}y^{5}z^{6})(1+x^{4}y^{5}z^{7})(1+x^{5}y^{6}z^{7})$
$\displaystyle(1+x^{1}y^{4}z^{5})(1+x^{2}y^{4}z^{6})(1+x^{3}y^{4}z^{7})(1+x^{4}y^{6}z^{7})(1+x^{5}y^{6}z^{8})(1+x^{6}y^{7}z^{8})$
$\displaystyle(1+x^{1}y^{2}z^{6})(1+x^{2}y^{5}z^{6})(1+x^{3}y^{5}z^{7})(1+x^{4}y^{5}z^{8})(1+x^{5}y^{7}z^{8})(1+x^{6}y^{7}z^{9})(1+x^{7}y^{8}z^{9})$
etc.

where $p_{3}(\mathfrak{D};i,j,k)$ is the number of 3D partitions of
$\left\langle i,j,k\right\rangle$ into distinct vector parts of type
$\left\langle a,b,c\right\rangle$ such that $0<a<b<c$ with positive integer
values of $a$, $b$, and $c$. In other words, $p_{3}(\mathfrak{D};i,j,k)$ is
the number of 3D partitions into distinct parts from (8.2).

As a logical companion to theorem 8.1, also easily inferred from our reduction
of 3D tableau to 2D tableau (8.2) is the following

######  Theorem 8.2.

(8.4)
$\prod_{0<a<b<c}\frac{1}{(1-x^{a}y^{b}z^{c})}=1+\sum_{0<i<j<k}p_{3}(\mathfrak{U};i,j,k)x^{i}y^{j}z^{k}$
$\displaystyle=$ $\displaystyle\frac{1}{(1-x^{1}y^{2}z^{3})}$
$\displaystyle\frac{1}{(1-x^{1}y^{2}z^{4})(1-x^{2}y^{3}z^{4})}$
$\displaystyle\frac{1}{(1-x^{1}y^{3}z^{4})(1-x^{2}y^{3}z^{5})(1-x^{3}y^{4}z^{5})}$
$\displaystyle\frac{1}{(1-x^{1}y^{2}z^{5})(1-x^{2}y^{4}z^{5})(1-x^{3}y^{4}z^{6})(1-x^{4}y^{5}z^{6})}$
$\displaystyle\frac{1}{(1-x^{1}y^{3}z^{5})(1-x^{2}y^{3}z^{6})(1-x^{3}y^{5}z^{6})(1-x^{4}y^{5}z^{7})(1-x^{5}y^{6}z^{7})}$
$\displaystyle\frac{1}{(1-x^{1}y^{4}z^{5})(1-x^{2}y^{4}z^{6})(1-x^{3}y^{4}z^{7})(1-x^{4}y^{6}z^{7})(1-x^{5}y^{6}z^{8})(1-x^{6}y^{7}z^{8})}$
$\displaystyle\frac{1}{(1-x^{1}y^{2}z^{6})(1-x^{2}y^{5}z^{6})(1-x^{3}y^{5}z^{7})(1-x^{4}y^{5}z^{8})(1-x^{5}y^{7}z^{8})(1-x^{6}y^{7}z^{9})(1-x^{7}y^{8}z^{9})}$
etc.

where $p_{3}(\mathfrak{U};i,j,k)$ is the number of 3D partitions of
$\left\langle i,j,k\right\rangle$ into unrestricted vector parts of type
$\left\langle a,b,c\right\rangle$ such that $0<a<b<c$ with positive integer
values of $a$, $b$, and $c$. In other words, $p_{3}(\mathfrak{U};i,j,k)$ is
the number of 3D partitions into unrestricted parts from (8.2).

##  9\. 2D and 3D Upper Radial Regions of vectors for partition identities

In the next sections we study 2D and 3D sets of vectors in regions radial from
the origin. A 2D or 3D vector partition is essentially:

  1. (a)

a bunch of vector arrows

  2. (b)

in order of non-increasing length,

  3. (c)

each with positive rational gradients in 2D or in 3D wrt the relevant axes,

  4. (d)

joined head to tail

  5. (e)

starting at the origin

  6. (f)

with ending destination an integer co-ordinate point,

  7. (g)

say $\left\langle a,b\right\rangle$ with $a$ and $b$ both non-negative
integers in 2D, or

  8. (h)

say $\left\langle a,b,c\right\rangle$ with $a$, $b$ and $c$ all non-negative
integers in 3D.

The number of arrows in this vector partition is the number of parts of the
partition.

Since a vector partition is a partially ordered set summed in a defined way,
this means there are potentially many ways to define the vector partition.
Rules used in defining a vector partition are possibly one or more of:

  * •

non-increasing arrow length (or non-decreasing),

  * •

non-increasing co-ordinate numbering (or non-decreasing) or co-ordinate
orderings, say X then Y; or Y then X axes,

  * •

non-increasing (respectively non-decreasing) angular measure from an axis as
polar co-ordinate systemic,

  * •

non-increasing (respectively non-decreasing) color gradations.

We consider 2D vectors $\left\langle a,b\right\rangle$ where $0<a<b$. We
mainly look into two types of such vectors for our vector partition
aggregations; both types being in ”radial from the origin” regions:

  1. (a)

All of the 2D vectors $\left\langle a,b\right\rangle$ where $0\leq a<b$ are in
the infinitely extended region radial from the origin. This may involve
examination of defined ”subset finite patches” of that radial region.

  2. (b)

All VPV 2D vectors are $\left\langle a,b\right\rangle$ where $0<a<b$ and
$\gcd(a,b)=1$ are in the infinitely extended region radial from the origin.
This also may involve examination of defined ”subset finite patches” of that
radial region.

Using ”All 2D first quadrant vectors” we have the examples

(9.1) $\prod_{j,k\geq 1}\frac{1}{1-x^{j}y^{k}}=\prod_{j,k\geq
1}\left\\{1+x^{j}y^{k}+x^{2j}y^{2k}+x^{3j}y^{3k}+\ldots\right\\}=\sum_{a,b\geq
0}\mathbf{p}_{2}(a,b)x^{a}y^{b}.$

Here we generate $\mathbf{p}_{2}(a,b)$ as the number of partitions of $\langle
j,k\rangle$ into unrestricted parts $\langle a,b\rangle$ with $a,b\geq 0$. In
this case each partition has a value of 1, so the number of partitions is
unsurprisingly just a count of each partition.

(9.2) $\prod_{j,k\geq
1}\left(1+\frac{x^{j}y^{k}}{(1-x^{j}y^{k})^{2}}\right)=\prod_{j,k\geq
1}\left\\{1+x^{j}y^{k}+2x^{2j}y^{2k}+3x^{3j}y^{3k}+\ldots\right\\}=\sum_{a,b\geq
0}\mathbf{q}_{2}(a,b)x^{a}y^{b}.$

This generates $\mathbf{q}_{2}(a,b)$ as a weighted sum of partitions of
$\langle j,k\rangle$ into unrestricted parts $\langle a,b\rangle$ with
$a,b\geq 0$. This is a function of a partition applied across all possible
partitions of $\langle j,k\rangle$. In this case each partition is assigned a
value (called also a weight) of

$n_{1}^{a_{1}}n_{2}^{a_{2}}...n_{r}^{a_{r}}$

for a partition of $\langle j,k\rangle$

$(\langle n_{1}j,n_{1}k\rangle^{a_{1}}\langle
n_{2}j,n_{2}k\rangle^{a_{2}}...\langle n_{r}j,n_{r}k\rangle^{a_{r}}),$

so the function of partitions (as distinct from number of partitions) on
$\langle j,k\rangle$ is derived as a combined function of each partition. So
$\mathbf{q}_{2}(a,b)$ is a sum of products $\sum
n_{1}^{a_{1}}n_{2}^{a_{2}}...n_{r}^{a_{r}}$ taken over every partition of
$\langle j,k\rangle$.

(9.3) $\prod_{j,k\geq
1}\left(1+\frac{x^{j}y^{k}(1+x^{j}y^{k})}{(1-x^{j}y^{k})^{3}}\right)=\prod_{j,k\geq
1}\left\\{1+1^{2}x^{j}y^{k}+2^{2}x^{2j}y^{2k}+3^{2}x^{3j}y^{3k}+\ldots\right\\}$
$=\sum_{a,b\geq 0}\mathbf{r}_{2}(a,b)x^{a}y^{b}.$

In this instance $\mathbf{r}_{2}(a,b)$ is a sum of products $\sum
n_{1}^{2a_{1}}n_{2}^{2a_{2}}...n_{r}^{2a_{r}}$ taken over every partition
$(\langle n_{1}j,n_{1}k\rangle^{a_{1}}\langle
n_{2}j,n_{2}k\rangle^{a_{2}}...\langle n_{r}j,n_{r}k\rangle^{a_{r}})$ of
$\langle j,k\rangle$.

(9.4) $\prod_{j,k\geq
1}\left(1+\frac{x^{j}y^{k}}{1-x^{2j}y^{2k}}\right)=\prod_{j,k\geq
1}\left\\{1+x^{j}y^{k}+3x^{3j}y^{3k}+5x^{5j}y^{5k}+\ldots\right\\}$
$=\sum_{a,b\geq 0}\mathbf{s}_{2}(a,b)x^{a}y^{b}.$

In this case $\mathbf{x}_{2}(a,b)$ is a sum of products
$\sum(2n_{1}+1)^{a_{1}}(2n_{2}+1)^{a_{2}}...(2n_{r}+1)^{a_{r}}$ taken over
every partition $(\langle n_{1}j,n_{1}k\rangle^{a_{1}}\langle
n_{3}j,n_{3}k\rangle^{a_{2}}...\langle n_{2r+1}j,n_{2r+1}k\rangle^{a_{r}})$ of
$\langle j,k\rangle$.

In the examples that follow here we use $\gcd(j,k)=1$ abbreviated to
$(j,k)=1$. It will become evident that when applying these generating function
forms to radial-from-origin regions over vector partitions, the theory of
$\langle j,k\rangle$ where $(j,k)=1$ simplifies things.

(9.5) $\prod_{\begin{subarray}{c}j,k\geq 1;\\\
(j,k)=1\end{subarray}}\frac{1}{1-x^{j}y^{k}}=\prod_{\begin{subarray}{c}j,k\geq
1;\\\
(j,k)=1\end{subarray}}\left\\{1+x^{j}y^{k}+x^{2j}y^{2k}+x^{3j}y^{3k}+\ldots\right\\}=\sum_{a,b\geq
0}\mathbf{v}_{2}(a,b)x^{a}y^{b}.$

Note that the curly-bracketed term in (9.5) contributes vector parts of
rational gradient $j/k$ with $(j,k)=1$ in the first 2D quadrant exactly one
part per partition. So $\mathbf{v}_{2}(a,b)$ is the number of partitions of
$\left\langle a,b\right\rangle$ into vectors each part of distinct positive
rational gradient. In traditional integer partition parlance this would say
$\mathbf{v}_{2}(a,b)$ is the number of partitions into unrestricted Visible
Point Vector (VPV) parts.

(9.6) $\prod_{\begin{subarray}{c}j,k\geq 1;\\\
(j,k)=1\end{subarray}}\left(1+\frac{x^{j}y^{k}}{(1-x^{j}y^{k})^{2}}\right)=\prod_{\begin{subarray}{c}j,k\geq
1;\\\
(j,k)=1\end{subarray}}\left\\{1+x^{j}y^{k}+2x^{2j}y^{2k}+3x^{3j}y^{3k}+\ldots\right\\}$
$=\sum_{a,b\geq 0}\mathbf{w}_{2}(a,b)x^{a}y^{b}.$

(9.7) $\prod_{\begin{subarray}{c}j,k\geq 1;\\\
(j,k)=1\end{subarray}}\left(1+\frac{x^{j}y^{k}(1+x^{j}y^{k})}{(1-x^{j}y^{k})^{3}}\right)=\prod_{\begin{subarray}{c}j,k\geq
1;\\\
(j,k)=1\end{subarray}}\left\\{1+1^{2}x^{j}y^{k}+2^{2}x^{2j}y^{2k}+3^{2}x^{3j}y^{3k}+\ldots\right\\}$
$=\sum_{a,b\geq 0}\mathbf{x}_{2}(a,b)x^{a}y^{b}.$

(9.8) $\prod_{\begin{subarray}{c}j,k\geq 1;\\\
\gcd(j,k)=1\end{subarray}}\left(1+\frac{x^{j}y^{k}}{1-x^{2j}y^{2k}}\right)=\prod_{\begin{subarray}{c}j,k\geq
1;\\\
\gcd(j,k)=1\end{subarray}}\left\\{1+x^{j}y^{k}+3x^{3j}y^{3k}+5x^{5j}y^{5k}+\ldots\right\\}$
$=\sum_{a,b\geq 0}\mathbf{y}_{2}(a,b)x^{a}y^{b}.$

We also consider 3D vectors $\left\langle a,b,c\right\rangle$ where $0<a<b<c$.
Again we mainly look into two types of such vectors for our vector partition
aggregations:

  1. (a)

All of the 3D vectors $\left\langle a,b,c\right\rangle$ where $0<a<b<c$ are in
the infinitely extended region radial from the origin. This on occasion means
examination of defined ”finite patches” of that radial region.

