Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Asymptotic expansion of the sum $ \sum\limits_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ The situation :
I am looking for an asymptotic expansion of the sum $\displaystyle a_n=\sum_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ when $n \to \infty$.
(The $ B_k $ are the Bernoulli numbers defined by $ \displaystyle ... | Let $n\in\mathbb{N}_{\ge 1}$, from the identity
\begin{align}
(k+1)^{n+1}-k^{n+1}=(n+1)k^n+\sum_{\ell=0}^{n-1}\binom{n+1}{\ell}k^{\ell}
\end{align}
we have
\begin{align}
a_n&=\sum_{p\ge 1}\left(\frac{1}{3^{p(n+1)}}\sum_{k=0}^{3^{p}-1}\left((k+1)^{n+1}-k^{n+1}-\sum_{\ell=0}^{n-1}\binom{n+1}{\ell}k^{\ell}\right)-1\right)... | {
"language": "en",
"url": "https://mathoverflow.net/questions/273001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Existence of Pillai equations with Catalan type solutions? In Catalan's conjecture we have $$x^m-y^n=1$$ having solution $(3,2,1,1)$ and $(3,2,2,3)$.
Call $$ax^m-by^n=k$$ to be Pillai Diophantine equation.
*
*Is it true no Pillai Diophantine equation exists with integer solutions $(x,y,m,n)$ and $(x,y,m+1,n+r)$ with... | The answer to questions 1 and 2 are both no. In the example you consider, where $m = 1$, $n=1$ and $r = 2$, you are seeking solutions to $x-y = x^{2} - y^{3} = k$. The equation $x-y = x^{2} - y^{3}$ is an elliptic curve and the largest integral point on this curve gives you a solution for $k = -20$, namely
$$
-20 = (... | {
"language": "en",
"url": "https://mathoverflow.net/questions/280320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For which finite projective planes can the incidence structure be written as a circulant matrix? It is well known that the projective plane of order $2$ can be represented by the circulant matrix $M_2:=circ(x,x,1,x,1,1,1)= \begin{pmatrix}
x&x&1&x&1&1&1\\
1&x&x&1&x&1&1\\
1&1&x&x&1&x&1\\
1&1&1&x&x&1&x\\
x&1&1&1&x&x&1\... | To answer your questions:
1) A projective plane admits a circulant incidence matrix if and only if the automorphism group contains a cyclic group acting regularly on points and regularly on blocks. Equivalently, the projective plane comes from a difference set. The automorphism group of the projective plane coming from... | {
"language": "en",
"url": "https://mathoverflow.net/questions/291495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Elliptic-type integral with nested radical Let:
$$Q(A,Y) = Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1$$
I’m curious as to whether it’s possible to find a closed-form solution for:
$$I(A)=\int_{Y_1(A)}^{Y_2(A)} \frac{1}{\sqrt{\left(1-Y^2\right) Q(A,Y)}}\,dY\\
= \int_{\sin^{-1}(Y_1(A))}^{\sin^{-1}(Y_2(A))}\frac{2 \sqrt{2}}{\sqrt... | If we define:
$$\begin{array}{rcl}
g(a)&=&a^2(a+2)^2-3\\
k(a)&=&(a+1)^3\sqrt{(1-a)(a+3)}\\
\end{array}$$
then, by taking limits of the antiderivative provided by Mathematica at the endpoints of the range of integration, we obtain:
$$I_0(a) = \frac{2\sqrt{2}\,(a+1)\,K\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\sqrt{a(... | {
"language": "en",
"url": "https://mathoverflow.net/questions/306100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Strings of consecutive integers divisible by 1, 2, 3, ..., N For each n, let $a_n$ be the least integer, greater than n, such that the numbers $a_n$, $a_n$+ 1, $a_n$+ 2, ..., $a_n$+ (n – 1) are divisible, in some order, by 1, 2, 3, ..., n. For example $a_{12}$ = 110.
What are the best estimates known for $a_n$?
| If $a_n=a \gt n,$ then there can only be one prime
among $a,a+1,\cdots, a+n-1.$ So the well studied topic of gaps between primes could provide upper bounds and might be the major factor.
Here are the gaps between the first few primes
$ 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6$
$, 2, 6, 4, 2, 6, 4, 6, 8$
$4 , 2,... | {
"language": "en",
"url": "https://mathoverflow.net/questions/326659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ I am looking for positive integer solutions to the Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ for distinct values of $(a,b,c,d)$.
There are many solutions with $a=b$ such as $(7,7,11,13)$ and $(13,13,22,23)$... | The equation you specify defines a surface $X$ in $\mathbb{P}^{3}$, and this surface is a K3 surface. It is conjectured that if $X$ is a K3 surface, there is a field extension $K/\mathbb{Q}$ over which the rational points on $X$ become Zariski dense. It's not clear that this happens over $\mathbb{Q}$.
However, one can ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/378229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Inequality of two variables
Let $a$ and $b$ be positive numbers. Prove that:
$$\ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b}\geq\ln^2\frac{(a+1)(b+1)}{2(a+b)}.$$
Since the inequality is not changed after replacing $a$ on $\frac{1}{a}$ and $b$ on $\frac{1}{b}$ and $\ln^2\frac{(a+1)(b+1)}{2(a+b)}\geq\ln^2\frac{(a+1)(b+1)}... | Remarks: @Fedor Petrov's proof is very nice. Here we give an alternative proof.
Using the identity
$$\ln (1 + u) = \int_0^1 \frac{u}{1 + ut}\, \mathrm{d} t,$$
the desired inequality is written as
$$\int_0^1 \frac{(1 - a)^2}{t(1 - a)^2 + 4a}\, \mathrm{d} t
\cdot \int_0^1 \frac{(1 - b)^2}{t(1 - b)^2 + 4b}\, \mathrm{d} t... | {
"language": "en",
"url": "https://mathoverflow.net/questions/386705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Real rootedness of a polynomial with binomial coefficients It is possible to show using diverse techniques that the following polynomial:
$$P_n(x)=1 + \binom{n}{2} x + \binom{n}{4} x^2 + \binom{n}{6} x^3 + \binom{n}{8} x^4 +\ldots + \binom{n}{2\lfloor\tfrac{n}{2}\rfloor} x^{\lfloor \frac{n}{2}\rfloor},$$
is real-rooted... | Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.
We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n+2nx$$ has $\lfloor n/2\rfloor$... | {
"language": "en",
"url": "https://mathoverflow.net/questions/394944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Deriving a relation in a group based on a presentation Suppose I have the group presentation $G=\langle x,y\ |\ x^3=y^5=(yx)^2\rangle$. Now, $G$ is isomorphic to $SL(2,5)$ (see my proof here). This means the relation $x^6=1$ should hold in $G$. I was wondering if anyone knows how to derive that simply from the group ... | OK, here is the derivation, based completely on the amazing information provided by Victor Miller (who I should also thank for letting me know about kbmag).
First, some identities:
(1) From $x^3=xyxy$ we get: (a) $x^2=yxy$; (b) $xyx^{-1}=y^{-1}x$; (c) $x^{-1}yx=xy^{-1}$.
(2) From $y^5=xyxy$ we get: (a) $y^4=xyx$; (b) $... | {
"language": "en",
"url": "https://mathoverflow.net/questions/15180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Is there any finitely-long sequence of digits which is not found in the digits of pi? I know it's likely that, given a finite sequence of digits, one can find that sequence in the digits of pi, but is there a proof that this is possible for all finite sequences? Or is it just very probable?
| This is an expansion for Pi in base 16 numeric system:
$$\pi = \sum_{k = 0}^{\infty}\frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6} \right)$$
So to get k-th digit you have to get one term and take account for possible translation from a neighboring digit.
Thus to find the... | {
"language": "en",
"url": "https://mathoverflow.net/questions/18375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Eigenvectors of a certain big upper triangular matrix I'm looking at this matrix:
$$
\begin{pmatrix}
1 & 1/2 & 1/8 & 1/48 & 1/384 & \dots \\
0 & 1/2 & 1/4 & 1/16 & 1/96 & \dots \\
0 & 0 & 1/8 & 1/16 & 1/64 & \dots \\
0 & 0 & 0 & 1/48 & 1/96 & \dots \\
0 & 0 & 0 & 0 & 1/384 & \dots \\
\vdots & \vdots & \vdots & \vdots &... | Might be a wild intuition, I'd say the eigenvalues are the entries of the first row, and that the eigenvector coresponding to the $n$th eigenvalue, $k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n\times n$ minor of the ma... | {
"language": "en",
"url": "https://mathoverflow.net/questions/26389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric identity needed for sums involving secants I am looking for a closed-form formula for the following sum:
$\displaystyle \sum_{k=0}^{N}{\frac{\sin^{2}(\frac{k\pi}{N})}{a \cdot \sin^{2}(\frac{k\pi}{N})+1}}=\sum_{k=0}^{N}{\frac{1}{a+\csc^{2}(\frac{k\pi}{N})}}$.
Is such a formula known?
| Two other references to similar sums are
Bruce C. Berndt and Boon Pin Yeap, Explicit evaluations and reciprocity theorems for finite trigonometric sums, Advances in Applied Mathematics
Volume 29, Issue 3, October 2002, Pages 358--385
and
Ira Gessel, Generating Functions and Generalized Dedekind Sums, Electronic J. Co... | {
"language": "en",
"url": "https://mathoverflow.net/questions/122231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$ for all $n > 1$? An equivalent inequality for integers follows:
$$(3n^2-5)\left\lceil n/\sqrt{3} \right\rceil^2 > n^4.$$
This has been checked for n = 2 to 60000. Perhaps there is some connection to the convergents to $\sqrt{3}$.
$\lceil \frac{n}... | The inequality is true for $n\geq 2$. Let $m:=\lceil n/\sqrt{3}\rceil$, then
$$ m^2-\frac{n^2}{3} = \frac{3m^2-n^2}{3} \geq \frac{2}{3}, $$
because $3m^2-n^2>0$ and $3m^2-n^2\not\equiv 1\pmod{3}$. Hence the inequality follows from
$$ \frac{2}{3}>\frac{n^4}{3n^2-5}-\frac{n^2}{3}=\frac{5n^2}{3(3n^2-5)}. $$
The latter is ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/186419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Nontrivial solutions for $\sum x_i = \sum x_i^3 = 0$ For $x_i \in \mathbb{Z}$, let $\{x_i\}$ be a fundamental solution to the equations:
$$
\sum_{i= 1}^N x_i = \sum_{i=1}^N x_i^3 = 0
$$
if $x \in \{x_i\} \Rightarrow -x \notin \{x_i\}$.