  2. (b)

All of the 3D vectors $\left\langle a,b,c\right\rangle$ where $0<a<b<c$ and
$gcd(a,b,c)=1$ are in the infinitely extended region radial from the origin.
This on occasion means examination of defined ”finite patches” of that radial
region.

##  10\. Defining 2D Upper Visible Point Vectors in origin-radial regions

These lattice point vectors count the visible from the origin points in the
infinite radial region of the first quadrant bounded by the positive $y$ axis
and the line $y=x$. The visible lattice points in 3D space are similarly a
countable set, and in fact in nD space with $n$ a positive integer greater
than unity. In 3D the set of visible points can be counted as shown below. In
order to count the entire first 3D hyperquadrant it suffices to count the
following vectors $\left\langle a,b,c\right\rangle$ such that $0<a<b<c$ has

A set of identities involving $n$ dimensional visible lattice points was
discovered by Campbell (1994). The visible lattice points in the 2 dimensional
first quadrant (see Weisstein, Eric W. ”Visible Point.” From MathWorld–A
Wolfram Web Resource.

http://mathworld.wolfram.com/VisiblePoint.html) correspond to the countable
set of rational numbers between 0 and 1, listed here as

$\begin{array}[]{ccccccccc}\frac{1}{2}&&&&&&&&\\\
\frac{1}{3}&\frac{2}{3}&&&&&&&\\\ \frac{1}{4}&&\frac{3}{4}&&&&&&\\\
\frac{1}{5}&\frac{2}{5}&\frac{3}{5}&\frac{4}{5}&&&&&\\\
\frac{1}{6}&&&&\frac{5}{6}&&&&\\\
\frac{1}{7}&\frac{2}{7}&\frac{3}{7}&\frac{4}{7}&\frac{5}{7}&\frac{6}{7}&&&\\\
\frac{1}{8}&&\frac{3}{8}&&\frac{5}{8}&&\frac{7}{8}&&\\\
\frac{1}{9}&\frac{2}{9}&&\frac{4}{9}&\frac{5}{9}&&\frac{7}{9}&\frac{8}{9}&\\\
\frac{1}{10}&&\frac{3}{10}&&&&\frac{7}{10}&&\frac{9}{10}\\\
&&&&etc.&&&&\end{array}$

giving rise to the visible from the origin vector (VPV) lattice points
$\langle x,y\rangle$ with $x<y$ and $\gcd(x,y)=1$ in the first 2D quadrant.

$\begin{array}[]{ccccccccc}&&&&etc.&&&&\\\ \langle 1,10\rangle&&\langle
3,10\rangle&&&&\langle 7,10\rangle&&\langle 9,10\rangle\\\ \langle
1,9\rangle&\langle 2,9\rangle&&\langle 4,9\rangle&\langle 5,9\rangle&&\langle
7,9\rangle&\langle 8,9\rangle&\\\ \langle 1,8\rangle&&\langle
3,8\rangle&&\langle 5,8\rangle&&\langle 7,8\rangle&&\\\ \langle
1,7\rangle&\langle 2,7\rangle&\langle 3,7\rangle&\langle 4,7\rangle&\langle
5,7\rangle&\langle 6,7\rangle&&&\\\ \langle 1,6\rangle&&&&\langle
5,6\rangle&&&&\\\ \langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle&&&&&\\\ \langle 1,4\rangle&&\langle
3,4\rangle&&&&&&\\\ \langle 1,3\rangle&\langle 2,3\rangle&&&&&&&\\\ \langle
1,2\rangle&&&&&&&&\end{array}$

These are the countable list of upper first quadrant 2D Visible Point Vectors.
That is, the VPVs in the infinitely extended radial region of the 2D first
quadrant above the line $y=x$.

##  11\. Examples of 2D VPV finite generating functions

###  11.1. 2D Distinct Upper VPV Coefficients - Order 2

(11.1) $(1+xy^{2})\quad\quad\quad and\quad\quad\quad(1-xy^{2})$ $3$ |  |  |  |  | $\sum r$
---|---|---|---|---|---
$2$ |  | $1$ |  |  | $1$
$1$ |  |  |  |  | $0$
$0$ | $1$ |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ |  |
$\sum c$ | $1$ | $1$ |  |  |
$3$ |  |  |  |  | $\sum r$
---|---|---|---|---|---
$2$ |  | $-1$ |  |  | $-1$
$1$ |  |  |  |  | $0$
$0$ | $1$ |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ |  |
$\sum c$ | $1$ | $-1$ |  |  |

###  11.2. 2D Distinct Upper VPV Coefficients - Order 3

(11.2) $(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})\quad\quad and\quad\quad(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})$ |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---
$8$ |  |  |  |  | $1$ |  | $1$
$7$ |  |  |  |  |  |  | $0$
$6$ |  |  |  | $1$ |  |  | $1$
$5$ |  |  | $1$ | $1$ |  |  | $2$
$4$ |  |  |  |  |  |  | $0$
$3$ |  | $1$ | $1$ |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  | $1$
$1$ |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ |  |
$\sum c$ | $1$ | $2$ | $2$ | $2$ | $1$ |  |
|  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---
$8$ |  |  |  |  | $-1$ |  | $-1$
$7$ |  |  |  |  |  |  | $0$
$6$ |  |  |  | $1$ |  |  | $1$
$5$ |  |  | $1$ | $1$ |  |  | $2$
$4$ |  |  |  |  |  |  | $0$
$3$ |  | $-1$ | $-1$ |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  | $-1$
$1$ |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ |  |
$\sum c$ | $1$ | $-2$ | $0$ | $2$ | $-1$ |  |

Note that row totals in the left side grid generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})\quad\textmd{where}\quad x=1,$

expanding to

$y^{8}+y^{6}+2y^{5}+2y^{3}+y^{2}+1.$

Also row totals in the right side grid generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})\quad\textmd{where}\quad x=1,$

expanding to

$-y^{8}+y^{6}+2y^{5}-2y^{3}-y^{2}+1.$

Likewise, the left side grid column totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})\quad\textmd{where}\quad y=1,$

expanding to

$x^{4}+2x^{3}+2x^{2}+2x+1;$

whilst the right side grid column totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})\quad\textmd{where}\quad y=1,$

expanding to

$-x^{4}+2x^{3}-2x+1.$

Combinatorial interpretation: Consider the three vectors

(11.3) $\begin{array}[]{cc}\langle 1,3\rangle&\langle 2,3\rangle\\\ \langle
1,2\rangle&\end{array}$

Let $p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv},\langle a,b\rangle)$, denote the
number of partitions of $\langle a,b\rangle$ into distinct parts from (11.3).
Then

(11.4) $\prod_{k=2}^{3}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{8}\sum_{a=1}^{b-1}p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv},\langle
a,b\rangle)x^{a}y^{b},$

and each entry in the left side grid after (11.2) gives each numerical value.

Let $p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv,\textmd{odd}},\langle
a,b\rangle)$, denote the number of partitions of $\langle a,b\rangle$ into an
odd number of distinct parts from (11.3). Let
$p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv,\textmd{even}},\langle a,b\rangle)$,
denote the number of partitions of $\langle a,b\rangle$ into an even number of
distinct parts from (11.3).Then

(11.5) $\prod_{k=2}^{3}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{8}\sum_{a=1}^{b-1}[p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv,\textmd{odd}},\langle
a,b\rangle)-p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv,\textmd{even}},\langle
a,b\rangle)]x^{a}y^{b},$

and each entry in the right side grid after (11.2) gives each numerical value.

###  11.3. 2D Distinct Upper VPV Coefficients - Order 4

(11.6) $(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})$ |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---
$16$ |  |  |  |  |  |  |  |  | $1$ |  | $1$
$15$ |  |  |  |  |  |  |  |  |  |  | $0$
$14$ |  |  |  |  |  |  |  | $1$ |  |  | $1$
$13$ |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$12$ |  |  |  |  |  | $1$ |  | $1$ |  |  | $2$
$11$ |  |  |  |  |  | $1$ | $1$ |  |  |  | $2$
$10$ |  |  |  |  | $1$ | $1$ | $1$ |  |  |  | $3$
$9$ |  |  |  | $1$ | $1$ | $1$ | $1$ |  |  |  | $4$
$8$ |  |  |  |  | $2$ |  |  |  |  |  | $2$
$7$ |  |  | $1$ | $1$ | $1$ | $1$ |  |  |  |  | $4$
$6$ |  |  | $1$ | $1$ | $1$ |  |  |  |  |  | $3$
$5$ |  |  | $1$ | $1$ |  |  |  |  |  |  | $2$
$4$ |  | $1$ |  | $1$ |  |  |  |  |  |  | $2$
$3$ |  | $1$ | $1$ |  |  |  |  |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  |  |  |  |  | $1$
$1$ |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |  |
$\sum c$ | $1$ | $3$ | $4$ | $5$ | $6$ | $5$ | $4$ | $3$ | $1$ |  |

Note that column totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})\quad\textmd{where}\quad
y=1,$

expanding to

$x^{8}+3x^{7}+4x^{6}+5x^{5}+6x^{4}+5x^{3}+4x^{2}+3x+1.$

Likewise, row totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})\quad\textmd{where}\quad
x=1,$

expanding to

$y^{16}+y^{14}+2y^{13}+2y^{12}+2y^{11}+3y^{10}+4y^{9}$
$+2y^{8}+4y^{7}+3y^{6}+2y^{5}+2y^{4}+2y^{3}+y^{2}+1.$

Combinatorial interpretation: Consider the aggregate of vectors

(11.7) $\begin{array}[]{ccc}\langle 1,4\rangle&&\langle 3,4\rangle\\\ \langle
1,3\rangle&\langle 2,3\rangle&\\\ \langle 1,2\rangle&&\end{array}$

Let $p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv},\langle a,b\rangle)$, denote the
number of partitions of $\langle a,b\rangle$ into distinct parts from (11.7).
Then

(11.8) $\prod_{k=2}^{4}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}p_{\,2,\,\textmd{up}}(\mathfrak{D}_{vpv},\langle
a,b\rangle)x^{a}y^{b},$

and each entry in the grid after (11.6) above gives each numerical value.

We continue in this way by considering the generating function

(11.9) $(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})$

which expanded gives us the grid,

|  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---
$16$ |  |  |  |  |  |  |  |  | $-1$ |  | $-1$
$15$ |  |  |  |  |  |  |  |  |  |  | $0$
$14$ |  |  |  |  |  |  |  | $1$ |  |  | $1$
$13$ |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$12$ |  |  |  |  |  | $1$ |  | $1$ |  |  | $2$
$11$ |  |  |  |  |  | $-1$ | $-1$ |  |  |  | $-2$
$10$ |  |  |  |  | $-1$ | $-1$ | $-1$ |  |  |  | $-3$
$9$ |  |  |  | $-1$ | $-1$ | $-1$ | $-1$ |  |  |  | $-4$
$8$ |  |  |  |  | $2$ |  |  |  |  |  | $2$
$7$ |  |  | $1$ | $1$ | $1$ | $1$ |  |  |  |  | $4$
$6$ |  |  | $1$ | $1$ | $1$ |  |  |  |  |  | $3$
$5$ |  |  | $1$ | $1$ |  |  |  |  |  |  | $2$
$4$ |  | $-1$ |  | $-1$ |  |  |  |  |  |  | $-2$
$3$ |  | $-1$ | $-1$ |  |  |  |  |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  |  |  |  |  | $-1$
$1$ |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |  |
$\sum c$ | $1$ | $-3$ | $2$ | $1$ | $2$ | $-1$ | $-2$ | $3$ | $-1$ |  |

Note that column totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})\quad\textmd{where}\quad
y=1,$

expanding to

$-x^{8}+3x^{7}-2x^{6}-x^{5}+x^{3}+2x^{2}-3x+1.$

Likewise, row totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})\quad\textmd{where}\quad
x=1,$

expanding to

$-y^{16}+y^{14}+2y^{13}+2y^{12}-2y^{11}-3y^{10}-4y^{9}+4y^{7}+3y^{6}+2y^{5}-2y^{4}-2y^{3}-y^{2}+1.$

Combinatorial interpretation: Consider the aggregate of vectors

(11.10) $\begin{array}[]{ccc}\langle 1,4\rangle&&\langle 3,4\rangle\\\ \langle
1,3\rangle&\langle 2,3\rangle&\\\ \langle 1,2\rangle&&\end{array}$

Let $p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{odd})};\langle
a,b\rangle)$, denote the number of partitions of $\langle a,b\rangle$ into
distinct odd number of parts from (11.7). Also let
$p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{even})};\langle
a,b\rangle)$, denote the number of partitions of $\langle a,b\rangle$ into
distinct even number of parts from (11.7).Then

(11.11) $\prod_{k=2}^{4}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1-x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}[p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{odd})};\langle
a,b\rangle)-p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{even})};\langle
a,b\rangle)]x^{a}y^{b},$

and each entry in the grid after (11.9) above gives each numerical value.

###  11.4. 2D Distinct Upper VPV Coefficients - Order 5

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

generates the grid

|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$36$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$
$35$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$34$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$
$33$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$32$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$ |  |  | $2$
$31$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  | $6$
$30$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ | $1$ |  |  |  | $3$
$29$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$ | $2$ | $2$ | $2$ |  |  |  | $8$
$28$ |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $4$ | $2$ | $1$ |  |  |  | $10$
$27$ |  |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $3$ | $3$ | $2$ | $1$ |  |  |  | $12$
$26$ |  |  |  |  |  |  |  |  |  |  |  | $2$ | $4$ | $5$ | $4$ | $2$ |  |  |  |  | $17$
$25$ |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $4$ | $4$ | $2$ | $1$ |  |  |  |  | $14$
$24$ |  |  |  |  |  |  |  |  |  | $1$ | $3$ | $5$ | $6$ | $5$ | $3$ | $1$ |  |  |  |  | $24$
$23$ |  |  |  |  |  |  |  |  |  | $1$ | $4$ | $6$ | $6$ | $4$ | $1$ |  |  |  |  |  | $22$
$22$ |  |  |  |  |  |  |  |  | $2$ | $3$ | $6$ | $7$ | $6$ | $3$ | $2$ |  |  |  |  |  | $29$
$21$ |  |  |  |  |  |  |  |  | $2$ | $5$ | $7$ | $7$ | $5$ | $2$ |  |  |  |  |  |  | $28$
$20$ |  |  |  |  |  |  |  | $1$ | $3$ | $6$ | $7$ | $6$ | $3$ | $1$ |  |  |  |  |  |  | $27$
$19$ |  |  |  |  |  |  | $1$ | $3$ | $6$ | $8$ | $8$ | $6$ | $3$ | $1$ |  |  |  |  |  |  | $36$
$18$ |  |  |  |  |  |  |  | $4$ | $6$ | $8$ | $6$ | $4$ |  |  |  |  |  |  |  |  | $28$
$17$ |  |  |  |  |  | $1$ | $3$ | $6$ | $8$ | $8$ | $6$ | $3$ | $1$ |  |  |  |  |  |  |  | $36$
$16$ |  |  |  |  |  | $1$ | $3$ | $6$ | $7$ | $6$ | $3$ | $1$ |  |  |  |  |  |  |  |  | $27$
$15$ |  |  |  |  |  | $2$ | $5$ | $7$ | $7$ | $5$ | $2$ |  |  |  |  |  |  |  |  |  | $28$
$14$ |  |  |  |  | $2$ | $3$ | $6$ | $7$ | $6$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  | $29$
$13$ |  |  |  |  | $1$ | $4$ | $6$ | $6$ | $4$ | $1$ |  |  |  |  |  |  |  |  |  |  | $22$
$12$ |  |  |  | $1$ | $3$ | $5$ | $6$ | $5$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  |  | $24$
$11$ |  |  |  | $1$ | $2$ | $4$ | $4$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  | $14$
$10$ |  |  |  | $2$ | $4$ | $5$ | $4$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  | $17$
$9$ |  |  | $1$ | $2$ | $3$ | $3$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $12$
$8$ |  |  | $1$ | $2$ | $4$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $10$
$7$ |  |  | $2$ | $2$ | $2$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $8$
$6$ |  |  | $1$ | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $3$
$5$ |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $6$
$4$ |  | $1$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$
$3$ |  | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$ | $18$ |  |
$\sum c$ | $1$ | $4$ | $8$ | $14$ | $23$ | $32$ | $41$ | $50$ | $55$ | $56$ | $55$ | $50$ | $41$ | $32$ | $23$ | $14$ | $8$ | $4$ | $1$ |  |

Note that column totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

where $y=1$, expanding to

$1+4x+8x^{2}+14x^{3}+23x^{4}+32x^{5}+41x^{6}+50x^{7}+55x^{8}+56x^{9}+55x^{10}+50x^{11}$
$+41x^{12}+32x^{13}+23x^{14}+14x^{15}+8x^{16}+4x^{17}+x^{18}.$

Likewise, row totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

where $x=1$, expanding to

$1+y^{2}+2y^{3}+2y^{4}+6y^{5}+3y^{6}+8y^{7}+10y^{8}+12y^{9}+17y^{10}+14y^{11}$
$+24y^{12}+22y^{13}+29y^{14}+28y^{15}+27y^{16}+36y^{17}+28y^{18}+36y^{19}$
$+27y^{20}+28y^{21}+29y^{22}+22y^{23}+24y^{24}+14y^{25}+17y^{26}+12y^{27}$
$+10y^{28}+8y^{29}+3y^{30}+6y^{31}+2y^{32}+2y^{33}+y^{34}+y^{36}.$

Combinatorial interpretation: Consider the aggregate of vectors

(11.12) $\begin{array}[]{cccc}\langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle\\\ \langle 1,4\rangle&&\langle 3,4\rangle&\\\
\langle 1,3\rangle&\langle 2,3\rangle&&\\\ \langle 1,2\rangle&&&\end{array}$

Let $p_{\,2,\textmd{vpv}}(\mathfrak{D},\langle a,b\rangle)$, denote the number
of partitions of $\langle a,b\rangle$ into distinct parts from (11.12). Then
we have the generating function

(11.13) $\prod_{k=2}^{5}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{36}\sum_{a=1}^{b-1}p_{\,2,\textmd{vpv}}(\mathfrak{D},\langle
a,b\rangle)x^{a}y^{b},$

encoding all possible partitions of this kind, so each entry in the previous
page $18\times 36$ grid gives all numerical values of
$p_{2}(\mathfrak{D},\langle a,b\rangle)$.

Next we see that

(11.14)
$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5})$

is a generating function for the grid

|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$36$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ |  | $-1$
$35$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$34$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$
$33$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$32$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$ |  |  | $2$
$31$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$ |  |  | $2$
$30$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ | $-1$ | $-1$ |  |  |  | $-3$
$29$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$ | $-2$ | $-2$ | $-2$ |  |  |  | $-8$
$28$ |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ | $-2$ | $-2$ | $-2$ | $-1$ |  |  |  | $-8$
$27$ |  |  |  |  |  |  |  |  |  |  |  | $-1$ |  | $-1$ | $-1$ |  | $-1$ |  |  |  | $-4$
$26$ |  |  |  |  |  |  |  |  |  |  |  |  | $2$ | $1$ | $2$ |  |  |  |  |  | $5$
$25$ |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $4$ | $4$ | $2$ | $1$ |  |  |  |  | $14$
$24$ |  |  |  |  |  |  |  |  |  | $1$ | $3$ | $3$ | $6$ | $3$ | $3$ | $1$ |  |  |  |  | $20$
$23$ |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $2$ | $2$ | $2$ | $1$ |  |  |  |  |  | $10$
$22$ |  |  |  |  |  |  |  |  | $-1$ |  | $-3$ |  | $-1$ |  |  |  |  |  |  |  | $-5$
$21$ |  |  |  |  |  |  |  |  | $-2$ | $-3$ | $-5$ | $-5$ | $-3$ | $-2$ |  |  |  |  |  |  | $-20$
$20$ |  |  |  |  |  |  |  | $-1$ | $-3$ | $-6$ | $-5$ | $-6$ | $-3$ | $-1$ |  |  |  |  |  |  | $-25$
$19$ |  |  |  |  |  |  | $-1$ | $-1$ | $4$ | $-4$ | $-4$ | $-4$ | $-1$ | $-1$ |  |  |  |  |  |  | $-20$
$18$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$17$ |  |  |  |  |  | $1$ | $1$ | $4$ | $4$ | $4$ | $4$ | $1$ | $1$ |  |  |  |  |  |  |  | $20$
$16$ |  |  |  |  |  | $1$ | $3$ | $6$ | $5$ | $6$ | $3$ | $1$ |  |  |  |  |  |  |  |  | $25$
$15$ |  |  |  |  |  | $2$ | $3$ | $5$ | $5$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  | $20$
$14$ |  |  |  |  |  | $1$ |  | $3$ |  | $1$ |  |  |  |  |  |  |  |  |  |  | $5$
$13$ |  |  |  |  | $-1$ | $-2$ | $-2$ | $-2$ | $-2$ | $-1$ |  |  |  |  |  |  |  |  |  |  | $-10$
$12$ |  |  |  | $-1$ | $-3$ | $-3$ | $-6$ | $-3$ | $-3$ | $-1$ |  |  |  |  |  |  |  |  |  |  | $-20$
$11$ |  |  |  | $-1$ | $-2$ | $-4$ | $-4$ | $-2$ | $-1$ |  |  |  |  |  |  |  |  |  |  |  | $-14$
$10$ |  |  |  |  | $-2$ | $-1$ | $-2$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $-5$
$9$ |  |  | $1$ |  | $1$ | $1$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $4$
$8$ |  |  | $1$ | $2$ | $2$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $8$
$7$ |  |  | $2$ | $2$ | $2$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $8$
$6$ |  |  | $1$ | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $3$
$5$ |  | $-1$ |  |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$4$ |  | $-1$ |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$3$ |  | $-1$ | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$ | $18$ |  |
$\sum c$ | $1$ | $-4$ | $4$ | $2$ | $-3$ | $0$ | $-7$ | $10$ | $-1$ | $0$ | $1$ | $-10$ | $7$ | $0$ | $3$ | $-2$ | $-4$ | $4$ | $-1$ |  |

Note that column totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5}),$

where $y=1$ expanding to

$1-4x+4x^{2}+2x^{3}-3x^{4}-7x^{6}+10x^{7}-x^{8}+x^{10}-10x^{11}$
$+7x^{12}+3x^{14}-2x^{15}-4x^{16}+4x^{17}-x^{18}.$

Likewise, row totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5}),$

where $x=1$ expanding to

$1-y^{2}-2y^{3}-2y^{4}-2y^{5}+3y^{6}+8y^{7}+8y^{8}+4y^{9}-5y^{10}-14y^{11}$
$-20y^{12}-10y^{13}+5y^{14}+20y^{15}+25y^{16}+20y^{17}-20y^{19}-25y^{20}$
$-20y^{21}-5y^{22}+10y^{23}+20y^{24}+14y^{25}+5y^{26}-4y^{27}-8y^{28}$
$-8y^{29}-3y^{30}+2y^{31}+2y^{32}+2y^{33}+y^{34}-y^{36}.$

Combinatorial interpretation: Consider the aggregate of vectors

(11.15) $\begin{array}[]{cccc}\langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle\\\ \langle 1,4\rangle&&\langle 3,4\rangle&\\\
\langle 1,3\rangle&\langle 2,3\rangle&&\\\ \langle 1,2\rangle&&&\end{array}$

Let $p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{odd})};\langle
a,b\rangle)$, denote the number of partitions of $\langle a,b\rangle$ into
distinct odd number of parts from (11.15). Also let
$p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{even})};\langle
a,b\rangle)$, denote the number of partitions of $\langle a,b\rangle$ into
distinct even number of parts from (11.15).Then

(11.16) $\prod_{k=2}^{5}\prod_{\begin{subarray}{c}j=1;\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1-x^{j}y^{k})=\sum_{b=2}^{36}\sum_{a=1}^{b-1}[p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{odd})};\langle
a,b\rangle)-p_{\,2,\,\textmd{up}}(\mathfrak{D}_{(vpv,\textmd{even})};\langle
a,b\rangle)]x^{a}y^{b},$

and each entry in the grid after (11.14) above gives each numerical value.