For instance, a fundamental solution with $N=7$ is given by
$$
x_1 = 4, \quad x_2 ... | We can get $N=6$ from $1+5+5=2+3+6$, $1^3+5^3+5^3=2^3+3^3+6^3$.
We can get $N=5$ from $2+4+10=7+9$, $2^3+4^3+10^3=7^3+9^3$.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/203453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Realization of numbers as a sum of three squares via right-angled tetrahedra De Gua's theorem
is a $3$-dimensional analog of the Pythagorean theorem:
The square of the area of the diagonal face of a right-angled tetrahedron
is the sum of the squares of the areas of the other three faces.
For certain tetrahedra, this p... | These numbers are not very prevalent and the ratio in question goes to zero. Note first that by Legendre's theorem, a positive proportion of the numbers below $n$ may be expressed as a sum of three squares. Now consider $N_T(n)$. This amounts to counting (with parity restrictions on $x$, $y$, $z$, and all three pos... | {
"language": "en",
"url": "https://mathoverflow.net/questions/226068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions of (x+1)(xy+1)=z^3 Consider the equation
$$(x+1)(xy+1)=z^3,$$
where $x,y$ and $z$ are positive integers with $x$ and $y$ both at least $2$ (and so $z$ is necessarily at least $3$). For every $z\geq 3$, there exists the solution
$$x=z-1 \quad \text{and} \quad y=z+1.$$
My question is, if one imposes the... | There are other solutions. Try $x+1=a^3$, $xy+1=b^3$, where $b$ is chosen so that $b^2+b+1$ is divisible by $a^3-1$. Such $b$ is always possible to choose provided that $a-1$ is not divisible by 3 and by primes of the form $3k-1$. Moreover, $b$ may be replaced to its remainder modulo $a^3-1=x$, in this case $xy<xy+1=b^... | {
"language": "en",
"url": "https://mathoverflow.net/questions/235539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A monotonicity proof of a sequence summation I meet a problem:
I have a conjecture that the following sequence summation is monotony increasing in N and tends to 1 as N tends to $\infty$
$$\sum_{k=0}^{2N+1}C_{2N+1}^{k}\pi^{k}(1-\pi)^{2N+1-k}\frac{(1-q)\pi^{k}(1-\pi)^{2N+1-k}}{(1-q)\pi^{k}(1-\pi)^{2N+1-k}+q(1-\pi)^{k}\p... | Let $0<b:=\frac{q}{1-q}<1$ and $0<x:=\frac{\pi}{1-\pi}<1$. So, your sum now takes the form
$$B(N):=\sum_{k=0}^{2N+1}\binom{2N+1}k\frac{\pi^k(1-\pi)^{2N+1-k}}{1+b\cdot x^{2N+1-2k}}.$$
We show monotonicity: $B(N+1)-B(N) > 0$.
Applying Pascal's recurrence twice, we get $\binom{2N+3}k=\binom{2N+1}k+2\binom{2N+1}{k-1}+\bino... | {
"language": "en",
"url": "https://mathoverflow.net/questions/257763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The average of reciprocal binomials This question is motivated by the MO problem here. Perhaps it is not that difficult.
Question. Here is an cute formula.
$$\frac1n\sum_{k=0}^{n-1}\frac1{\binom{n-1}k}=\sum_{k=1}^n\frac1{k2^{n-k}}.$$
I've one justification along the lines of Wilf-Zeilberger (see below). Can you ... | As in the question
$$a_n:=\sum_{k=0}^{n-1}\frac{2^n}{n\binom{n-1}k} \qquad \text{and} \qquad
b_n:=\sum_{k=1}^n\frac{2^k}k.$$
It is clear that $a_1=b_1$ and $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$. But we have the same recursive relation for $a_n$ because
\begin{align}
a_n&=2^n\sum_{k=0}^{n-1}\frac{k!(n-1-k)!(k+1+n-k)}{n!(n+1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/262578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Is there an explicit expression for Chebyshev polynomials modulo $x^r-1$? This is an immediate successor of Chebyshev polynomials of the first kind and primality testing and does not have any other motivation - although original motivation seems to be huge since a positive answer (if not too complicated) would give a v... | The coefficient of $x^j$ in $(T_n(x)\bmod (x^r-1))$ equals the coefficient of $t^{n+r-j-1}$ in
$$\frac{(1+t^2)^{r-j}}{2^{r-j}} \frac{((1+t^2)^{r-1}t - 2^{r-1}t^{r-1})}{((1+t^2)^r - 2^rt^r)}.$$
This coefficient can be explicitly computed as
$$\sum_{k\geq 0} 2^{rk-r+j} \left( \binom{r-1-j-rk}{\frac{n+r-j-2-rk}{2}} - 2^{r... | {
"language": "en",
"url": "https://mathoverflow.net/questions/286626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
Polynomials for which $f''$ divides $f$ Let $n \geq 2$ and let $a < b$ be real numbers. Then it is easy to see that there is a unique up to scale polynomial $f(x)$ of degree $n$ such that
$$f(x) = \frac{(x-a)(x-b)}{n(n-1)} f''(x).$$
*
*Have these polynomials been studied? Do they have a standard name?
*Is it true t... | Recording a CW-answer to take this off the unanswered list. The Gegenbauer polynomials are defined by the differential equation
$$(1-x^2) g'' - (2 \alpha+1) x g' +n(n+2 \alpha) g =0.$$
Putting $\alpha = -1/2$, we get
$$g=\frac{(x+1)(x-1)}{n(n-1)} g''.$$
If we want some other values for $a$ and $b$, we can put $f(x) = g... | {
"language": "en",
"url": "https://mathoverflow.net/questions/292703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Elliptic-type integral with nested radical Let:
$$Q(A,Y) = Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1$$
I’m curious as to whether it’s possible to find a closed-form solution for:
$$I(A)=\int_{Y_1(A)}^{Y_2(A)} \frac{1}{\sqrt{\left(1-Y^2\right) Q(A,Y)}}\,dY\\
= \int_{\sin^{-1}(Y_1(A))}^{\sin^{-1}(Y_2(A))}\frac{2 \sqrt{2}}{\sqrt... | If the purpose of this calculation is to test the geometry conjecture, comparing expansions in powers of $A$ or $a$ should be effective. The indefinite integral can readily be evaluated to any order in $A$, but for the definite integral I run into a difficulty. Take the zeroth order term $A=0=a$, when $Y_{1,2}=\pm 1$, ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/306100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Does the equation $(xy+1)(xy+x+2)=n^2$ have a positive integer solution? Does there exist a positive integral solution $(x, y, n)$ to $(xy+1)(xy+x+2)=n^2$? If there doesn't, how does one prove that?
| $4n^2=4(xy+1)(xy+x+2)=(3+x+2xy)^2-(1+x)^2$.
For positive even $k$ and $ka^2+b^2=c^2$, when $a,b,c$ is positive integers, and $a=2m,b=km^2-1,c=km^2+1$, when $m$ is any positive integer.
Then $n=2m,1+x=4m^2-1,3+x+2xy=4m^2+1$, and then $2m^2=1$, i.e. $m$ is not integer - contradiction. Means equation $n^2=(xy+1)(xy+x+2)$ ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/313339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Formula for a sum of product of binomials We know that equation $$s_1+s_2+s_3=n-1 \quad \mbox{$s_1,s_2,s_3$}\geq 1$$
has $\binom{n-2}{2}$ solution.
I want to find any good formulae for the following form :
$$\sum_{(s_1,s_2,s_3)}\prod_{i=1}^3\binom{s_i+s_{i-1}-1}{s_i}=?$$
where, $s_0=1$ and each $(s_1,s_2,s_3)$ is the ... | The generating function is $$ \sum_{s_1,s_2, s_3} {s_1 + s_2-1 \choose s_2} {s_2+ s_3-1 \choose s_3} x^{s_1+s_2+s_3}.$$
$$ \sum_{s_3} {s_2+ s_3-1 \choose s_3} x^{s_3} = \left( \frac{1}{1-x}\right)^{s_2}$$.
Then the sum over the $s_2$ variable is
$$ \sum_{s_2} {s_1 + s_2-1 \choose s_2}\left( \frac{x}{1-x}\right)^{s_2} =... | {
"language": "en",
"url": "https://mathoverflow.net/questions/317499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Can you make an identity from this product? Start with the product
$$(1+x+x^2) (1+x^2)(1+x^3)(1+x^4)\cdots$$
(The first polynomial is a trinomial..The others are binomials..)
Is it possible by changing some of the signs to get a series all of whose coefficients are $ -1,0,$or $1$?
A simple computer search should suffic... | Proof of Doriano Brogloli's answer:
Call $a(n)$ the $n$th coefficient of $A(x)=(1-x)(1-x^2)\cdots$. By Euler's pentagonal theorem
we have $a(n)=0$ unless $n=m(3m\pm1)/2$ for some $m$, in which case $a(m)=(-1)^m$.
Call $b(n)$ the $n$th coefficient of $B(x)=(1-x^2)(1-x^3)...$. Since $A(x)=(1-x)B(x)$
we have $b(n)-b(n-1)=... | {
"language": "en",
"url": "https://mathoverflow.net/questions/333548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Solutions of $y^2=\binom{x}{0}+2\binom{x}{1}+4\binom{x}{2}+8\binom{x}{3}$ for positive integers $x$ and $y$ I was interested in create and solve a Diophantine equation similar than was proposed in the section D3 of [1]. I would like to know what theorems or
techniques can be applied to prove or refute that the Diophan... | The curve $y^2 = 1+\frac{8}{3} x - 2 x^2 + \frac{4}{3} x^3$ is elliptic. Siegel's theorem says it has only finitely many integral points.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/337036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Difference of two integer sequences: all zeros and ones? Suppose that $c$ is a nonnegative integer and $A_c = (a_n)$ and $B_c = (b_n)$ are strictly increasing complementary sequences satisfying
$$a_n = b_{2n} + b_{4n} + c,$$
where $b_0 = 1.$ Can someone prove that the sequence $A_1-A_0$ consists entirely of zeros and ... | The same method as in this answer to a previous your question works as well.