###  11.5. 2D weighted Upper VPV Coefficients - Order 5

The following product is five of the factors in the known VPV infinite product
for $\left(\frac{1-y}{1-xy}\right)^{\frac{1}{1-x}}$.

(11.17) $(1-xy^{2})^{1/2}$ $(1-xy^{3})^{1/3}(1-x^{2}y^{3})^{1/3}$
$(1-xy^{4})^{1/4}(1-x^{3}y^{4})^{1/4}$
$(1-xy^{5})^{1/5}(1-x^{2}y^{5})^{1/5}(1-x^{3}y^{5})^{1/5}(1-x^{4}y^{5})^{1/5}.$

This product is encoded by the grid

|  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---
$13$ |  |  |  | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  | $\vdots$
$12$ |  |  |  | $\frac{-301}{9600}$ | $\frac{-163279}{1555200}$ | $\frac{86923}{3110400}$ | $\frac{360869}{2073600}$ | $\frac{86923}{3110400}$ | $\frac{-163279}{1555200}$ | $\frac{-301}{9600}$ |  | $\frac{-2431243}{6220800}$
$11$ |  |  |  | $\frac{41}{1440}$ | $\frac{127}{3240}$ | $\frac{-899}{17280}$ | $\frac{-899}{17280}$ | $\frac{127}{3240}$ | $\frac{41}{1440}$ |  |  | $\frac{811}{25920}$
$10$ |  |  | $\frac{-2}{25}$ | $\frac{1171}{14400}$ | $\frac{-1511}{14400}$ | $\frac{23}{6400}$ | $\frac{-1511}{14400}$ | $\frac{1171}{14400}$ | $\frac{-2}{25}$ |  |  | $\frac{-12143}{57600}$
$9$ |  |  | $\frac{1}{20}$ | $\frac{-23}{810}$ | $\frac{61}{432}$ | $\frac{61}{432}$ | $\frac{-23}{810}$ | $\frac{1}{20}$ |  |  |  | $211/648$
$8$ |  |  | $\frac{-13}{480}$ | $\frac{317}{1440}$ | $\frac{583}{5760}$ | $\frac{317}{1440}$ | $\frac{-13}{480}$ |  |  |  |  | $2807/5760$
$7$ |  |  | $\frac{11}{60}$ | $\frac{9}{40}$ | $\frac{9}{40}$ | $\frac{11}{60}$ |  |  |  |  |  | $49/60$
$6$ |  |  | $\frac{1}{72}$ | $\frac{1}{144}$ | $\frac{1}{72}$ |  |  |  |  |  |  | $11/144$
$5$ |  | $\frac{-1}{5}$ | $\frac{-1}{30}$ | $\frac{-1}{30}$ | $\frac{-1}{5}$ |  |  |  |  |  |  | $-7/15$
$4$ |  | $\frac{1}{4}$ | $\frac{1}{8}$ | $\frac{1}{4}$ |  |  |  |  |  |  |  | $-5/8$
$3$ |  | $\frac{-1}{3}$ | $\frac{-1}{3}$ |  |  |  |  |  |  |  |  | $-2/3$
$2$ |  | $\frac{-1}{2}$ |  |  |  |  |  |  |  |  |  | $-1/2$
$1$ |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ |  |
$\sum c$ | $1$ | $\frac{-77}{60}$ | $\frac{-2531}{7200}$ | $\ldots$ | $see$ | $below$ |  |  |  |  |  |

Note that column totals generate from

$(1-xy^{2})^{1/2}(1-xy^{3})^{1/3}(1-x^{2}y^{3})^{1/3}(1-xy^{4})^{1/4}(1-x^{2}y^{4})^{1/4}(1-x^{3}y^{4})^{1/4}$
$(1-xy^{5})^{1/5}(1-x^{2}y^{5})^{1/5}(1-x^{3}y^{5})^{1/5}(1-x^{4}y^{5})^{1/5}\quad\textmd{where}\quad
y=1,$

expanding to

$1-\frac{77}{60}x-\frac{2531}{7200}x^{2}+\frac{360127}{1296000}x^{3}+\frac{54348601}{311040000}x^{4}+\frac{52396725643}{93312000000}x^{5}-\frac{15725249008811}{33592320000000}x^{6}$
$+\frac{536095858573681}{2015539200000000}x^{7}-\frac{221227534655582777}{967458816000000000}x^{8}-\frac{15078454659730017851}{522427760640000000000}x^{9}$
$+\frac{55218367281675862901707}{313456656384000000000000}x^{10}+O(x^{11})$

However, only the first three grid columns are complete. Likewise, row totals
generate from

$(1-xy^{2})^{1/2}(1-xy^{3})^{1/3}(1-x^{2}y^{3})^{1/3}(1-xy^{4})^{1/4}(1-x^{2}y^{4})^{1/4}(1-x^{3}y^{4})^{1/4}$
$(1-xy^{5})^{1/5}(1-x^{2}y^{5})^{1/5}(1-x^{3}y^{5})^{1/5}(1-x^{4}y^{5})^{1/5}\quad\textmd{where}\quad
x=1,$

expanding to

$1-\frac{1}{2}y^{2}-\frac{2}{3}y^{3}-\frac{5}{8}y^{4}-\frac{7}{15}y^{5}+\frac{11}{144}y^{6}+\frac{49}{60}y^{7}+\frac{2807}{5760}y^{8}+\frac{211}{648}y^{9}-\frac{12143}{57600}y^{10}$
$+\frac{811}{25920}y^{11}-\frac{2431243}{6220800}y^{12}+\frac{19889}{259200}y^{13}+\frac{626597}{2488320}y^{14}-\frac{7647377}{46656000}y^{15}+O(y^{15}).$

Combinatorial interpretation: Consider the sub-aggregate of the countable list
of upper first quadrant 2D Visible Point Vectors.

(11.18) $\begin{array}[]{cccc}\langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle\\\ \langle 1,4\rangle&&\langle 3,4\rangle&\\\
\langle 1,3\rangle&\langle 2,3\rangle&&\\\ \langle 1,2\rangle&&&\end{array}$

Let $p_{\,2,\textmd{up}}(\mathfrak{W},\langle a,b\rangle)$, denote the
weighted sum of partitions of $\langle a,b\rangle$ into parts of the form
$\binom{1/k}{n}\langle j,k\rangle$ for integers $n\geq 0$, with $1\leq j<k\leq
5$, $\gcd(j,k)=1$, so that each $\langle j,k\rangle$ from (11.18). Then

(11.19) $\prod_{k=2}^{5}\prod_{\begin{subarray}{c}j=1,\\\
\gcd(j,k)=1\end{subarray}}^{k-1}(1-x^{j}y^{k})^{1/k}=\sum_{b=2}^{\infty}\sum_{a=1}^{b-1}p_{\,2,\textmd{up}}(\mathfrak{W},\langle
a,b\rangle)x^{a}y^{b},$

and each entry in the grid after (11.17) gives each numerical value of
$p_{\,2,\textmd{up}}(\mathfrak{W},\langle a,b\rangle)$.

While the above argument of partitions of vectors looks a bit disheveled and
onerous, a priori there is another interpretation which appears more natural.

The Light Diffusion Model: Consider the following mapping of the (11.18)
vectors plus the origin point $\langle 0,0\rangle$, where each $\circ$ is
considered as an infinitesimal small lens that receives light intensity
$L_{n}$, and radiates it out at intensity $L_{n+1}$ with each
$L_{n+1}=f(L_{n})$. ie. $L_{n+1}$ is a function of $L_{n}$.

(11.20)
$\begin{array}[]{cccccc}\lx@intercol\hfil$5$\hfil\lx@intercol\vrule\lx@intercol&&\langle
1,5\rangle&\langle 2,5\rangle&\langle 3,5\rangle&\langle 4,5\rangle\\\
\cline{1-1}\cr\lx@intercol\hfil$4$\hfil\lx@intercol\vrule\lx@intercol&&\langle
1,4\rangle&&\langle 3,4\rangle&\\\
\cline{1-1}\cr\lx@intercol\hfil$3$\hfil\lx@intercol\vrule\lx@intercol&&\langle
1,3\rangle&\langle 2,3\rangle&&\\\
\cline{1-1}\cr\lx@intercol\hfil$2$\hfil\lx@intercol\vrule\lx@intercol&&\langle
1,2\rangle&&&\\\
\cline{1-1}\cr\lx@intercol\hfil$1$\hfil\lx@intercol\vrule\lx@intercol&&&&&\\\
\cline{1-1}\cr\lx@intercol\hfil$0$\hfil\lx@intercol\vrule\lx@intercol&\langle
0,0\rangle&&&&\\\
\cline{1-6}\cr\lx@intercol\hfil$y/x$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$0$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$1$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$2$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$3$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$4$\hfil\lx@intercol\vrule\lx@intercol\end{array}\mapsto\begin{array}[]{cccccc}\lx@intercol\hfil$5$\hfil\lx@intercol\vrule\lx@intercol&\uparrow&\circ&\circ&\circ&\circ\\\
\cline{1-1}\cr\lx@intercol\hfil$4$\hfil\lx@intercol\vrule\lx@intercol&\uparrow&\circ&&\circ&\nearrow\\\
\cline{1-1}\cr\lx@intercol\hfil$3$\hfil\lx@intercol\vrule\lx@intercol&\uparrow&\circ&\circ&\nearrow&\\\
\cline{1-1}\cr\lx@intercol\hfil$2$\hfil\lx@intercol\vrule\lx@intercol&\uparrow&\circ&\nearrow&&\\\
\cline{1-1}\cr\lx@intercol\hfil$1$\hfil\lx@intercol\vrule\lx@intercol&\uparrow&\nearrow&&&\\\
\cline{1-1}\cr\lx@intercol\hfil$0$\hfil\lx@intercol\vrule\lx@intercol&\circ&&&&\\\
\cline{1-6}\cr\lx@intercol\hfil$y/x$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$0$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$1$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$2$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$3$\hfil\lx@intercol\vrule\lx@intercol&\vrule\lx@intercol\hfil$4$\hfil\lx@intercol\vrule\lx@intercol\end{array}$

Suppose the origin lens $\circ$ radiates with intensity $\Upsilon$ upward
along $x=0$ and across to the diagonal $y=x$ in the radial region between
lines. The arrows in the right side grid here depict this phenomenon. Each
visible point designated by vectors $\langle a,b\rangle$ (with $\gcd(a,b)=1$
and $1\leq a<b\leq 5$) will receive the light ray at intensity $\Upsilon$,
then in turn radiate that light onward in the same shaped radial upward sector
around to the diagonal with intensity $f(\Upsilon)$, where $f$ is some defined
function. Each lens receiving light transmits that light further to a point
visible to it with a new intensity of $f(f(\Upsilon))$.

The product generating function (11.17) for our present exercise gives us a
means to examine light intensity radiating from an initial point, the origin,
and extending via visible points in a radial region from the origin. The
Visible Point Vector (VPV) identities can be construed as encoding this type
of light diffusion.

##  12\. Defining radial from origin region 2D Upper All Vectors aggregates

These are all lattice point vectors count in the infinite radial region of the
first quadrant bounded by the positive $y$ axis and the line $y=x$. The
visible lattice points in 3D space are similarly a countable set, and so also
are the n-component Vectors in nD space with $n$ a positive integer greater
than unity. In order to count the entire first 3D hyperquadrant comprising all
$a$, $b$, and $c$ positive integer components of $\langle a,b,c\rangle$ it
suffices to count the following vectors $\left\langle a,b,c\right\rangle$ such
that $0<a<b<c$ and apply symmetries in ways to be shown later in this paper.

A set of identities involving $n$ dimensional visible lattice points were
discovered by Campbell (1994). However, there has not been any emphasis on the
all vector analogues of the VPV identities, despite it seeming to be an
important area of research requiring a clear framework base from which to
launch researches.

The Upper All Vector (UAV) lattice points $\langle x,y\rangle$ with $x<y$ in
the first 2D quadrant are:

$\begin{array}[]{ccccccccc}&&&&etc.&&&&\\\ \langle 1,10\rangle&\langle
2,10\rangle&\langle 3,10\rangle&\langle 4,10\rangle&\langle
5,10\rangle&\langle 6,10\rangle&\langle 7,10\rangle&\langle
8,10\rangle&\langle 9,10\rangle\\\ \langle 1,9\rangle&\langle
2,9\rangle&\langle 3,9\rangle&\langle 4,9\rangle&\langle 5,9\rangle&\langle
6,9\rangle&\langle 7,9\rangle&\langle 8,9\rangle&\\\ \langle
1,8\rangle&\langle 2,8\rangle&\langle 3,8\rangle&\langle 4,8\rangle&\langle
5,8\rangle&\langle 6,8\rangle&\langle 7,8\rangle&&\\\ \langle
1,7\rangle&\langle 2,7\rangle&\langle 3,7\rangle&\langle 4,7\rangle&\langle
5,7\rangle&\langle 6,7\rangle&&&\\\ \langle 1,6\rangle&\langle
2,6\rangle&\langle 3,6\rangle&\langle 4,6\rangle&\langle 5,6\rangle&&&&\\\
\langle 1,5\rangle&\langle 2,5\rangle&\langle 3,5\rangle&\langle
4,5\rangle&&&&&\\\ \langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle&&&&&&\\\ \langle 1,3\rangle&\langle 2,3\rangle&&&&&&&\\\ \langle
1,2\rangle&&&&&&&&\end{array}$

These are the countable list of upper first quadrant 2D Vectors. That is, the
vectors in the infinitely extended radial region of the 2D first quadrant
between the $y$-axis and the line $y=x$.

##  13\. Examples of 2D Upper All Vectors finite generating functions

###  13.1. 2D Distinct Upper All Vectors Coefficients - Order 2

Note that for Orders 2 and 3 the Upper All Vectors set is identical to the
Upper VPSs.

(13.1) $(1+xy^{2})\quad\quad\quad and\quad\quad\quad(1-xy^{2})$ $3$ |  |  |  |  | $\sum r$
---|---|---|---|---|---
$2$ |  | $1$ |  |  | $1$
$1$ |  |  |  |  | $0$
$0$ | $1$ |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ |  |
$\sum c$ | $1$ | $1$ |  |  |
$3$ |  |  |  |  | $\sum r$
---|---|---|---|---|---
$2$ |  | $-1$ |  |  | $-1$
$1$ |  |  |  |  | $0$
$0$ | $1$ |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ |  |
$\sum c$ | $1$ | $-1$ |  |  |

###  13.2. 2D Distinct Upper All Vectors Coefficients - Order 3

Note that for Order 3 the Upper All Vectors set is identical to the Upper VPVs
set.

(13.2) $(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})\quad\quad and\quad\quad(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})$ |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---
$8$ |  |  |  |  | $1$ |  | $1$
$7$ |  |  |  |  |  |  | $0$
$6$ |  |  |  | $1$ |  |  | $1$
$5$ |  |  | $1$ | $1$ |  |  | $2$
$4$ |  |  |  |  |  |  | $0$
$3$ |  | $1$ | $1$ |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  | $1$
$1$ |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ |  |
$\sum c$ | $1$ | $2$ | $2$ | $2$ | $1$ |  |
|  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---
$8$ |  |  |  |  | $-1$ |  | $-1$
$7$ |  |  |  |  |  |  | $0$
$6$ |  |  |  | $1$ |  |  | $1$
$5$ |  |  | $1$ | $1$ |  |  | $2$
$4$ |  |  |  |  |  |  | $0$
$3$ |  | $-1$ | $-1$ |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  | $1$
$1$ |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ |  |
$\sum c$ | $1$ | $-2$ | $0$ | $2$ | $-1$ |  |

###  13.3. 2D Distinct Upper All Vectors Coefficients - Order 4

(13.3) $(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})$ |  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
$20$ |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$
$19$ |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$18$ |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$
$17$ |  |  |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$16$ |  |  |  |  |  |  |  | $1$ | $1$ | $1$ |  |  | $3$
$15$ |  |  |  |  |  |  |  | $1$ | $1$ |  |  |  | $2$
$14$ |  |  |  |  |  |  | $1$ | $2$ | $1$ |  |  |  | $4$
$13$ |  |  |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  | $6$
$12$ |  |  |  |  |  | $1$ | $2$ | $1$ |  |  |  |  | $4$
$11$ |  |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  | $6$
$10$ |  |  |  |  | $2$ | $2$ | $2$ |  |  |  |  |  | $6$
$9$ |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  |  | $6$
$8$ |  |  |  | $1$ | $2$ | $1$ |  |  |  |  |  |  | $4$
$7$ |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  |  |  | $6$
$6$ |  |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  | $4$
$5$ |  |  | $1$ | $1$ |  |  |  |  |  |  |  |  | $2$
$4$ |  | $1$ | $1$ | $1$ |  |  |  |  |  |  |  |  | $3$
$3$ |  | $1$ | $1$ |  |  |  |  |  |  |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  |  |  |  |  |  |  | $1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ |  |
$\sum c$ | $1$ | $3$ | $5$ | $8$ | $10$ | $10$ | $10$ | $8$ | $5$ | $3$ | $1$ |  |

Note that column totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})\quad\textmd{where}\quad
y=1,$

expanding to

$x^{10}+3x^{9}+5x^{8}+8x^{7}+10x^{6}+10x^{5}+10x^{4}+8x^{3}+5x^{2}+3x+1.$

Likewise, row totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})\quad\textmd{where}\quad
x=1,$

expanding to

$y^{20}+y^{18}+2y^{17}+3y^{16}+2y^{15}+4y^{14}+6y^{13}+4y^{12}+6y^{11}$
$+6y^{10}+6y^{9}+4y^{8}+6y^{7}+4y^{6}+2y^{5}+3y^{4}+2y^{3}+y^{2}+1.$

Combinatorial interpretation: Consider the aggregate of vectors

(13.4) $\begin{array}[]{ccc}\langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle\\\ \langle 1,3\rangle&\langle 2,3\rangle&\\\ \langle
1,2\rangle&&\end{array}$

Let $p_{2}(\mathfrak{D},\langle a,b\rangle)$, denote the number of partitions
of $\langle a,b\rangle$ into distinct parts from (13.4). Then

(13.5)
$\prod_{k=2}^{4}\prod_{j=1}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}p_{2}(\mathfrak{D},\langle
a,b\rangle)x^{a}y^{b},$

and each entry in the grid after (13.3) above gives each numerical value.

Next we consider the following product

(13.6)
$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})$

which is the generating function for the grid

|  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
$20$ |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$
$19$ |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$18$ |  |  |  |  |  |  |  |  |  | $-1$ |  |  | $-1$
$17$ |  |  |  |  |  |  |  |  | $-1$ | $-1$ |  |  | $-2$
$16$ |  |  |  |  |  |  |  | $-1$ | $-1$ | $1$ |  |  | $-3$
$15$ |  |  |  |  |  |  |  | $1$ | $1$ |  |  |  | $2$
$14$ |  |  |  |  |  |  | $1$ | $2$ | $1$ |  |  |  | $4$
$13$ |  |  |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  | $6$
$12$ |  |  |  |  |  | $1$ |  | $1$ |  |  |  |  | $2$
$11$ |  |  |  |  | $-1$ | $-2$ | $-2$ | $-1$ |  |  |  |  | $-6$
$10$ |  |  |  |  | $-2$ | $-2$ | $-2$ |  |  |  |  |  | $-6$
$9$ |  |  |  | $-1$ | $-2$ | $-2$ | $-1$ |  |  |  |  |  | $-6$
$8$ |  |  |  | $1$ |  | $1$ |  |  |  |  |  |  | $2$
$7$ |  |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  |  |  | $6$
$6$ |  |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  | $4$
$5$ |  |  | $1$ | $1$ |  |  |  |  |  |  |  |  | $2$
$4$ |  | $-1$ | $-1$ | $-1$ |  |  |  |  |  |  |  |  | $-3$
$3$ |  | $-1$ | $-1$ |  |  |  |  |  |  |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  |  |  |  |  |  |  | $-1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ |  |
$\sum c$ | $1$ | $-3$ | $1$ | $4$ | $-2$ | $-2$ | $-2$ | $4$ | $1$ | $-3$ | $1$ |  |

We see that column totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})\quad\textmd{where}\quad
y=1,$

expanding to

$x^{10}-3x^{9}+x^{8}+4x^{7}-2x^{6}-2x^{5}-2x^{4}+4x^{3}+x^{2}-3x+1.$

Likewise, row totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})\quad\textmd{where}\quad
x=1,$

expanding to

$y^{20}-y^{18}-2y^{17}-3y^{16}+2y^{15}+4y^{14}+6y^{13}+2y^{12}-6y^{11}$
$-6y^{10}-6y^{9}+2y^{8}+6y^{7}+4y^{6}+2y^{5}-3y^{4}-2y^{3}-y^{2}+1.$

Combinatorial interpretation: Consider the aggregate of vectors

(13.7) $\begin{array}[]{ccc}\langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle\\\ \langle 1,3\rangle&\langle 2,3\rangle&\\\ \langle
1,2\rangle&&\end{array}$

Let $p_{2,\,e}(\mathfrak{D},\langle a,b\rangle)$, respectively
$p_{2,\,o}(\mathfrak{D},\langle a,b\rangle)$, denote the number of partitions
of $\langle a,b\rangle$ from (13.7) into an even respectively odd number of
distinct parts. Then

(13.8)
$\prod_{k=2}^{4}\prod_{j=1}^{k-1}(1-x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}\left[p_{2,\,e}(\mathfrak{D},\langle
a,b\rangle)-p_{2,\,o}(\mathfrak{D},\langle a,b\rangle)\right]x^{a}y^{b},$

and each entry in the grid after (13.6) above gives each numerical value.

###  13.4. 2D Unrestricted Upper All Vectors Partitions - Order 4

(13.9)
$\frac{1}{(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})}$

encodes the following part sector of the infinitely extended radial grid

|  |  |  |  |  |  | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$20$ |  |  |  |  |  | $1$ | $4$ | $11$ | $21$ | $31$ | $37$ | $31$ | $21$ | $11$ | $4$ |  | $173$
$19$ |  |  |  |  |  | $1$ | $1$ | $5$ | $12$ | $21$ | $28$ | $21$ | $12$ | $5$ | $1$ |  | $134$
$18$ |  |  |  |  |  | $2$ | $7$ | $15$ | $23$ | $28$ | $23$ | $15$ | $7$ | $2$ |  |  | $122$
$17$ |  |  |  |  |  | $2$ | $7$ | $14$ | $20$ | $20$ | $14$ | $7$ | $2$ |  |  |  | $86$
$16$ |  |  |  |  | $1$ | $4$ | $10$ | $17$ | $21$ | $17$ | $10$ | $4$ | $1$ |  |  |  | $85$
$15$ |  |  |  |  | $1$ | $5$ | $10$ | $15$ | $15$ | $10$ | $5$ | $1$ |  |  |  |  | $62$
$14$ |  |  |  |  | $2$ | $6$ | $12$ | $15$ | $12$ | $6$ | $2$ |  |  |  |  |  | $55$
$13$ |  |  |  |  | $2$ | $6$ | $10$ | $10$ | $6$ | $2$ |  |  |  |  |  |  | $36$
$12$ |  |  |  | $1$ | $4$ | $8$ | $11$ | $8$ | $4$ | $1$ |  |  |  |  |  |  | $37$
$11$ |  |  |  | $1$ | $4$ | $7$ | $7$ | $4$ | $1$ |  |  |  |  |  |  |  | $24$
$10$ |  |  |  | $2$ | $5$ | $8$ | $5$ | $2$ |  |  |  |  |  |  |  |  | $22$
$9$ |  |  |  | $2$ | $4$ | $4$ | $2$ |  |  |  |  |  |  |  |  |  | $12$
$8$ |  |  | $1$ | $3$ | $5$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  | $13$
$7$ |  |  | $1$ | $3$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  |  | $8$
$6$ |  |  | $2$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  |  |  | $7$
$5$ |  |  | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $2$
$4$ |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $4$
$3$ |  | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ |  |
$\sum c$ | $1$ | $3$ | $8$ | $17$ | $33$ | $58$ | $97$ | $153$ | $233$ | $342$ | $489$ | $681$ | $930$ | $1245$ | $1641$ |  |