As for $(A_0)$. Starting with a guess
$$
7n+2\leq a_n\leq 7n+3,
$$
and trying to prove it inductionally, we arrive at $b_{6n+2}\geq 7n+4$ and $b_{6n}\leq 7n+1$, hence
$$
t+\left\lfloor\frac{t+4}6\right\rfloor+1\leq b_t\leq t+\left\lceil\fr... | {
"language": "en",
"url": "https://mathoverflow.net/questions/342867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
an identity between two elliptic integrals I would like a direct change of variable proof of the identity
$$\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}d\phi=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}d\phi\,.$$
I n... | Since the bountied question has changed substantially, now asking for the application of an identity in Legendre's Traite des fonctions elliptiques, I am starting a new answer. Legendre defines
\begin{align}
&F(\phi,k)=\int_0^{\phi}\frac{d\phi'}{\sqrt{1-k^2\sin^2\phi'}},\\
&\sin\phi=\frac{\sin(\theta/2)}{\sqrt{\tfrac{1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/349272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Cancellation in a particular sum In an attempt to compute cycle counts in an of a certain number theoretic graph, the following estimate was needed.
It is true that
$$\bigg|\sum_{a,b,c\in \mathbb{Z}/p\mathbb{Z}}\bigg(\sum_{d=1}^{p-1}\bigg(\frac{d}{p}\bigg)w^{-d-(a^2+b^2+c^2)/d}\bigg)^3\bigg| = o(p^{9/2}).$$
$w$ here is... | We can actually get an explicit formula for the whole sum (I will assume $p \ne 2,3$ throughout). We start with the Salié sum:
$$\sum_{d=1}^{p-1}\bigg(\frac{d}{p}\bigg)w^{-d-(a^2+b^2+c^2)/d} = \bigg(\sum_d \bigg(\frac{d}{p}\bigg)w^{-d}\bigg) \sum_{x^2 \equiv 4(a^2+b^2+c^2)} w^x.$$
The first factor is a Gauss sum, and h... | {
"language": "en",
"url": "https://mathoverflow.net/questions/356759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternative proofs sought after for a certain identity Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so
QUESTION. can you provide another verification for the problem below?
Problem. Prove that
$$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$
... | \begin{align*}
\sum_{k=1}^{n} \frac{2^k-1}{k}
&=\sum_{k=1}^{n} \frac{1}{k}\left(\sum_{j=1}^{k} \binom{k}{j}\right) \\
&=\sum_{j=1}^{n} \sum_{k=j}^{n} \binom{k}{j}\frac{1}{k} \\
&=\sum_{j=1}^{n} \frac{1}{j}\left(\sum_{k=j}^{n} \binom{k-1}{j-1}\right) \\
&=\sum_{j=1}^{n} \frac{1}{j} \binom{n}{j}.
\end{align*}
| {
"language": "en",
"url": "https://mathoverflow.net/questions/379248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 4
} |
Inequality of two variables
Let $a$ and $b$ be positive numbers. Prove that:
$$\ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b}\geq\ln^2\frac{(a+1)(b+1)}{2(a+b)}.$$
Since the inequality is not changed after replacing $a$ on $\frac{1}{a}$ and $b$ on $\frac{1}{b}$ and $\ln^2\frac{(a+1)(b+1)}{2(a+b)}\geq\ln^2\frac{(a+1)(b+1)}... | $$
\ln\frac{(a+1)^2}{4a}\ln\frac{(b+1)^2}{4b}
=\ln\left(1-\left(\frac{a-1}{a+1}\right)^2\right)\ln\left(1-\left(\frac{b-1}{b+1}\right)^2\right)\\=
\left(\sum_{n=1}^\infty\frac1n \left(\frac{a-1}{a+1}\right)^{2n}\right)\times
\left(\sum_{n=1}^\infty\frac1n \left(\frac{b-1}{b+1}\right)^{2n}\right)\\
\geqslant
\left(\su... | {
"language": "en",
"url": "https://mathoverflow.net/questions/386705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving a binomial sum identity
QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true?
$$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$
POSTSCRIPT. I like to record this presentable form by Alexander Burstein:
$$\sum_{n=0}^{\infty}\frac{\binom{... | Let, $f(x)=\frac{\sqrt{\pi}}{\Gamma(x)}$, The values of $f(x)$ at $x=-1,-2,...-N$ points are $y_{-n}=\frac{\binom{2n}{n}}{(-4)^n}$.
Now, for $N+1$ points $n=0,-1,-2,-3,..,-N$ we define $F_N(x):=\frac{\sqrt{\pi}}{\Gamma(x+\frac{1}{2})}N^x$ and $y_N(-n)=\frac{n!\binom{2n}{n}N^{-n}}{(-4)^n}$.
Hence, from Lagrange's interp... | {
"language": "en",
"url": "https://mathoverflow.net/questions/398317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Upper bound on double series We consider the sum
$$ \sum_{m \in \mathbb Z^2} \frac{1}{(3 m_1^2+3m_2^2+3(m_1+m_1m_2+m_2)+1)^2}. $$
Numerically, it is not particularly hard to see that the value of this series is well below $4$, indeed one gets numerically an upper bound of roughly $3.43$
I wonder if there is analyticall... | As noted in the comment by Beni Bogosel, the sum in question is
\begin{equation}
s:=\sum_{x=-\infty}^\infty\sum_{y=-\infty}^\infty\frac1{f(x,y)^2},
\end{equation}
where
\begin{equation}
f(x,y):=\frac32\, ((x + 1)^2 + (y + 1)^2 + (x + y)^2) - 2.
\end{equation}
Note that
\begin{equation}
f(x,y)\ge x^2+y^2+... | {
"language": "en",
"url": "https://mathoverflow.net/questions/418629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Upper bound for a+b+c in terms of ab+bc+ac I am given a triple of positive integers $a,b,c$ such that $a \geq 1$ and $b,c \geq 2$.
I would like to find an upper bound for $a+b+c$ in terms of $n = ab+bc+ac$. Clearly $a+b+c < ab+bc+ac = n$.
Is there any sharper upper bound that could be obtained (perhaps asimptotically)?... | The bound $n/2 + 1$ is tight. First, it is a bound because
$a + b + c \leq a b/2 + b c /2 + a c \leq n/2 + a c / 2 \leq n/2 + 1$
Equality is achieved when $a = 1$ and $b = c = 2$, since then we get $a + b + c = 5$ and $n/2 + 1 = (2+4+2)/2 + 1 = 5$.
Note that there might be other tight bounds that are also functions o... | {
"language": "en",
"url": "https://mathoverflow.net/questions/21088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
The difference of two sums of unit fractions I had this question bothering me for a while, but I can't come up with a meaningful answer.
The problem is the following:
Let integers $a_i,b_j\in${$1,\ldots,n$} and $K_1,K_2\in$ {$1,\ldots,K$}, then how small (as a function of $K$ and $n$), but strictly positive, can the f... | We can prove Christian Elsholtz's conjecture using only linear functions.
Let $m_1, ..., m_{2K}$ and $\alpha_1, ..., \alpha_{2K}$ be integers satisfying $\sum_i \frac{\alpha_i^j}{m_i} = 0$ for $j = 0, ..., 2K-2$. Then for large $x$ we have
$\sum_i \frac{1}{m_ix - m_i\alpha_i} = \frac{1}{x}\sum_i\frac{1}{m_i}\sum_j\frac... | {
"language": "en",
"url": "https://mathoverflow.net/questions/40819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Perfect numbers $n$ such that $2^k(n+1)$ is also perfect The smallest two perfect numbers $n=6$ and $m=28$ satisfy
$$
\frac{m}{n+1} = 2^k
$$
with $k=2.$
Question: Are there more pairs of perfect numbers $n,m$ with $n < m$
and such that
$$
\frac{m}{n+1} = 2^k
$$
for some positive integer $k>0.$
Observe that the perfect... | If $m$ is odd, it's clearly impossible.
If $m$ is even and $n$ is odd, I don't know.
So suppose $m$, $n$ both even. Then $m=2^{r-1}p$ where $p=2^r-1$ is prime, and $n=2^{s-1}q$ where $q=2^s-1$ is prime, and $s\lt r$.
The equation becomes $$2^k(n+1)=2^k(2^{s-1}q+1)=2^{k+s-1}q+2^k=2^{r-1}p$$ Now $2^k$ divides the seco... | {
"language": "en",
"url": "https://mathoverflow.net/questions/64340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
antiderivative involving modified bessel function This little integral has been holding me up for weeks. Has anyone come across a similar integral in their work.
$\int {\frac{d x}{c-I_0(x)}} $
| I doubt that there is a closed-form formula. Neither Maple nor Mathematica could find one. They don't even get formulas for the case $c=0$. Would a series expansion in $x$ be of any use? It is
$$
\frac{1}{c-1} x + \frac{1}{12 (c-1)^2} x^{3} + \frac{c+3}{320 (c-1)^3} x^{5} +
$$
$$
+ \frac{c^2+16 c+19}{16128 (c-1)^4}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/68529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of height-limited rational points on a circle Consider origin-centered circles $C(r)$ of radius $r \le 1$.
I am seeking to learn how many rational points might lie on $C(r)$,
where each rational point coordinate has height $\le h$.
For example, these are the rationals in $[0,1]$ with $h \le 5$:
$$
\left(
0,\frac... | I'll content myself with counting the number of points on $C(1)$ (which should surely be close to the maximum) -- the answer is quite nice, it is about $ \frac{4}{\pi } h$.
To see this, note that we are counting essentially Pythagorean triples $u^2-v^2, 2uv, u^2+v^2$, with $u^2+v^2\le h$ and we may suppose that $u$ a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/219465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions of (x+1)(xy+1)=z^3 Consider the equation
$$(x+1)(xy+1)=z^3,$$
where $x,y$ and $z$ are positive integers with $x$ and $y$ both at least $2$ (and so $z$ is necessarily at least $3$). For every $z\geq 3$, there exists the solution
$$x=z-1 \quad \text{and} \quad y=z+1.$$
My question is, if one imposes the... | Here is another family of solutions.