Note that column totals generate from

$\frac{1}{(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})}\quad\textmd{where}\quad
y=1,$

expanding to

$1+3x+8x^{2}+17x^{3}+33x^{4}+58x^{5}+97x^{6}+153x^{7}+233x^{8}+342x^{9}$
$+489x^{10}+681x^{11}+930x^{12}+1245x^{13}+1641x^{14}+2130x^{15}+2730x^{16}$
$+3456x^{17}+4330x^{18}+O(x^{19})$

Likewise, row totals generate from

$\frac{1}{(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})}\quad\textmd{where}\quad
x=1,$

expanding to

$1+y^{2}+2y^{3}+4y^{4}+2y^{5}+7y^{6}+8y^{7}+13y^{8}+12y^{9}+22y^{10}+24y^{11}+37y^{12}$
$+36y^{13}+55y^{14}+62y^{15}+85y^{16}+86y^{17}+122y^{18}+134y^{19}+173y^{20}+182y^{21}$
$+239y^{22}+260y^{23}+326y^{24}+344y^{25}+431y^{26}+470y^{27}+569y^{28}+602y^{29}$
$+734y^{30}+794y^{31}+938y^{32}+O(y^{33})$

Combinatorial interpretation: Consider the aggregate of vectors

(13.10) $\begin{array}[]{ccc}\langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle\\\ \langle 1,3\rangle&\langle 2,3\rangle&\\\ \langle
1,2\rangle&&\end{array}$

Let $p_{2,\textmd{up}}(\mathfrak{U},\langle a,b\rangle)$, denote the number of
partitions of $\langle a,b\rangle$ into unrestricted parts from (13.10). Then

(13.11)
$\prod_{k=2}^{4}\prod_{j=1}^{k-1}\frac{1}{(1-x^{j}y^{k})}=\sum_{b=2}^{\infty}\sum_{a=1}^{b-1}p_{2,\textmd{up}}(\mathfrak{U},\langle
a,b\rangle)x^{a}y^{b},$

and each entry in the grid after (13.9) above gives each numerical value.

Observable Features. Just quickly inspecting the grid suggests the following
points for the 2D Unrestricted Upper All Vectors Partitions of Order 4:

  1. (a)

The grid table extends indefinitely and therefore is not simply a finite patch
as are most of the distinct partitions grids so far given.

  2. (b)

The grid bounds are best defined by the radial region between the lines $y=4x$
and $y=x$ with some of the vectors actually on $y=4x$.

  3. (c)

Each row of number entries is a symmetric finite sequence.

###  13.5. 2D Distinct Upper All Vectors Coefficients - Order 5

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

is encoded by the grid

|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$40$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$
$39$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$38$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$
$37$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$36$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ | $1$ |  |  | $3$
$35$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $2$ | $1$ |  |  | $6$
$34$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $1$ |  |  |  | $4$
$33$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$ | $3$ | $3$ | $2$ |  |  |  | $10$
$32$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $3$ | $4$ | $3$ | $1$ |  |  |  | $12$
$31$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $3$ | $5$ | $5$ | $3$ | $1$ |  |  |  | $18$
$30$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$ | $5$ | $6$ | $5$ | $2$ |  |  |  |  | $20$
$29$ |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $4$ | $6$ | $6$ | $4$ | $1$ |  |  |  |  | $8$
$28$ |  |  |  |  |  |  |  |  |  |  |  | $1$ | $4$ | $7$ | $10$ | $7$ | $4$ | $1$ |  |  |  |  | $34$
$27$ |  |  |  |  |  |  |  |  |  |  |  | $2$ | $6$ | $9$ | $9$ | $6$ | $2$ |  |  |  |  |  | $34$
$26$ |  |  |  |  |  |  |  |  |  |  | $2$ | $5$ | $10$ | $12$ | $10$ | $5$ | $2$ |  |  |  |  |  | $46$
$25$ |  |  |  |  |  |  |  |  |  |  | $3$ | $7$ | $11$ | $11$ | $7$ | $3$ |  |  |  |  |  |  | $42$
$24$ |  |  |  |  |  |  |  |  |  | $2$ | $6$ | $11$ | $13$ | $11$ | $6$ | $2$ |  |  |  |  |  |  | $51$
$23$ |  |  |  |  |  |  |  |  | $1$ | $4$ | $10$ | $14$ | $14$ | $10$ | $4$ | $1$ |  |  |  |  |  |  | $58$
$22$ |  |  |  |  |  |  |  |  | $2$ | $7$ | $12$ | $15$ | $12$ | $7$ | $2$ |  |  |  |  |  |  |  | $53$
$21$ |  |  |  |  |  |  |  | $1$ | $5$ | $11$ | $15$ | $15$ | $11$ | $5$ | $1$ |  |  |  |  |  |  |  | $64$
$20$ |  |  |  |  |  |  |  | $2$ | $6$ | $12$ | $14$ | $12$ | $6$ | $2$ |  |  |  |  |  |  |  |  | $54$
$19$ |  |  |  |  |  |  | $1$ | $5$ | $11$ | $15$ | $15$ | $11$ | $5$ | $1$ |  |  |  |  |  |  |  |  | $64$
$18$ |  |  |  |  |  |  | $2$ | $7$ | $12$ | $15$ | $12$ | $7$ | $2$ |  |  |  |  |  |  |  |  |  | $57$
$17$ |  |  |  |  |  | $1$ | $4$ | $10$ | $14$ | $14$ | $10$ | $4$ | $1$ |  |  |  |  |  |  |  |  |  | $58$
$16$ |  |  |  |  |  | $2$ | $6$ | $11$ | $13$ | $11$ | $6$ | $2$ |  |  |  |  |  |  |  |  |  |  | $51$
$15$ |  |  |  |  |  | $3$ | $7$ | $11$ | $11$ | $7$ | $3$ |  |  |  |  |  |  |  |  |  |  |  | $42$
$14$ |  |  |  |  | $2$ | $5$ | $10$ | $12$ | $10$ | $5$ | $2$ |  |  |  |  |  |  |  |  |  |  |  | $46$
$13$ |  |  |  |  | $2$ | $6$ | $9$ | $9$ | $6$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  | $34$
$12$ |  |  |  | $1$ | $4$ | $7$ | $10$ | $7$ | $4$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $34$
$11$ |  |  |  | $1$ | $4$ | $6$ | $6$ | $4$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $22$
$10$ |  |  |  | $2$ | $5$ | $6$ | $5$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $20$
$9$ |  |  | $1$ | $3$ | $5$ | $5$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $18$
$8$ |  |  | $1$ | $3$ | $4$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $12$
$7$ |  |  | $2$ | $3$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $10$
$6$ |  |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $4$
$5$ |  | $1$ | $2$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $6$
$4$ |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $4$
$3$ |  | $1$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $2$
$2$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$ | $18$ | $19$ | $20$ |  |
$\sum c$ | $1$ | $4$ | $9$ | $18$ | $31$ | $46$ | $64$ | $82$ | $96$ | $106$ | $110$ | $106$ | $96$ | $82$ | $64$ | $46$ | $31$ | $18$ | $9$ | $4$ | $1$ |  |

Note that column totals generate from

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

where $y=1$, expanding to

$x^{20}+4x^{19}+9x^{18}+18x^{17}+31x^{16}+46x^{15}+64x^{14}+82x^{13}+96x^{12}+106x^{11}$
$+110x^{10}+106x^{9}+96x^{8}+82x^{7}+64x^{6}+46x^{5}+31x^{4}+18x^{3}+9x^{2}+4x+1.$

Likewise, row totals generate from again

$(1+xy^{2})(1+xy^{3})(1+x^{2}y^{3})(1+xy^{4})(1+x^{2}y^{4})(1+x^{3}y^{4})(1+xy^{5})(1+x^{2}y^{5})(1+x^{3}y^{5})(1+x^{4}y^{5})$

where $x=1$, expanding to

$1+y^{2}+2y^{3}+3y^{4}+6y^{5}+4y^{6}+10y^{7}+12y^{8}+18y^{9}+20y^{10}+22y^{11}+34y^{12}$
$+34y^{13}+46y^{14}+42y^{15}+51y^{16}+58y^{17}+57y^{18}+64y^{19}+54y^{20}+64y^{21}$
$+57y^{22}+58y^{23}+51y^{24}+42y^{25}+46y^{26}+34y^{27}+34y^{28}+22y^{29}+20y^{30}$
$+18y^{31}+12y^{32}+10y^{33}+4y^{34}+6y^{35}+3y^{36}+2y^{37}+y^{38}+y^{40}.$

Combinatorial interpretation: Consider the aggregate of vectors

(13.12) $\begin{array}[]{cccc}\langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle\\\ \langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle&\\\ \langle 1,3\rangle&\langle 2,3\rangle&&\\\ \langle
1,2\rangle&&&\end{array}$

Let $p_{2}(\mathfrak{D},\langle a,b\rangle)$, denote the number of partitions
of $\langle a,b\rangle$ into distinct parts from (13.12). Then we have the
generating function

(13.13)
$\prod_{k=2}^{4}\prod_{j=1}^{k-1}(1+x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}p_{2}(\mathfrak{D},\langle
a,b\rangle)x^{a}y^{b},$

encoding all possible partitions of this kind, so each entry in the previous
page $20\times 40$ grid gives all numerical values of
$p_{2}(\mathfrak{D},\langle a,b\rangle)$.

Next we see that

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5})$

is encoded by the grid

|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\sum r$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$36$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ |  | $-1$
$35$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$34$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$
$33$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ | $1$ |  |  | $2$
$32$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  | $1$ |  |  | $2$
$31$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$ |  |  | $1$ |  |  | $2$
$30$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ | $-2$ | $-1$ |  |  |  | $-3$
$29$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$ | $-2$ | $-2$ | $-2$ |  |  |  | $-8$
$28$ |  |  |  |  |  |  |  |  |  |  |  |  | $-1$ | $-2$ | $-2$ | $-2$ | $-1$ |  |  |  | $-8$
$27$ |  |  |  |  |  |  |  |  |  |  |  | $-1$ |  | $-1$ | $-1$ |  | $-1$ |  |  |  | $-4$
$26$ |  |  |  |  |  |  |  |  |  |  |  |  | $2$ | $1$ | $2$ |  |  |  |  |  | $5$
$25$ |  |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $4$ | $4$ | $2$ | $1$ |  |  |  |  | $14$
$24$ |  |  |  |  |  |  |  |  |  | $1$ | $3$ | $3$ | $6$ | $3$ | $3$ | $1$ |  |  |  |  | $20$
$23$ |  |  |  |  |  |  |  |  |  | $1$ | $2$ | $2$ | $2$ | $2$ | $1$ |  |  |  |  |  | $10$
$22$ |  |  |  |  |  |  |  |  | $-1$ |  | $-3$ |  | $-1$ |  |  |  |  |  |  |  | $-5$
$21$ |  |  |  |  |  |  |  |  | $-2$ | $-3$ | $-5$ | $-5$ | $-3$ | $-2$ |  |  |  |  |  |  | $-20$
$20$ |  |  |  |  |  |  |  | $-1$ | $-3$ | $-6$ | $-5$ | $-6$ | $-3$ | $-1$ |  |  |  |  |  |  | $-25$
$19$ |  |  |  |  |  |  | $-1$ | $-1$ | $4$ | $-4$ | $-4$ | $-4$ | $-1$ | $-1$ |  |  |  |  |  |  | $-20$
$18$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$17$ |  |  |  |  |  | $1$ | $1$ | $4$ | $4$ | $4$ | $4$ | $1$ | $1$ |  |  |  |  |  |  |  | $20$
$16$ |  |  |  |  |  | $1$ | $3$ | $6$ | $5$ | $6$ | $3$ | $1$ |  |  |  |  |  |  |  |  | $25$
$15$ |  |  |  |  |  | $2$ | $3$ | $5$ | $5$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  | $20$
$14$ |  |  |  |  |  | $1$ |  | $3$ |  | $1$ |  |  |  |  |  |  |  |  |  |  | $5$
$13$ |  |  |  |  | $-1$ | $-2$ | $-2$ | $-2$ | $-2$ | $-1$ |  |  |  |  |  |  |  |  |  |  | $-10$
$12$ |  |  |  | $-1$ | $-3$ | $-3$ | $-6$ | $-3$ | $-3$ | $-1$ |  |  |  |  |  |  |  |  |  |  | $-20$
$11$ |  |  |  | $-1$ | $-2$ | $-4$ | $-4$ | $-2$ | $-1$ |  |  |  |  |  |  |  |  |  |  |  | $-14$
$10$ |  |  |  |  | $-2$ | $-1$ | $-2$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $-5$
$9$ |  |  | $1$ |  | $1$ | $1$ |  | $1$ |  |  |  |  |  |  |  |  |  |  |  |  | $4$
$8$ |  |  | $1$ | $3$ | $2$ | $3$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  | $10$
$7$ |  |  | $2$ | $3$ | $3$ | $2$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $8$
$6$ |  |  | $1$ | $2$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $3$
$5$ |  | $-1$ |  |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$4$ |  | $-1$ |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$3$ |  | $-1$ | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-2$
$2$ |  | $-1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $-1$
$1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $0$
$0$ | $1$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $1$
$y/x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$ | $18$ |  |
$\sum c$ | $1$ | $-4$ | $4$ | $2$ | $-3$ | $0$ | $-7$ | $10$ | $-1$ | $0$ | $1$ | $-10$ | $7$ | $0$ | $3$ | $-2$ | $-4$ | $4$ | $-1$ |  |

Note that column totals generate from

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5})$

where $y=1$, expanding to

$1-4x+3x^{2}+6x^{3}-7x^{4}-2x^{5}-4x^{6}+10x^{7}+6x^{8}-10x^{9}+2x^{10}-10x^{11}$
$+6x^{12}+10x^{13}-4x^{14}-2x^{15}-7x^{16}+6x^{17}+3x^{18}-4x^{19}+x^{20}.$

Likewise, row totals generate from again

$(1-xy^{2})(1-xy^{3})(1-x^{2}y^{3})(1-xy^{4})(1-x^{2}y^{4})(1-x^{3}y^{4})(1-xy^{5})(1-x^{2}y^{5})(1-x^{3}y^{5})(1-x^{4}y^{5})$

where $x=1$, expanding to

$1-y^{2}-2y^{3}-3y^{4}-2y^{5}+4y^{6}+10y^{7}+10y^{8}+6y^{9}-8y^{10}-22y^{11}-28y^{12}$
$-14y^{13}+10y^{14}+34y^{15}+45y^{16}+30y^{17}-5y^{18}-40y^{19}-50y^{20}-40y^{21}$
$-5y^{22}+30y^{23}+45y^{24}+34y^{25}+10y^{26}-14y^{27}-28y^{28}-22y^{29}-8y^{30}$
$+6y^{31}+10y^{32}+10y^{33}+4y^{34}-2y^{35}-3y^{36}-2y^{37}-y^{38}+y^{40}.$

Combinatorial interpretation: Consider the aggregate of vectors

(13.14) $\begin{array}[]{cccc}\langle 1,5\rangle&\langle 2,5\rangle&\langle
3,5\rangle&\langle 4,5\rangle\\\ \langle 1,4\rangle&\langle 2,4\rangle&\langle
3,4\rangle&\\\ \langle 1,3\rangle&\langle 2,3\rangle&&\\\ \langle
1,2\rangle&&&\end{array}$

Let $p_{(2,e)}(\mathfrak{D},\langle a,b\rangle)$, denote the number of
partitions of $\langle a,b\rangle$ into an even number of distinct parts from
(13.14). Let $p_{(2,o)}(\mathfrak{D},\langle a,b\rangle)$, denote the number
of partitions of $\langle a,b\rangle$ into an odd number of distinct parts
from (13.14). Then we have the generating function

(13.15)
$\prod_{k=2}^{4}\prod_{j=1}^{k-1}(1-x^{j}y^{k})=\sum_{b=2}^{20}\sum_{a=1}^{b-1}[p_{(2,e)}(\mathfrak{D},\langle
a,b\rangle)-p_{(2,o)}(\mathfrak{D},\langle a,b\rangle)]x^{a}y^{b},$

encoding all $20\times 40$ grid numerical values of
$p_{(2,e)}(\mathfrak{D},\langle a,b\rangle)-p_{(2,o)}(\mathfrak{D},\langle
a,b\rangle)$, with respect to $\langle a,b\rangle)$.

##  14\. Introducing the VPV identities

In the 1990s and up to 2000 the author published a series of papers
introducing the so-called Visible Point Vector (VPV) identities. In these
papers such as for example Campbell [13], [14], [16], the identities of the
present note were given. They attracted some interest, but most people did not
see them as much more than curiosities. Perhaps the proofs in the early papers
were too obscure or cryptic. So, our approach here is to introduce the deeper
$n$-dimensional identities by first trying to simply derive the easier
2-dimensional VPV identities.

##  15\. Deriving the $2D$ first quadrant VPV identity.

We will derive the 2D VPV identity for the first quadrant integer lattice
points. Let $(j,k)=1$ mean that $j$ and $k$ are coprime integers. We see that

$\left(\sum_{m=1}^{\infty}\frac{y^{m}}{m^{a}}\right)\left(\sum_{n=1}^{\infty}\frac{z^{n}}{n^{b}}\right)$

$=\left(\frac{y^{1}}{1^{a}}+\frac{y^{2}}{2^{a}}+\frac{y^{3}}{3^{a}}+\frac{y^{4}}{4^{a}}+\frac{y^{5}}{5^{a}}+\frac{y^{6}}{6^{a}}+\frac{y^{7}}{7^{a}}+\cdots\right)\left(\frac{z^{1}}{1^{b}}+\frac{z^{2}}{2^{b}}+\frac{z^{3}}{3^{b}}+\frac{z^{4}}{4^{b}}+\frac{z^{5}}{5^{b}}+\frac{z^{6}}{6^{b}}+\frac{z^{7}}{7^{b}}+\cdots\right)$

$=\frac{y^{1}z^{1}}{1^{a}1^{b}}+\frac{y^{2}z^{1}}{2^{a}1^{b}}+\frac{y^{3}z^{1}}{3^{a}1^{b}}+\frac{y^{4}z^{1}}{4^{a}1^{b}}+\frac{y^{5}z^{1}}{5^{a}1^{b}}+\frac{y^{6}z^{1}}{6^{a}1^{b}}+\frac{y^{7}z^{1}}{7^{a}1^{b}}+\cdots$
$+\frac{y^{1}z^{2}}{1^{a}2^{b}}+\frac{y^{2}z^{2}}{2^{a}2^{b}}+\frac{y^{3}z^{2}}{3^{a}2^{b}}+\frac{y^{4}z^{2}}{4^{a}2^{b}}+\frac{y^{5}z^{2}}{5^{a}2^{b}}+\frac{y^{6}z^{2}}{6^{a}2^{b}}+\frac{y^{7}z^{2}}{7^{a}2^{b}}+\cdots$
$+\frac{y^{1}z^{3}}{1^{a}3^{b}}+\frac{y^{2}z^{3}}{2^{a}3^{b}}+\frac{y^{3}z^{3}}{3^{a}3^{b}}+\frac{y^{4}z^{3}}{4^{a}3^{b}}+\frac{y^{5}z^{3}}{5^{a}3^{b}}+\frac{y^{6}z^{3}}{6^{a}3^{b}}+\frac{y^{7}z^{3}}{7^{a}3^{b}}+\cdots$
$+\frac{y^{1}z^{4}}{1^{a}4^{b}}+\frac{y^{2}z^{4}}{2^{a}4^{b}}+\frac{y^{3}z^{4}}{3^{a}4^{b}}+\frac{y^{4}z^{4}}{4^{a}4^{b}}+\frac{y^{5}z^{4}}{5^{a}4^{b}}+\frac{y^{6}z^{4}}{6^{a}4^{b}}+\frac{y^{7}z^{4}}{7^{a}4^{b}}+\cdots$
$+\frac{y^{1}z^{5}}{1^{a}5^{b}}+\frac{y^{2}z^{5}}{2^{a}5^{b}}+\frac{y^{3}z^{5}}{3^{a}5^{b}}+\frac{y^{4}z^{5}}{4^{a}5^{b}}+\frac{y^{5}z^{5}}{5^{a}5^{b}}+\frac{y^{6}z^{5}}{6^{a}5^{b}}+\frac{y^{7}z^{5}}{7^{a}5^{b}}+\cdots$
$+\frac{y^{1}z^{6}}{1^{a}6^{b}}+\frac{y^{2}z^{6}}{2^{a}6^{b}}+\frac{y^{3}z^{6}}{3^{a}6^{b}}+\frac{y^{4}z^{6}}{4^{a}6^{b}}+\frac{y^{5}z^{6}}{5^{a}6^{b}}+\frac{y^{6}z^{6}}{6^{a}6^{b}}+\frac{y^{7}z^{6}}{7^{a}6^{b}}+\cdots$
$+\frac{y^{1}z^{7}}{1^{a}7^{b}}+\frac{y^{2}z^{7}}{2^{a}7^{b}}+\frac{y^{3}z^{7}}{3^{a}7^{b}}+\frac{y^{4}z^{7}}{4^{a}7^{b}}+\frac{y^{5}z^{7}}{5^{a}7^{b}}+\frac{y^{6}z^{7}}{6^{a}7^{b}}+\frac{y^{7}z^{7}}{7^{a}7^{b}}+\cdots$
$+\quad\vdots\quad+\quad\vdots\quad+\quad\vdots\quad+\quad\vdots\quad+\quad\vdots\quad+\quad\vdots\quad+\quad\vdots\quad\ddots$
$=\sum_{m,n\geq 1}^{\infty}\frac{y^{m}z^{n}}{m^{a}n^{b}}$
$=\sum_{\begin{subarray}{c}h\geq 1\\\
(j,k)=1\end{subarray}}^{\infty}\frac{(y^{j}z^{k})^{h}}{h^{a+b}(j^{a}k^{b})}$
$=\sum_{(j,k)=1}^{\infty}\frac{1}{(j^{a}k^{b})}\sum_{h=1}^{\infty}\frac{(y^{j}z^{k})^{h}}{h^{a+b}}$
$=\sum_{(j,k)=1}^{\infty}\frac{1}{(j^{a}k^{b})}\log\left(\frac{1}{1-y^{j}z^{k}}\right)\quad
if\quad a+b=1.$

Therefore, we have shown that

$\left(\sum_{m=1}^{\infty}\frac{y^{m}}{m^{a}}\right)\left(\sum_{n=1}^{\infty}\frac{z^{n}}{n^{b}}\right)=\sum_{(j,k)=1}^{\infty}\frac{1}{(j^{a}k^{b})}\log\left(\frac{1}{1-y^{j}z^{k}}\right)\quad
if\quad a+b=1.$

Exponentiating both sides gives us the 2D first quadrant VPV identity,

(15.1) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-y^{j}z^{k}}\right)^{\frac{1}{j^{a}k^{b}}}=\exp\left\\{\left(\sum_{j=1}^{\infty}\frac{y^{j}}{j^{a}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{k}}{k^{b}}\right)\right\\}\quad
if\quad a+b=1.$

This approach will be useful to return to when considering derivation of 2D
identities from first principles. Clearly, we have summed on the 2D first
quadrant lattice points, that is on points with positive integer coordinates.
We give immediate interesting cases of (15.1) now.