For natural $q$:
$x=3,y=144 q^{3} + 360 q^{2} + 300 q + 83,z=12q+10$.
Yours is linear in $y$, so solutions are positive integers
of the form $y=\frac{z^{3} - x - 1}{x^{2} + x}$.
Plugging small values of $x$, more solutions may come from
integers:
x= 2 y= (1/6) * (z^3 - 3)
x= 3 ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/235539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Is there an explicit expression for Chebyshev polynomials modulo $x^r-1$? This is an immediate successor of Chebyshev polynomials of the first kind and primality testing and does not have any other motivation - although original motivation seems to be huge since a positive answer (if not too complicated) would give a v... | There's a rapid algorithm to compute $T_n(x)$ modulo $(n,x^r-1)$. Note that
$$
\pmatrix{T_n(x) \\ T_{n-1}(x)} = \pmatrix { 2x & -1 \\ 1&0} \pmatrix{T_{n-1}(x) \\ T_{n-2}(x)} = \pmatrix { 2x & -1 \\ 1&0}^{n-1} \pmatrix{ x\\ 1}.
$$
Now you can compute these matrix powers all modulo $(n, x^{r}-1)$ rapidly by repeated ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/286626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
Bounding an elliptic-type integral Let $K>L>0$. I would like to find a good upper bound for the integral
$$\int_0^L \sqrt{x \left(1 + \frac{1}{K-x}\right)} \,dx.$$
An explicit expression for the antiderivative would have to involve elliptic functions; thus, if we want to stick to simple expressions, a bound is the best... | In general one has the following bound on your integral
$$
\frac{2}{3}L^{3/2}+ \frac{L^{1/2}}{\sqrt{2}}\ln \left(\frac{1+2K}{2+2(K-L)}\right)\leq \mathrm{integral}\leq \frac{2}{3}L^{3/2}+ L^{1/2} \ln\left(\frac{3+4K}{1+4(K-L)}\right)
$$
Let us make change of variables $x=Ly$. Then we want to estimate from above
$$
L^... | {
"language": "en",
"url": "https://mathoverflow.net/questions/299025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Number of numbers in $n$th difference sequence Suppose that $r$ is an irrational number with fractional part between $1/3$ and $2/3$. Let $D_n$ be the number of distinct $n$th differences of the sequence $(\lfloor{kr}\rfloor)$. It appears that
$$D_n=(2,3,3,5,4,7,5,9,6,11,7,13,8,\ldots),$$
essentially A029579. Can so... | It's easy to see that $D_n\leq \texttt{A029579}(n)$. Indeed, $\Delta^1$ is a Sturmian word, which is known to have exactly $n+1$ factors of length $n$.
Now, $\Delta^n$ is formed by values of the $(n-1)$-th difference operator on the factors of $\Delta^1$ of length $n$, i.e.,
$$\Delta^n = \left(\sum_{i=0}^{n-1} \binom{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/331724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
an identity between two elliptic integrals I would like a direct change of variable proof of the identity
$$\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}d\phi=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}d\phi\,.$$
I n... | Not yet an answer, but a bit too long for a comment. The Legendre normal form of these elliptic integrals might be a first step, at least by introducing simpler integration limits:
\begin{align}
&I_1=\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{d\phi}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}=\int_0^{\sqrt{3}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/349272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that $x_{k+1}+x_{k+2} + \cdots + x_n < 2m$? Let $k \le n$ be positive integers and let $m$ be a positive integer. Assume that $x_1, \ldots, x_n$ are non-negative integers and
\begin{align}
& x_1^2 + x_2^2 + \cdots + x_n^2 - (k-2) m^2=2, \\
& x_1 + \cdots + x_n = k m, \\
& x_1 \ge x_2 \ge \cdots \ge x_n.
\e... | I am afraid it is not true. Test the situation when $x_2=x_3=\ldots=1$, equations read as $x_1^2+(n-1)=(k-2)m^2+2$, $x_1+(n-1)=km$, that gives $x_1^2-x_1=(k-2)m^2-km+2$. Let's think that both $k$ and $m$ are large (say greater than 1000). Then $x_1$ is something like $\sqrt{k}m$, $x_{k+1}+\ldots+x_n=n-k=km+1-k-x_1$ is ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/351162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficients of $(2+x+x^2)^n$ from trinomial coefficients I would like to be able to express the coefficients of $(2+x+x^2)^n$ in terms of the trinomial coefficients studied by Euler, ${n \choose \ell}_2 = [x^\ell](1+x+x^2)^n$ where $[x^\ell]$ denotes the coefficient of $x^\ell$. The triangle of these numbers is given... | Using Abdelmalek's tip in the comments, here's a solution to a more general version of the "larger program" mentioned at the end. For an arbitrary constant $c$,
\begin{align}
[x^\ell](c+x+\cdots+x^k)^n & = [x^\ell] \left((c-1) + (1+x+\cdots+x^k)\right)^n \\
& = \sum_{m=0}^n {n \choose m}(c-1)^{n-m}[x^\ell](1+x+\cdots+... | {
"language": "en",
"url": "https://mathoverflow.net/questions/362041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The edge precoloring extension problem for complete graphs Consider coloring the edges of a complete graph on even order. This can be seen as the completion of an order $n$ symmetric Latin square except the leading diagonal. My question pertains to whether we can always complete the edge coloring in $n-1$ colors give... | Assuming that you mean to precolour $k$ subdiagonals and have no further constraints on the precolouring, the answer to both of your questions is no.
For every $n$ there is a precolouring which cannot be extended: choose colours $1, \dots n/2$ in the first row and colours $n/2+1, \dots, n-1$ in the second row (and thus... | {
"language": "en",
"url": "https://mathoverflow.net/questions/366312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Looking for a combinatorial proof for a Catalan identity Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers.
QUESTION. Is there a combinatorial or conceptual justification for this identity?
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2=C_{2n-1}.$$
| Let, $$\sum_{k=1}^{n} (\frac{k}{n}\binom{2n}{n-k})^2 =A_{n}$$
Now, using the fact that $(n+k)^2+(n-k)^2=2(n^2+k^2)$ and $\binom{2n}{n-k}=\binom{2n}{n+k}$, we get the following
$$\frac{1}{n^2}(\sum_{a=0}^{2n} a^2\binom{2n}{a}^2 -n^2\binom{2n}{n}^2)=\frac{1}{n^2}[\sum_{k=1}^n (n-k)^2\binom{2n}{n-k}^2+(n+k)^2\binom{2n}{n... | {
"language": "en",
"url": "https://mathoverflow.net/questions/383314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
The devil's playground On the $\mathbb{R}^2$ plane, the devil has trapped the angel in an equilateral triangle of firewalls.
The devil
*
*starts at the apex of the triangle.
*can move at speed $1$ to leave a trajectory of firewall behind, as this
*
*can teleport from one point to another along the firewall.
Th... | Here is an upper bound for the triangle of side length $1$.
First, divide in half into two triangles of angles $\frac{\pi}{6},\frac{\pi}{3}, \frac{\pi}{2}$ triangle and side lengths $1, \frac{\sqrt{3}}{2},\frac{1}{2}$. This requires drawing an edge of length $\frac{\sqrt{3}}{2}$. It doesn't matter which one the angel g... | {
"language": "en",
"url": "https://mathoverflow.net/questions/403396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
An inequality involving binomial coefficients and the powers of two I came across the following inequality, which should hold for any integer $k\geq 1$:
$$\sum_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j}\leq
\frac{1}{3}.$$
I have been struggling with this statement for a while. It looks valid for small... | For $j=0,\dots,k-1$,
\begin{equation*}
\frac1{2k+1-j}=\int_0^1 x^{2k-j}\,dx.
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
s:=&\sum_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j} \\
&=\int_0^1 dx\,\sum_{j=0}^{k-1}(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)x^{2k-j} \\
&=\int_0^1 dx\,kx^{k+... | {
"language": "en",
"url": "https://mathoverflow.net/questions/417283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Perfect numbers $n$ such that $2^k(n+1)$ is also perfect The smallest two perfect numbers $n=6$ and $m=28$ satisfy
$$
\frac{m}{n+1} = 2^k
$$
with $k=2.$
Question: Are there more pairs of perfect numbers $n,m$ with $n < m$
and such that
$$
\frac{m}{n+1} = 2^k
$$
for some positive integer $k>0.$
Observe that the perfect... | [After typing out this attempt at a "partial" answer, I realized that the details have already been worked out by Luis, Gerhard and Todd. I am posting it as an answer for anybody else who might be interested in how the final result is obtained. - Arnie]
Suppose $m$ is even and $n$ is odd.
Then if $m$ and $n$ are perfe... | {
"language": "en",
"url": "https://mathoverflow.net/questions/64340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric identity needed for sums involving secants I am looking for a closed-form formula for the following sum:
$\displaystyle \sum_{k=0}^{N}{\frac{\sin^{2}(\frac{k\pi}{N})}{a \cdot \sin^{2}(\frac{k\pi}{N})+1}}=\sum_{k=0}^{N}{\frac{1}{a+\csc^{2}(\frac{k\pi}{N})}}$.
Is such a formula known?
| This may not be of much (any!) help, but Mathematica 7 gives a closed-form solution in terms of QPolyGamma functions:
$\frac{\psi _{e^{-\frac{2 i \pi }{n}}}^{(0)}\left(1-\frac{\log
\left(\frac{a-2 \sqrt{a+1}+2}{a}\right)}{\log \left(e^{-\frac{2 i
\pi }{n}}\right)}\right)-\psi _{e^{-\frac{2 i \pi
}{n}}}^{(0)}\l... | {
"language": "en",
"url": "https://mathoverflow.net/questions/122231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Weiestrass Form How to convert this to weiestrass form?
$x^{2}y^{2}-2\left( 1+2\rho \right) xy^{2}+y^{2}-x^{2}-2\left( 1+2\rho
\right) x-1=0$
| You can rewrite the form as
\begin{equation*}
y^2=\frac{x^2+2(2\rho+1)x+1}{x^2-2(2\rho+1)x+1}
\end{equation*}
so, for rational solutions (which I presume you want), there exists $z \in \mathbb{Q}$ with
\begin{equation*}
z^2=(x^2+2(2\rho+1)x+1)(x^2-2(\rho+1)x+1)=x^4-2(8\rho^2+8\rho+1)x^2+1
\end{equation*}
This quartic c... | {
"language": "en",
"url": "https://mathoverflow.net/questions/130893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many solutions to $2^a + 3^b = 2^c + 3^d$? A few weeks ago, I asked on math.stackexchange.com how many quadruples of non-negative integers $(a,b,c,d)$ satisfy the following equation:
$$2^a + 3^b = 2^c + 3^d \quad (a \neq c)$$
I found 5 quadruples: $5 = 2^2 + 3^0 = 2^1 + 3^1$, $11 = 2^3 + 3^1 = 2^1 + 3^2$, $17 = 2... | Here is a proof of finiteness, along the lines of Jordan's answer. Assume without loss that $a>c$ and $d>b$. Then
$$2^a-2^c=3^d-3^b,$$
or equivalently
$$\frac{1-3^{b-d}}{1-2^{c-a}}=\frac{2^a}{3^d}.$$
Since $a>d$, Matveev's explicit bound for linear forms in logarithms implies that
$$\Bigl\lvert\frac{2^{c-a}-3^{b-d}}{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/164624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
Combinatorial identity involving the square of $\binom{2n}{n}$ Is there any closed formula for
$$
\sum_{k=0}^n\frac{\binom{2k}{k}^2}{2^{4k}}
$$
?
This sum of is made out of the square of terms $a_{k}:=\frac{\binom{2k}{k}}{2^{2k}}$
I have been trying to verify that $$
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\sum_{... | The limit I want to verify is
$$
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{\left(\sum_{j=0}^k a^2_{j}\right)}{(2k+1)(2k+2)}\right] -\frac{1}{2}{\sum_{k=0}^na^2_{k}}=\frac{1}{2\pi}
$$
For this it is sufficient to prove that the above expression under the limit is bounded above. I know this is tr... | {
"language": "en",
"url": "https://mathoverflow.net/questions/167355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Number of zeros of a polynomial in the unit disk Suppose $m$ and $n$ are two nonnegative integers. What is the number of zeros of the polynomial $(1+z)^{m+n}-z^n$ in the unit ball $|z|<1$?
Some calculations for small values of $m$ and $n$ suggests the following formula:
$$n-1+2\left\lfloor\frac{m-n+5}{6} \right\rfloor... | Here is my solution of this problem. So we have next equation for the zeros:
$$
(1+z)^{n+m}=z^n
$$
We can modify it like that:
$$
\Bigl(1+\frac{1}{z}\Bigr)^n \Bigl(1+z\Bigr)^m = 1
$$
Next we can mark the first factor as $re^{i\varphi}$, so the equation splits into two ones:
$$
\Bigl(1+z\Bigr)^m = re^{i\varphi} \\
\Bigl... | {
"language": "en",
"url": "https://mathoverflow.net/questions/171895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Number of intervals needed to cross, Brownian motion Let $B_t$ be a standard Brownian motion. Let $E_{j, n}$ denote the event$$\left\{B_t = 0 \text{ for some }{{j-1}\over{2^n}} \le t \le {j\over{2^n}}\right\},$$and let$$K_n = \sum_{j = 2^n + 1}^{2^{2n}} 1_{E_{j,n}},$$where $1$ denotes indicator function. I have three q... | I'll address the second question on the expected value of the sum $K_n$.
Let $\phi(x)$ and $\Phi(x)$ be the probability density function and cumulative distribution functions for a standard normal distribution.
Let $h(a,b)$ be the probability that a Brownian motion without drift returns to $0$ at some time in $[a,b]$. ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/221115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Number of all different $n\times n$ matrices where sum of rows and columns is $3$ For a given positive integer $n$, I need to learn the number of $n\times n$ matrices of nonnegative integers with the following restrictions:
*
*The sum of each row and column is equal to $3$.
*Two matrices are considered equal if one... | This problem was solved by Ron Read in his PhD thesis (University of London, 1958). Without requirement 2 there is a summation which isn't too horrible. With requirement 2 added as well, Read's solution needs the cycle index polynomial of a group and you could reasonably call it a solution in principle.
The problem is ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/251916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Efficiently lifting $a^2+b^2 \equiv c^2 \pmod{n}$ to coprime integers Let $n$ be integer with unknown factorization. Assume factoring $n$
is inefficient.
Let $a,b,c$ satisfy $a^2+b^2 \equiv c^2 \bmod{n}, 0 \le a,b,c \le n-1$.
Is it possibly to lift the above
congruence to coprime integers in $O(\mathsf{polylog}(n))$ ti... | Unless I'm mistaken, if we had an efficient algorithm we would be able to factor integers efficiently. We may suppose $n$ is odd.
Randomly choose
coprime integers $X, Y$, not both odd, in some large interval. Then $A = X^2 - Y^2$, $B = 2 X Y$,
$C = X^2 + Y^2$ are a primitive Pythagorean triple. Now compute reduced r... | {
"language": "en",
"url": "https://mathoverflow.net/questions/252523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A four-variable maximization problem We let function
\begin{equation}
\begin{aligned}
f(x_1,~x_2,~x_3,~x_4) ~&=~ \sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} \\
&+ \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} \\
&+ \sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} \\
&+ \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)},
\end{aligned}
\end{equat... | Using the inequality of the means for 2 variables, $(\sqrt{ab}\leq\frac{a+b}{2})$ for positive $a$ and $b$, equality only when $a=b$ we have
\begin{equation}
\begin{aligned}
&\sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} + \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} \\&=\sqrt{(x_1+x_2)}(\sqrt{(x_1+x_3)(x_1+x_4)}+\sqrt{(x_2+x_3)(x_2+x_4)})... | {
"language": "en",
"url": "https://mathoverflow.net/questions/346297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of alternated row and column normalizations Let $E_0$ be a matrix with non-negative entries.
Given $E_n$, we apply the following two operations in sequence to produce $E_{n+1}$.
A. Divide every entry by the sum of all entries in its column (to make the matrix column-stochastic).
B. Divide every entry by the sum o... | When $E_0$ is square (i.e., $r = c$) this procedure is called Sinkhorn iteration or the Sinkhorn-Knopp algorithm (see this Wikipedia page). You can find a wealth of results by Googling those terms, the most well-known of which is that if $E_0$ has strictly positive entries (and again, is square) then the limit of $E_n$... | {
"language": "en",
"url": "https://mathoverflow.net/questions/349274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
$\varepsilon$-net of a $d$-dimensional unit ball formed by power set of $V = \{+1, 0 -1\}^d$ I have a set of $d$-dimensional vectors $V = \{+1, 0, -1\}^d $. Then $P(V)$ constitutes the power set of $V$. I now construct a set of unit vectors $V_{\mathrm{sum}}$ from the power set $P(V)$ such that
$$
V_{\mathrm{sum}} = \l... | Since the notation quickly becomes cumbersome for any $S \in 2^{V}$ define
\begin{equation}
v_S \overset{\text{def}}{=} \sum_{v \in S} v,
\end{equation}
and let
\begin{equation}
\hat v_S \overset{\text{def}}{=} \frac{v_S}{\|v_S\| },
\end{equation}
If we fix a $v_S \in \text{span}(V)$
then the goal is to find/bound
\beg... | {
"language": "en",
"url": "https://mathoverflow.net/questions/360487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Looking for a combinatorial proof for a Catalan identity Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers.
QUESTION. Is there a combinatorial or conceptual justification for this identity?
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2=C_{2n-1}.$$
| Expanding my previous comment into an answer at the OP's request. We can write
$$
\frac{k}{n}\binom{2n}{n-k}=A_k-B_k,
$$
where
$$
A_k=\binom{2n-1}{n-k}=\binom{2n-1}{n+k-1}, \qquad B_k=\binom{2n-1}{n+k}=\binom{2n-1}{n-k-1}.
$$
Then
$$
\sum_{k=1}^{n}\left(\frac{k}{n}\binom{2n}{n-k}\right)^2=\sum_{k=1}^{n}(A_k^2+B_k^2)-\s... | {
"language": "en",
"url": "https://mathoverflow.net/questions/383314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Asymptotic analysis of $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2$
Problem: Let $x_1 = 1$ and $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2, \ n\ge 1$.
Find the third term in the asymptotic expansion of $x_n$.
I have posted it in MSE six months ago without solution for the third term
https://math.stackexchan... | Consider the substitutions
\begin{equation*}
x_n=n+1/2+y_n/n,\quad y_n=u_n+5/8.
\end{equation*}
Then $u_1=-9/8$ and
\begin{equation*}
u_{n+1}=f_n(u_n)
\end{equation*}
for $n\ge1$, where
\begin{equation*}
f_n(u):=\frac{-64 n^4 u-8 n^3 (4 u-13)+n^2 (56 u+115)+n (96 u+76)+4 (8 u+5)}{8 n^2 \left(8 n^2+4 n+8 u+... | {
"language": "en",
"url": "https://mathoverflow.net/questions/384047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Decoupling a double integral I came across this question while making some calculations.
QUESTION. Can you find some transformation to "decouple" the double integral as follows?
$$\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2}\frac{d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}}
=\frac14\int_0^{\frac{\pi}2}\frac{d\theta}... | (Thanks go to Etanche and Jandri)
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\
&\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\th... | {
"language": "en",
"url": "https://mathoverflow.net/questions/384145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
A counterexample to: $\frac{1-f(x)^2}{1-x^2}\le f'(x)$ — revisited Can we find a counterexample to the following assertion?
Assume that $f:[-1,1]\to [-1,1]$ an odd function of class $C^3$, and assume thaht $f$ is a concave increasing diffeomorphism of $[0,1]$ onto itself. Then my examples say that $$\frac{1-f(x)^2}{1-x... | I complete the reformulation given by Andrea Marino and give another counterexample.