Taking $a=0$, $b=1$ gives

(15.2) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-y^{j}z^{k}}\right)^{\frac{1}{k}}=\left(\frac{1}{1-z}\right)^{\frac{y}{1-y}}$
$=\sum_{n=0}^{\infty}z^{n}(-1)^{n}\binom{\frac{y}{y-1}}{n}$
$=1-\frac{yz}{y-1}+\frac{yz^{2}}{2!(y-1)^{2}}+\frac{(y-2)yz^{3}}{3!(y-1)^{3}}+\frac{(y-2)y(2y-3)z^{4}}{4!(y-1)^{4}}$
$+\frac{(y-2)y(2y-3)(3y-4)z^{5}}{5!(y-1)^{5}}+O(z^{6}).$

A partition grid part for $\left(\frac{1}{1-z}\right)^{\frac{y}{1-y}}$ for
coefficients of $y^{a}z^{b}$ with $0\leq a\leq 9,\;0\leq b\leq 10$ is

$\begin{array}[]{c|cccccccccc}\textbf{10}&&\frac{362880}{10!}&\frac{1389456}{10!}&\frac{3588732}{10!}&\frac{7684388}{10!}&\frac{14669429}{10!}&\frac{25869458}{10!}&\frac{43015399}{10!}&\frac{68326540}{10!}&\frac{104604811}{10!}\\\
\textbf{9}&&\frac{40320}{9!}&\frac{149904}{9!}&\frac{377612}{9!}&\frac{790728}{9!}&\frac{1478985}{9!}&\frac{2559101}{9!}&\frac{4179861}{9!}&\frac{6527781}{9!}&\frac{9833391}{9!}\\\
\textbf{8}&&\frac{5040}{8!}&\frac{18108}{8!}&\frac{44308}{8!}&\frac{90409}{8!}&\frac{165140}{8!}&\frac{279512}{8!}&\frac{447168}{8!}&\frac{684762}{8!}&\frac{1012368}{8!}\\\
\textbf{7}&&\frac{720}{7!}&\frac{2484}{7!}&\frac{5872}{7!}&\frac{11619}{7!}&\frac{20635}{7!}&\frac{34026}{7!}&\frac{53116}{7!}&\frac{79470}{7!}&\frac{114918}{7!}\\\
\textbf{6}&&\frac{120}{6!}&\frac{394}{6!}&\frac{893}{6!}&\frac{1702}{6!}&\frac{2921}{6!}&\frac{4666}{6!}&\frac{7070}{6!}&\frac{10284}{6!}&\frac{14478}{6!}\\\
\textbf{5}&&\frac{24}{5!}&\frac{74}{5!}&\frac{159}{5!}&\frac{289}{5!}&\frac{475}{5!}&\frac{729}{5!}&\frac{1064}{5!}&\frac{1494}{5!}&\frac{2034}{5!}\\\
\textbf{4}&&\frac{6}{4!}&\frac{17}{4!}&\frac{34}{4!}&\frac{58}{4!}&\frac{90}{4!}&\frac{131}{4!}&\frac{182}{4!}&\frac{244}{4!}&\frac{318}{4!}\\\
\textbf{3}&&\frac{2}{3!}&\frac{5}{3!}&\frac{9}{3!}&\frac{14}{3!}&\frac{20}{3!}&\frac{27}{3!}&\frac{35}{3!}&\frac{44}{3!}&\frac{54}{3!}\\\
\textbf{2}&&\frac{1}{2!}&\frac{2}{2!}&\frac{3}{2!}&\frac{4}{2!}&\frac{5}{2!}&\frac{6}{2!}&\frac{7}{2!}&\frac{8}{2!}&\frac{9}{2!}\\\
\textbf{1}&&1&1&1&1&1&1&1&1&1\\\ \textbf{0}&1&&&&&&&&&\\\
\hline\cr\;b/a&\textbf{0}&\textbf{1}&\textbf{2}&\textbf{3}&\textbf{4}&\textbf{5}&\textbf{6}&\textbf{7}&\textbf{8}&\textbf{9}\end{array}$

Or equivalent to (15.2) is it’s reciprocal of both sides,

(15.3) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-y^{j}z^{k}\right)^{\frac{1}{k}}=\left(1-z\right)^{\frac{y}{1-y}}$
$=\sum_{n=0}^{\infty}z^{n}(-1)^{n}\binom{\frac{y}{1-y}}{n}$
$=1+\frac{yz}{y-1}+\frac{y(2y-1)z^{2}}{2!(y-1)^{2}}+\frac{y(2y-1)(3y-2)z^{3}}{3!(y-1)^{3}}+\frac{y(2y-1)(3y-2)(4y-3)z^{4}}{4!(y-1)^{4}}$
$+\frac{y(2y-1)(3y-2)(4y-3)(5y-4)z^{5}}{5!(y-1)^{5}}+O(z^{6}).$

A partition grid part for $\left(1-z\right)^{\frac{y}{1-y}}$ for coefficients
of $y^{a}z^{b}$ with $0\leq a\leq 9,\;0\leq b\leq 10$ is

$\begin{array}[]{c|cccccccccc}\textbf{10}&&\frac{-362880}{10!}&\frac{663696}{10!}&\frac{517572}{10!}&\frac{-77572}{10!}&\frac{-667381}{10!}&\frac{-1003552}{10!}&\frac{-990011}{10!}&\frac{-637670}{10!}&\frac{-26939}{10!}\\\
\textbf{9}&&\frac{-40320}{9!}&\frac{69264}{9!}&\frac{60724}{9!}&\frac{1344}{9!}&\frac{-64041}{9!}&\frac{-108509}{9!}&\frac{-119061}{9!}&\frac{-93141}{9!}&\frac{-35631}{9!}\\\
\textbf{8}&&\frac{-5040}{8!}&\frac{8028}{8!}&\frac{7964}{8!}&\frac{1537}{8!}&\frac{-6444}{8!}&\frac{-12808}{8!}&\frac{-15728}{8!}&\frac{-14454}{8!}&\frac{-9072}{8!}\\\
\textbf{7}&&\frac{-720}{7!}&\frac{1044}{7!}&\frac{1184}{7!}&\frac{435}{7!}&\frac{-643}{7!}&\frac{-1644}{7!}&\frac{-2296}{7!}&\frac{-2442}{7!}&\frac{-2022}{7!}\\\
\textbf{6}&&\frac{-120}{6!}&\frac{154}{6!}&\frac{203}{6!}&\frac{112}{6!}&\frac{-49}{6!}&\frac{-224}{6!}&\frac{-370}{6!}&\frac{-456}{6!}&\frac{-462}{6!}\\\
\textbf{5}&&\frac{-24}{5!}&\frac{26}{5!}&\frac{42}{5!}&\frac{31}{5!}&\frac{5}{5!}&\frac{-29}{5!}&\frac{-64}{5!}&\frac{-94}{5!}&\frac{-114}{5!}\\\
\textbf{4}&&\frac{-6}{4!}&\frac{5}{4!}&\frac{10}{4!}&\frac{10}{4!}&\frac{6}{4!}&\frac{-1}{4!}&\frac{-10}{4!}&\frac{-20}{4!}&\frac{-30}{4!}\\\
\textbf{3}&&\frac{-2}{3!}&\frac{1}{3!}&\frac{3}{3!}&\frac{4}{3!}&\frac{4}{3!}&\frac{3}{3!}&\frac{1}{3!}&\frac{-2}{3!}&\frac{-6}{3!}\\\
\textbf{2}&&\frac{1}{2!}&\frac{0}{2!}&\frac{-1}{2!}&\frac{-2}{2!}&\frac{-3}{2!}&\frac{-4}{2!}&\frac{-5}{2!}&\frac{-6}{2!}&\frac{-7}{2!}\\\
\textbf{1}&&1&1&1&1&1&1&1&1&1\\\ \textbf{0}&1&&&&&&&&&\\\
\hline\cr\;b/a&\textbf{0}&\textbf{1}&\textbf{2}&\textbf{3}&\textbf{4}&\textbf{5}&\textbf{6}&\textbf{7}&\textbf{8}&\textbf{9}\end{array}$

From the case of (15.3) with $y^{2}$ and $z^{2}$ replacing $y$ and $z$, and
dividing both sides of (15.3) into it we have,

(15.4) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1+y^{j}z^{k}\right)^{\frac{1}{k}}=\frac{\left(1-z^{2}\right)^{\frac{y^{2}}{1-y^{2}}}}{\left(1-z\right)^{\frac{y}{1-y}}}.$
$=\sum_{n=0}^{\infty}\frac{\Gamma\left(n+\frac{y}{1-y}\right)}{\Gamma\left(\frac{y}{1-y}\right)}\,_{3}F_{2}\left(\frac{1}{2}-\frac{n}{2},-\frac{n}{2},-\frac{y^{2}}{1-y^{2}};\frac{1}{2}-\frac{n}{2}-\frac{y}{2(1-y)},1-\frac{n}{2}-\frac{y}{2(1-y)};1\right)\frac{z^{n}}{n!}.$

So, in (15.1) we have derived the VPV 2D first quadrant identity summed upon
the visible points in that quadrant as depicted in yellow shaded area of
Figure 1.

We note that from the power series in $z$ for (15.3), the series is finite
when $y=\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\ldots$
leading us to the following identities,

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-\left(\frac{1}{2}\right)^{j}z^{k}\right)^{\frac{1}{k}}=1-z,$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-\left(\frac{2}{3}\right)^{j}z^{k}\right)^{\frac{1}{k}}=(1-z)^{2}=1-2z+z^{2},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-\left(\frac{3}{4}\right)^{j}z^{k}\right)^{\frac{1}{k}}=(1-z)^{3}=1-3z+3z^{2}-z^{3},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-\left(\frac{4}{5}\right)^{j}z^{k}\right)^{\frac{1}{k}}=(1-z)^{4}=1-4z+6z^{2}-4z^{3}+z^{4},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-\left(\frac{5}{6}\right)^{j}z^{k}\right)^{\frac{1}{k}}=(1-z)^{5}=1-5z+10z^{2}-10z^{3}+5z^{4}-z^{5},$

and so on. These infinite products tell an interesting story as they are a
product of infinite series that reduce to a finite polynomial.

Similarly, for (15.2), the series is finite when
$y=2,\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\ldots$ leading us to the
following identities,

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-2^{j}z^{k}}\right)^{\frac{1}{k}}=(1-z)^{2}=1-2z+z^{2},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-\left(\frac{3}{2}\right)^{j}z^{k}}\right)^{\frac{1}{k}}=(1-z)^{3}=1-3z+3z^{2}-z^{3},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-\left(\frac{4}{3}\right)^{j}z^{k}}\right)^{\frac{1}{k}}=(1-z)^{4}=1-4z+6z^{2}-4z^{3}+z^{4},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-\left(\frac{5}{4}\right)^{j}z^{k}}\right)^{\frac{1}{k}}=(1-z)^{5}=1-5z+10z^{2}-10z^{3}+5z^{4}-z^{5},$

$\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-\left(\frac{6}{5}\right)^{j}z^{k}}\right)^{\frac{1}{k}}=(1-z)^{6}=1-6z+15z^{2}-20z^{3}+15z^{4}-6z^{5}+z^{6},$

and so on.

Another obvious case of (15.1) to consider is where $a=b=\frac{1}{2}$, so then

(15.5) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(\frac{1}{1-y^{j}z^{k}}\right)^{\frac{1}{\sqrt{jk}}}=\exp\left\\{\left(\sum_{j=1}^{\infty}\frac{y^{j}}{\sqrt{j}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{k}}{\sqrt{k}}\right)\right\\},$

and equivalently,

(15.6) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1-y^{j}z^{k}\right)^{\frac{1}{\sqrt{jk}}}=\exp\left\\{-\left(\sum_{j=1}^{\infty}\frac{y^{j}}{\sqrt{j}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{k}}{\sqrt{k}}\right)\right\\}.$

From the case of (15.5) with $y^{2}$ and $z^{2}$ replacing $y$ and $z$, and
dividing both sides of (15.5) into it we have

(15.7) $\prod_{\begin{subarray}{c}(j,k)=1\\\ j,k\geq
1\end{subarray}}\left(1+y^{j}z^{k}\right)^{\frac{1}{\sqrt{jk}}}=\exp\left\\{\left(\sum_{j=1}^{\infty}\frac{y^{j}}{\sqrt{j}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{k}}{\sqrt{k}}\right)-\left(\sum_{j=1}^{\infty}\frac{y^{2j}}{\sqrt{j}}\right)\left(\sum_{k=1}^{\infty}\frac{z^{2k}}{\sqrt{k}}\right)\right\\}.$

Figure 1. VPV first quadrant 2D lattice points region.

##  16\. Deriving the $n$-dimensional first hyperquadrant VPV identity.

They are however, generating functions for weighted vector partitions, and
evidently contribute a new branch to the literature on partitions of vectors.

So, we establish the following conventions for the VPV identities.

######  Definition 16.1.

We use the notation $\left(x_{1},x_{2},x_{3},...,x_{n}\right)$, to mean “the
greatest common divisor of all of $x_{1},x_{2},x_{3},...,x_{n}$ together; the
same as $\gcd\left(x_{1},x_{2},x_{3},...,x_{n}\right)$”. It is important to
distinguish between this and the ordered $n$-tuple utilized for the vector
$\langle x_{1},x_{2},x_{3},...,x_{n}\rangle$. In either case we will be
concerned with lattice points in the relevant Euclidean space, hence any
vector or gcd will be over integer coordinates.

######  Definition 16.2.

Any Euclidean vector $\langle x_{1},x_{2},x_{3},...,x_{n}\rangle$ for which
$\left(x_{1},x_{2},x_{3},...,x_{n}\right)=1$ we call a visible point vector,
abbreviated VPV.

######  Theorem 16.1.

The first hyperquadrant VPV identity. If $i=1,2,3,...,n$ then for each