First, the inequality of the beginning can be written
$$\forall x \in (0,1), \quad \frac{f'(x)}{1-f^2(x)} \ge \frac{1}{1-f^2(x)}$$
and means that the function $x \mapsto \arg\tanh(f(x))-\arg\tanh(x)$ is non-decreasing on $(0,1)$, or eq... | {
"language": "en",
"url": "https://mathoverflow.net/questions/421985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Parametrization of integral solutions of $3x^2+3y^2+z^2=t^2$ and rational solutions of $3a^2+3b^2-c^2=-1$ 1/ Is it known the parameterisation over $\mathbb{Q}^3$ of the solutions of
$3a^2+3b^2-c^2=-1$
2/ Is it known the parameterisation over $\mathbb{Z}^4$ of the solutions of
$3x^2+3y^2+z^2=t^2$
References, articles ... | For #1, we can take a particular solution such as $(a_0,b_0,c_0)=(0,0,1)$, and search parametric solution in the form: $(a,b,c)=(a_0+\alpha t, b_0+\beta t, c_0+t)$. Plugging it into the equation and solving for $t\ne 0$, we get:
$$t = \frac{2}{3\alpha^2 + 3\beta^2 - 1}.$$
So, we get rational parametrization with parame... | {
"language": "en",
"url": "https://mathoverflow.net/questions/422125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Irreducibility measure of integer polynomials (Partial) Question in short: Let $p$ be a monic integer polynomial of degree $n$. Is there a natural number $k$ with $0 \leq k \leq n$ such that $p+k$ is irreducible over the integers?
Longer version:
Let $p$ be a monic polynomial over the integers. Define the irreducibilit... | For $f=x^6 - 3x^5 - 2x^4 + 10x^3 + x^2 - 8x - 5$, one has
$$f=(x^3 - 2x^2 - 2x + 5)(x^3 - x^2 - 2x - 1),$$
$$f+1=(x-2)(x^5 - x^4 - 4x^3 + 2x^2 + 5x + 2),$$
$$f+2=(x^2-x-1)(x^4 - 2x^3 - 3x^2 + 5x + 3),$$
$$f+3=(x^2-2)(x^4 - 3x^3 + 4x + 1),$$
$$f+4=(x+1)(x^5 - 4x^4 + 2x^3 + 8x^2 - 7x - 1),$$
$$f+5=x(x^5 - 3x^4 - 2x^3 + 1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/441141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
What are fixed points of the Fourier Transform The obvious ones are 0 and $e^{-x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?
| $\bf{1.}$ A more complete list of particular self-reciprocal Fourier functions of the first kind, i.e. eigenfunctions of the cosine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$:
$1.$ $\displaystyle e^{-x^2/2}$ (more generally $e^{-x^2/2}H_{2n}(x)$, $H_n$ is Hermite polynomial)
$2.$ $\displa... | {
"language": "en",
"url": "https://mathoverflow.net/questions/12045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 4,
"answer_id": 1
} |
Product of sine For which $n\in \mathbb{N}$, can we find (reps. find explicitly) $n+1$ integers $0 < k_1 < k_2 <\cdots < k_n < q<2^{2n}$
such that
$$\prod_{i=1}^{n} \sin\left(\frac{k_i \pi}{q} \right) =\frac{1}{2^n} $$
P.S.: $n=2$ is obvious answer, $n=6 $ is less obvious but for instance we have $k_1 = 1$, $k_2 = 67$... | Consider the identity (as quoted by drvitek):
$$\prod_{k=1}^{n} \sin \left(\frac{(2k-1) \pi}{2n}\right) = \frac{2}{2^n},$$
This is completely correct but doesn't quite answer the question because the RHS is
not $1/2^n$ $\text{---}$ this is an issue related to the fact that $\zeta - \zeta^{-1}$ is not
a unit if $\zeta$ ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/44211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Binary representation of powers of 3 I asked this question at Mathematics Stack Exchange but since I didn't got a satisfactory answer I decided to ask it here as well.
We write a power of 3 in bits in binary representation as follows.
For example $3=(11)$, $3^2=(1001)$ which means that we let the $k$-th bit from the ri... | Here is an extended hint for proving 2 (an almost complete proof is in the Update below). If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/92397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Distance between the product of marginal distributions and the joint distribution Given a joint distribution $P(A,B,C)$, we can compute various marginal distributions. Now suppose:
\begin{align}
P1(A,B,C) &= P(A) P(B) P(C) \\
P2(A,B,C) &= P(A,B) P(C) \\
P3(A,B,C) &= P(A,B,C)
\end{align}
Is it true that $d(P1,P3) \geq... | I just find the following counter-example. Suppose $A,B,C$ are discrete variables. $A,B$ can each take two values while $C$ can take three values.
The joint distribution $P(A,B,C)$ is:
\begin{array}{cccc}
A & B & C & P(A,B,C) \\
1 & 1 & 1 & 0.1/3 \\
1 & 1 & 2 & 0.25/3 \\
1 & 1 & 3 & 0.25/3 \\
1 & 2 & 1 & 0.4/3 \\
1 & ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/138031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Impossible Range for Minkowski-Like Sum of Squares Given coprime positive integers M,N, and a corresponding integer z outside of the range (for all integers x,a,b,c) of $Mx^2-N(a^2+b^2+c^2)$, is there any such z which is "deceptive", meaning that it is modularly inside the range of the smaller components, i.e., inside ... | There are no deceptive integers for indefinite forms in at least 4 variables, this begins with Siegel (1951). I have the 1961 article by Martin Kneser, it is famous for a variety of reasons. This article is by J. W.S. Cassels, presented in the International Congress in Singapore in 1981, the article is pages 9-26 in th... | {
"language": "en",
"url": "https://mathoverflow.net/questions/138426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
State of knowledge of $a^n+b^n=c^n+d^n$ vs. $a^n+b^n+c^n=d^n+e^n+f^n$ As far as I understand, both of the Diophantine equations
$$a^5 + b^5 = c^5 + d^5$$
and
$$a^6 + b^6 = c^6 + d^6$$
have no known nontrivial solutions, but
$$24^5 + 28^5 + 67^5 = 3^5+64^5+62^5$$
and
$$3^6+19^6+22^6 = 10^6+15^6+23^6$$
among many other s... | Label the equation,
$$x_1^k+x_2^k+\dots+x_m^k = y_1^k+y_2^k+\dots+y_n^k$$
as a $(k,m,n)$. Let type of primitive solutions be polynomial identity $P(n)$, or elliptic curve $E$. Then results (mostly) for the balanced case $m=n$ are,
I. Table 1
$$\begin{array}{|c|c|c|}
(k,m,n)& \text{# of known solutions}& \text{Type}\\
3... | {
"language": "en",
"url": "https://mathoverflow.net/questions/150428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
} |
A Problem Concerning Odd Perfect Number Briefly, prove that every odd number having only three distinct prime factors cannot be a perfect number.
I know there are results much stronger than the one above, but I am looking for an answer without computer cracking (which means the computation can be carried out by a perso... | Here is a proof, perhaps not the simplest. Let $n$ be an odd perfect number with three distinct prime factors. As observed in the comments, $n$ is of the form $3^a5^bp^c$, where $p\in\{7,11,13\}$.
It is a simple fact (observed by Euler) that exactly one of the exponents $a$, $b$, $c$ is odd. Let $q^r\parallel n$ be the... | {
"language": "en",
"url": "https://mathoverflow.net/questions/179504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Failing of heuristics from circle method The heuristic from circle method for integral points on diagonal cubic surfaces $x^3+y^3+z^3=a$ ($a$ is a cubic-free integer) seems to fit well with numerical computations by ANDREAS-STEPHAN ELSENHANS AND JORG JAHNEL. The only known exception is the surface $x^3+y^3+z^3=2$. Circ... | Note as a caveat that the heuristics can fail in the other direction as well: Dietmann and Elsholtz ( http://www.math.tugraz.at/~elsholtz/WWW/papers/papers26de08.pdf ) have shown that if $p\equiv 7\pmod{8}$ is prime, then $p^2$ cannot be represented as $x^2+y^2+z^4$. This strongly contradicts the heuristic of the circl... | {
"language": "en",
"url": "https://mathoverflow.net/questions/180300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Gradual monotonic morphing between two natural numbers Let $a < b$ be two natural numbers. I will use these as an example:
\begin{align*}
a & = 2^5 \cdot 3^2 \cdot 5^2 = 7200\\\
b & = 2^3 \cdot 3^5 \cdot 7^1 = 13608
\end{align*}
I seek to "morph" $a$ to $b$ via $a{=}n_0,n_1,n_2,\ldots,n_k{=}b$
such that
*
*Each step... | If I understand the question right, then $k$ is simply the number of divisors $d$ of $\text{lcm}(a,b)$ such that $a\le d\le b$ and $d\mid\gcd(a,b)$. (So in finding a longest chain, we may assume that $a$ and $b$ are relatively prime.)
In your example, a longest chain would be
\begin{equation}
7200\to 7560\to 7776\to 90... | {
"language": "en",
"url": "https://mathoverflow.net/questions/224300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An inequality of a cyclic polygon I am looking for a proof of the inequality as follows:
Let $A_1A_2....A_n$ be the regular polygon incribed in a circle $(O)$ with radius $R$. Let $B_1B_2....B_n$ be a polygon incribed the circle $(O)$. We let $x_{ij}=A_iA_j$ and $y_{ij}=B_{i}B_{j}$. Let $f(x)=x^m$ (where $m=1, m=2$),... | As for the sum of squares (m=2), denoting the vectors $\overline{OB_i}=b_i$ we get $\sum_{i<j} y_{ij}^2=\sum_{i<j} (b_i-b_j)^2=n\sum_i b_i^2-(\sum b_i)^2=n^2R^2-(\sum b_i)^2$ that is maximal if and only if $\sum b_i=0$ --- so, in particular, for a regular polygon.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/304456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(x + y + z)....(x + y\omega_n^{n-1} + z\omega_n) = x^n + y^n + z^n - P(x,y,z)$ To find $P$ $$(x + y + z)(x + y\omega_n + z\omega_n^{n-1})(x + y\omega_n^2 + z\omega_n^{n-2})....(x + y\omega_n^{n-1} + z\omega_n) = x^n + y^n + z^n - P(x,y,z)$$ where $\omega_n$ is an nth root of unity.
The question is to find the polynomi... | As suggested in the first comment, computationally it looks like $$P\equiv P_n=\dfrac{x^n}{t^n}L_n(t)-x^n=\dfrac{x^n}{t^n}(L_n(t)-t^n), $$
where $L_n$ is the $n$th Lucas polynomial in $t:=\dfrac{{ix}}{\sqrt{yz^{\phantom l}}}$.
E.g. for $n=6$, $$P=6x^4yz-9x^2y^2z^2+2xy^3z^3$$ while $$L_6(t)=t^6+6t^4+9t^2+2.$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/313350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of hypergeometric function near $z=1$ Given the hypergeometric function $_2F_1[a,b,c,z]$
in the interval $z\in(1,\infty)$. What is the proper asymptotic expansion of the aforesaid function near $z=1$, when one is approaching from $z>1$.
| Mathematica says:
$$\left(\frac{\pi \csc ((-a-b+c) \pi ) \Gamma (c)}{\Gamma (a+b-c+1) \Gamma (c-a) \Gamma
(c-b)}-\frac{a b \pi \csc ((-a-b+c) \pi ) \Gamma (c) (z-1)}{\Gamma (a+b-c+2) \Gamma
(c-a) \Gamma (c-b)}+\frac{a (a+1) b (b+1) \pi \csc ((-a-b+c) \pi ) \Gamma (c) (z-1)^2}{2
\Gamma (a+b-c+3) \Gamma (c-a)... | {
"language": "en",
"url": "https://mathoverflow.net/questions/323608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Conjecture: $a^n+b^n+c^n\ge x^n+y^n+z^n$ let $a,b,c,x,y,z>0$ such $x+y+z=a+b+c,abc=xyz$,and $a>\max\{x,y,z\}$,
I conjecture $$a^n+b^n+c^n\ge x^n+y^n+z^n,\forall n\in N^{+}$$
Maybe this kind of thing has been studied, like I found something relevant, but I didn't find the same one.[Schur convexity and Schur multiplicati... | Yes, it is true. Without loss of generality, $a\geq b \geq c$. Let $a+b+c=s,\ abc=p$ then $b=\frac{(s \ - \ a) \ + \ d}{2}, \ c=\frac{(s \ - \ a) \ - \ d}{2}$ with $d=b-c=\sqrt{(s-a)^2- \frac{4p}{a}}$.
If we consider $b,c$ as variables in $a,s,p$, we find that
$$
\begin{eqnarray}
\frac{\partial (a^n+b^n+c^n)}{\partial... | {
"language": "en",
"url": "https://mathoverflow.net/questions/327017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
A four-variable maximization problem We let function
\begin{equation}
\begin{aligned}
f(x_1,~x_2,~x_3,~x_4) ~&=~ \sqrt{(x_1+x_2)(x_1+x_3)(x_1+x_4)} \\
&+ \sqrt{(x_2+x_1)(x_2+x_3)(x_2+x_4)} \\
&+ \sqrt{(x_3+x_1)(x_3+x_2)(x_3+x_4)} \\
&+ \sqrt{(x_4+x_1)(x_4+x_2)(x_4+x_3)},
\end{aligned}
\end{equat... | By Cauchy-Schwarz for sums (with $x_i=\sqrt{a+b}$ and $y_i=\sqrt{a+c}\sqrt{a+d}$, also, I am using cyclic sum notation):
\begin{split}
f(a,b,c,d)&=\sum_{\text{cyc}} \sqrt{(a+b)(a+c)(a+d)}\\&\le \left(\left(\sum_{\text{cyc}} a+b\right) \cdot \left(\sum_{\text{cyc}} (a+c)\cdot (a+d)\right)\right)^{\frac12}\\
&=\sqrt{\big... | {
"language": "en",
"url": "https://mathoverflow.net/questions/346297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of alternated row and column normalizations Let $E_0$ be a matrix with non-negative entries.
Given $E_n$, we apply the following two operations in sequence to produce $E_{n+1}$.
A. Divide every entry by the sum of all entries in its column (to make the matrix column-stochastic).
B. Divide every entry by the sum o... | The paragraph below applies to a different problem, where row normalization is split out from column normalization, so I have an $"E_{n+1/2}"$ which will be sometimes different from both $E_n$ and $E_{n+1}$. (If the starting matrix "looks like" an order k square stochastic matrix, then it will be invariant under both n... | {
"language": "en",
"url": "https://mathoverflow.net/questions/349274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Perfect squares between certain divisors of a number Let $n$ be a positive integer. We will call a divisor $d(<\sqrt{n})$ of $n$ special if there exists no perfect squares between $d$ and $\frac{n}{d}$. Prove that $n$ can have at-most one special divisor.
My progress: I boiled down the problem to the following:
Suppose... | Here is an elementary approach. We want to rule out finding distinct $a,b,c,d$ in short interval (specifically $[n^2+1,n^2+2n]$) with $ad=bc.$ We may assume that $a$ is the smallest and that $b<c$. Then $a<b<c<d.$ I will show that $a+2\sqrt{a}+1\leq d$ so that if $n^2 \leq a$ then $(n+1)^2 \leq d.$
Claim: There are in... | {
"language": "en",
"url": "https://mathoverflow.net/questions/355600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximal (minimal) value of $S=x_1^2x_2+x_2^2x_3+\cdots+x_{n-1}^2x_n+x_n^2x_1$ on condition that $x_1^2+x_2^2+\cdots+x_n^2=1$ since $x_1^2+x_2^2+\cdots+x_n^2=1$ is sphere,a compact set,so $S$ has a maximal(minimal) value. But when I try to solve it using the Lagrangian multiplier method, I don't know how to solve these ... | For $n=3$ we need to find $$\max_{a^2+b^2+c^2=1}(a^2b+b^2c+c^2a).$$
Indeed, let $\{|a|,|b|,|c|\}=\{x,y,z\}$, where $x\geq y\geq z\geq0$.
Thus, by Rearrangement and AM-GM we obtain:
$$\sum_{cyc}a^2b\leq|a|\cdot(|a||b|)+|b|\cdot(|b||c|)+|c|\cdot(|c||a|)\leq$$
$$\leq x\cdot xy+y\cdot xz+z\cdot yz=y(x^2+xz+z^2)\leq y\left(... | {
"language": "en",
"url": "https://mathoverflow.net/questions/383431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
In search of a $q$-analogue of a Catalan identity Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\begin{equation}
\label1
\sum_{k=0}^n\binom{2n-2k}{n-k}C_k=\binom{2n+1}n \... | Decided to make a cw post: it is sort of amusing.
Let $C_n(q)$ be defined by
$$\sum_{k=0}^n\binom{2n-2k}{n-k}_qC_k(q)q^{2n-2k}=\binom{2n+1}n_q,\qquad n=0,1,2,\dotsc.$$
Then
\begin{multline*}
C_n(q)=1+q+q^2+q^3+2 q^4+3 q^5+3 q^6+3 q^7+4 q^8+6 q^9+\dotsb\\\dotsb-7q^{(n+1)^2-6}-5q^{(n+1)^2-5}-3q^{(n+1)^2-4}-2q^{(n+1)^2-3}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/400793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Parametrization of integral solutions of $3x^2+3y^2+z^2=t^2$ and rational solutions of $3a^2+3b^2-c^2=-1$ 1/ Is it known the parameterisation over $\mathbb{Q}^3$ of the solutions of
$3a^2+3b^2-c^2=-1$
2/ Is it known the parameterisation over $\mathbb{Z}^4$ of the solutions of
$3x^2+3y^2+z^2=t^2$
References, articles ... | on the second question $3x^2 + 3 y^2 + z^2 = t^2$ in integers:
One description is this: if $x,y$ are integers and not both odd, then $3 (x^2 + y^2) \neq 2 \pmod 4.$ As a result it may be expressed a few ways as $t^2 - z^2 = (t+z)(t-z)$ When odd, we may take $t-z$ to be any factor of $3 (x^2 + y^2)$ because $t... | {
"language": "en",
"url": "https://mathoverflow.net/questions/422125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Analogue of Fermat's "little" theorem Let $p$ be a prime, and consider $$S_p(a)=\sum_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$
I have a rather complicated (15 lines) proof that $S_p(a)\equiv0\pmod{p}$. This must be
extremely classical: is there a simple direct proof ?
| Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's.
We have
$$
\sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}.
$$
Setting $j=(p-1)k$ and summing on $k$ gives
$$
\sum_{a=0}^\infty x^a \sum_{p-1\mid j}\binom{a}{j} = \frac{(1-x)^{p-1}}{(1-x)^p -x^{p-1}(1-x)}.
$$
We... | {
"language": "en",
"url": "https://mathoverflow.net/questions/424694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Is there a simple expression for $\sum_{k =1}^{(p-3)/2} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k\cdot(2k+1)} \bmod p$? Let $p \equiv 1 \pmod 4$ be a prime and $E_n$ denote the $n$-th Euler number. While investigating $E_{p-1} \pmod{p^2}$ I have encountered this summation (modulo $p$)
\begin{align*}
\sum_{k =1}^... | It is known that $$\sum_{k=0}^\infty\frac{\binom{2k}k}{(2k+1)4^k}=\frac{\pi}2.$$ Motivated by this, I proved in my paper On congruences related to central binomial coefficients [J. Number Theory 131(2011), 2219-2238] the following result (as part (i) of Theorem 1.1 in the paper):
Let $p$ be any odd prime. Then
$$\sum_{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/428183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$ Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $a(n,m)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)=m$. In other words, $a(n,m)$ is ... | Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have
$$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$
Then the sum in question can be easily computed:
\begin{split}
& a(1,2^{n-1}+n)+\sum_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)\\
&= \sum_{i=0}^{2^{n-1}+1} a(i+1,2^{n-1}+n) - (2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-1}) ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/430476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Transforming matrix to off-diagonal form I wonder if one can write the following matrix in the form $A = \begin{pmatrix} 0 & B \\ B^* & 0 \end{pmatrix}.$
The matrix I have is of the form
$$ C = \begin{pmatrix} 0 & a & b & 0 & 0 & 0 \\
\bar a & 0 & 0 &b & 0& 0\\
\bar b & 0 & 0 & a & f & 0 \\
0 & \bar b & \bar a & 0 & 0 ... | The $C$ is a particular block matrix, $C\in \mathbb{M}_3(\mathbb{M}_2(\mathbb{C}))$. For $V$ unitary let $V\begin{pmatrix}0&a\\\bar a&0\end{pmatrix}V^*=\begin{pmatrix}s&0\\0&-s\end{pmatrix}$, $P$ the perfect shuffle matrix (permutation), $D=\text{diag}(1,1,1,1,-1,1)$ and $U=\frac{1}{\sqrt{2}}\begin{pmatrix}I&I\\-I&I\en... | {
"language": "en",
"url": "https://mathoverflow.net/questions/436391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to solve this integral? I can't manage to solve this integral. Does it have an analytical solution?
$$\int\left(\frac{e^{x}(a-1)-1+\frac{1}{a}}{e^{x}(1-b)+1-\frac{1}{a}}\right)e^{-\frac{(x-(\mu-\frac{\sigma^{2}}{2})t)^{2}}{2t\sigma^{2}}}dx$$
| $$\int\left(\frac{e^{x}(a-1)-1+\frac{1}{a}}{e^{x}(1-b)+1-\frac{1}{a}}\right)e^{-\frac{(x-c)^2}{d^2}}\,dx$$
If $b=1$, the integral is
$$-\frac{ \sqrt{\pi }}{2} \, d \left(a e^{c+\frac{d^2}{4}} \text{erf}\left(\frac{2
c+d^2-2 x}{2 d}\right)+\text{erf}\left(\frac{x-c}{d}\right)\right)$$
If $b$ is "close" to $1$
$$\frac... | {
"language": "en",
"url": "https://mathoverflow.net/questions/440112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fibonacci identity using generating function There are many nice ways of showing that $f_0^2+f_1^2+\cdots+f_n^2=f_{n+1}f_n$. I was wondering if there is a way of showing this using the generating function $F(x)=\frac{1}{1-x-x^2}=\sum_{i\geq0}f_ix^i$. In other words, is there any operation (perhaps the Hadamard product)... | Here are the details on proving these identities using Hadamard products of generating functions. (You can find explanations of how to compute Hadamard products of rational functions here.)
I'll write $U(x)*V(x)$ for the Hadmard product of $U(x)$ and $V(x)$:
$$\sum_{n=0}^\infty u_n x^n *\sum_{n=0}^\infty v_n x^n
=\su... | {
"language": "en",
"url": "https://mathoverflow.net/questions/11972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$? The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content.
If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, sho... | Maybe there is a small trick yet.
For $a + b = 1$ we can write the sum as
$$a^{2(1-a)} + b^{2(1-b)} = (\frac{a}{a^a})^2+ (\frac{b}{b^b})^2 $$.
Obviously the sum is 1 for $(a, b) = (0,1), (\frac{1}{2},\frac{1}{2}), (1, 0)$. The question is whether at $(\frac{1}{2},\frac{1}{2})$ there is a maximum or a minimum, i.e. whe... | {
"language": "en",
"url": "https://mathoverflow.net/questions/17189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 6
} |
Diophantine question This came up when I did a brand-new (or maybe it's just "birdtracks" in disguise :-)
graph-based construction of the E8 family. x,y,z are dimensions and thus integer. x<0 actually doesn't hurt. (For $y=x,z=(3 (-2 + o) o)/(10 + o)$ the standard E8 setup results, o must now divide 360 etc. pp.) So he... | Let $P(x,y,z)=3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z$ be the product which we wish to be a square. As noted,
*
*$P(1,y,z)=\left(6yz\right)^2.$
A nice parametric solution is
*
*$P(2r,3r,r+1)=\left(12r(r+1)(2r^2-2r-1)\right)^2$
Also, for fixed $x,y$ or $x,z$ we are left with a Pell Equation.
... | {
"language": "en",
"url": "https://mathoverflow.net/questions/121017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is the $n$-th prime $p_n$ expressible as the difference of coprime $A, B$ such that the set of prime divisors of $AB$ is $\{p_1, \dots, p_{n-1}\}$? We define recursively
$p_1=2,p_2=3$
and
$$p_{n}= \min_{(A,B)\in F_{n-1}}|A-B| $$
Where
$$
\begin{split}
F_n=\{(A,B) |&\gcd (A,B)=1,\quad |A-B| \not =1,
\\\
&\text{both $A... | Barry Cipra has already computed the first few values.
The next couple of numbers $p_n$ are
$13 = 5 \cdot 11 - 2 \cdot 3 \cdot 7$,
$17 = 2 \cdot 7 \cdot 13 - 3 \cdot 5 \cdot 11$,
$19 = 2^2 \cdot 3 \cdot 5 \cdot 17 - 7 \cdot 11 \cdot 13 $,
$23 = 7 \cdot 11 \cdot 17^2 - 2 \cdot 3^2 \cdot 5 \cdot 13 \cdot 19$,
$29 = 3 \cd... | {
"language": "en",
"url": "https://mathoverflow.net/questions/124071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
A question of terminology regarding integer partitions I am wondering if there is a standard notation and name for the following. Let $\lambda$ be a partition $\lambda_1\geq \lambda_2\geq\cdots\geq \lambda_r\geq 1$ of $n$ into $r$ parts. Then we can form a partition $\mu$ of $r$ by keeping track of how many times eac... | Let $\mu=(\mu_1,\dots,\mu_m)$ and $\gamma_i$ be the number of parts equal $i$ in $\mu$. Then
$$\sum_{i=1}^r \gamma_i = m\quad\text{and}\quad\sum_{i=1}^r i\cdot\gamma_i = r.$$
Then $C_{n,\mu}$ equals $\frac{1}{\gamma_1!\cdots \gamma_r!}$ times the number of solutions to
$$(\star)\qquad \mu_1\cdot y_1 + \dots + \mu_m\cdo... | {
"language": "en",
"url": "https://mathoverflow.net/questions/203748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all solution $a,b,c$ with $(1-a^2)(1-b^2)(1-c^2)=8abc$ Two years ago, I made a conjecture on stackexchange:
Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$
I have found some solutions, such as
$$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4... | @Allan methods it's nice! here is my answer:
since
$$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$
so let
$$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{z}=\dfrac{1-b^2}{2b},\;\dfrac{z}{x}=\dfrac{1-c^2}{2c}$$
and solving for $a,b,c$,
$$a = \frac{-x+\sqrt{x^2+y^2}}{y},\;\;b... | {
"language": "en",
"url": "https://mathoverflow.net/questions/208485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Tricky two-dimensional recurrence relation I would like to obtain a closed form for the recurrence relation
$$a_{0,0} = 1,~~~~a_{0,m+1} = 0\\a_{n+1,0} = 2 + \frac 1 2 \cdot(a_{n,0} + a_{n,1})\\a_{n+1,m+1} = \frac 1 2 \cdot (a_{n,m} + a_{n,m+2}).$$
Even obtaining a generating function for that seems challenging. Is ther... | For $n\geqslant0$ let $F_n(t)=\sum_{m\in\mathbb Z}a_{n,m}t^m$, where we are going to define $a_{n,m}$ for negative $m$ in such a way that $a_{n+1,m}=\frac{a_{n,m-1}+a_{n,m+1}}2$ for all $n\geqslant0$ and all $m\in\mathbb Z$ (so that $F_{n+1}(t)=\frac{t+t^{-1}}2F_n(t)$) and moreover the remaining requirements $a_{0,0}=... | {
"language": "en",
"url": "https://mathoverflow.net/questions/235041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
For which $n$ is there a permutation such that the sum of any two adjacent elements is a prime? For which $n$ is it possible to find a permutation of the numbers from $1$ to $n$ such that the sum of any two adjacent elements of the permutation is a prime?
For example: For $n=4$ the permutation $(1,2,3,4)$ has sums $1+2... | Here are constructions with few sums.
The permutation $(12, 1, 10, 3, 8, 5, 6, 7, 4, 9, 2, 11)$ has adjacent sums equal to $11$, $13$, or $23$.
The permutation $(12, 1, 10, 3, 8, 5, 6, 7, 4, 9, 2, 11, 18, 13, 16, 15, 14, 17)$ has adjacent sums equal to $11$, $13$,$29$, or $31$.
If you have two pairs of twin primes $p... | {
"language": "en",
"url": "https://mathoverflow.net/questions/241569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 2
} |
Determinant of a specific $4 \times 4$ symmetric matrix In a recent research work, I have come across the following nice identity, where the entries $a,b,x$ belong to an arbitrary commutative unital ring:
$$\begin{vmatrix}
2 & a & b & ab-x \\
a & 2 & x & b \\
b & x & 2 & a \\
ab-x & b & a & 2
\end{vmatrix}=(x^2-abx+a^... | Here's a method for calculating the determinant, explaining at least why it ends up as a product. I don't know if there's any significance to your determinant being a square.
Define
$$H=
\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & -1 \\
\end{pmatrix}.
$$
(The tenso... | {
"language": "en",
"url": "https://mathoverflow.net/questions/247898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Upper triangular $2\times2$-matrices over a Baer *-ring Let $A$ be a Baer $*$-ring. Let us denote $B$ by the space of all upper triangular matrices
$\left(\begin{array}{cc}
a_1& a_2 \\
0 & a_4
\end{array}\right)$ where $a_i$'s are in $A$. Is $B$ a Baer *-ring too?
As for the involution on $B$, I mean any involution... | There exist two involutions on ring $B$ as follows:
$\begin{pmatrix}
a&b\\
0&c
\end{pmatrix}^{\ast_1}= \begin{pmatrix}
c^{\ast}&b^{\ast}\\
0&a^{\ast}
\end{pmatrix},$
and
$\begin{pmatrix}
a&b\\
0&c
\end{pmatrix}^{\ast_2}= \begin{pmatrix}
c^{\ast}&-b^{\ast}\\
0&a^{\ast}
\end{pmatrix}$.
An involution is proper if $aa^{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/307440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.