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Via two calculations of the same quantity within a probability model, Svante Janson and I observed a very indirect proof that for all $0 \le i \le n-2$, the identity $$ n \sum_{j = i+1}^{n-1} \frac{...
David Aldous's user avatar
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6 votes
2 answers
822 views

I need the following identity to prove something (regarding functions of two variables): $$\sum_{k=0}^{n-r} (-1)^k \binom{n-k}{k} \binom{n-2k}{n-k-r} = 1$$ Is this well-known? Is there a quick proof?
Peter M. Higgins's user avatar
  • 61
11 votes
2 answers
739 views

I conjecture the following identity is true for $a,b,c$ nonnegative integers with $a$ even: $$ \sum_{k,\ell,m} (-1)^k \frac{(k+\ell)!(a+b-k-\ell)!^2(a+b-m)!}{k!(a-k)!\ell!(b-\ell)!m!(c-m)!(a+b-k-\ell-...
Abdelmalek Abdesselam's user avatar
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3 votes
2 answers
504 views

Dirichlet's $L$-function plays a central role in analytic number theory. For any integer $d\equiv0,1\pmod4$, let $$L_d(2):=L\left(2,\left(\frac{d}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac dk)...
Zhi-Wei Sun's user avatar
  • 18k
5 votes
0 answers
460 views

In 1914 Ramanujan [Quart. J. Math. (Oxford) 45 (1914)] discovered the following irrational series for $1/\pi$: $$\sum_{k=0}^\infty\left(k+\frac{31}{270+48\sqrt5}\right)\binom{2k}k^3\left(\frac{(\sqrt5-...
Zhi-Wei Sun's user avatar
  • 18k
1 vote
0 answers
257 views

Recall that the Apéry numbers are given by $$A_n:=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2\ \ \ \ (n=0,1,2,\dotsc).$$ In 2002 T. Sato discovered the following series for $1/\pi$ involving Apéry numbers: $...
Zhi-Wei Sun's user avatar
  • 18k
3 votes
0 answers
255 views

As in Question 491655, Question 491762 and Question 491811, we define $$a_n(x):=\sum_{i,j=0}^n\binom ni^2\binom nj^2\binom{i+j}ix^{i+j}$$ for each nonnegative integer $n$. Using my own way (mentioned ...
Zhi-Wei Sun's user avatar
  • 18k
0 votes
2 answers
300 views

Let $t,d$ be positive integers. Is there a simple way to prove $$\sum_{j=1}^{td}(-1)^{td-j}{td \choose j}{{ j \choose t} \choose d}=\frac{1}{d!}\prod_{j=1}^d { jt \choose t}$$ (Added) I was actually ...
CHUAKS's user avatar
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0 votes
1 answer
183 views

Let's $a>0$ and $b>0$ be some, fixed real numbers. Let us consider sequence $S_m$, $m=0,1,\ldots$ defined as follow: $$ S_m = \sum_{n=0}^{\left\lfloor \frac{m-1}{2} \right\rfloor} \;\sum_{k=0}^{...
azonips's user avatar
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4 votes
0 answers
251 views

In Question 488044, I asked for a proof of the hypergeometric identity $$\sum_{k=0}^\infty (22k^2-92k+11)\frac{\binom{4k}k}{16^k}=-5.\tag{1}$$ Here I propose some further identities motivated by $(1)$....
Zhi-Wei Sun's user avatar
  • 18k
10 votes
1 answer
602 views

I have found the hypergeometric identity $$\sum_{k=0}^\infty(22k^2-92k+11)\frac{\binom{4k}k}{16^k}=-5. \tag{1}$$ As the series converges fast, one can easily check $(1)$ numerically by Mathematica. ...
Zhi-Wei Sun's user avatar
  • 18k
2 votes
0 answers
258 views

In a 2014 paper of Chu and Zhang, the authors gave several geometric series for some well known constants with converging rate $-1/27$. In particular, Example 93 of that paper gives the identity $$\...
Zhi-Wei Sun's user avatar
  • 18k
2 votes
0 answers
376 views

In 2023, I discovered the identity $$\sum_{k=0}^\infty\frac{(92k^3+54k^2+12k+1)\binom{2k}k^7}{(6k+1)256^k\binom{6k}{3k}\binom{3k}k}=\frac{12}{\pi^2},\tag{1}$$ which was later confirmed by K. C. Au in ...
Zhi-Wei Sun's user avatar
  • 18k
6 votes
0 answers
417 views

Series for $\pi^{10}$ with geometric converging rate are very rare. In Question 486518, I reported such a series. Now, I'd like to propose another series for $\pi^{10}$ with geometric converging rate. ...
Zhi-Wei Sun's user avatar
  • 18k
5 votes
0 answers
441 views

In the 2023 preprint [arXiv:2312.14051], K. C. Au used the WZ method to confirm the identity $$\sum_{k=1}^\infty\frac{(21k^3-22k^2+8k-1)256^k}{k^7\binom{2k}k^7}=\frac{\pi^4}{8}\tag{1}$$ conjectured by ...
Zhi-Wei Sun's user avatar
  • 18k
13 votes
2 answers
2k views

Last week I found the following curious hypergeometric identity: $$\sum_{k=1}^\infty\frac{(-1)^k(560k^4-640k^3+408k^2-136k+17)}{(2k-1)^4 k^5 \binom{2k}k\binom{3k}{k}}=180\zeta(5)-\frac{56}3\pi^2\zeta(...
Zhi-Wei Sun's user avatar
  • 18k
4 votes
0 answers
199 views

Assume $n$ and $m$ are positive integers. My eventual wish is in finding a $q$-analogue of the below identity but for now I wish to see alternative proofs. I can supply a justification using the Wilf-...
T. Amdeberhan's user avatar
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23 votes
3 answers
3k views

I was writing a research paper in Computer Science. I had to provide an upper bound for the number of steps of the algorithm I had found with my colleagues; the nature of the algorithm is totally ...
Melanzio's user avatar
  • 448
7 votes
1 answer
325 views

I stumbled upon a curious identity that seems to hold for all integers $n\ge3$: $$\sum_{i=0}^{\lfloor n/3\rfloor}\prod_{j=1}^i{-{n-3j\choose3}\over{n\choose3}-{n-3j\choose3}} ={n\over3}\sum_{i=0}^{\...
ho boon suan's user avatar
  • 1,017
0 votes
1 answer
305 views

How this can be proved? $$ E = {}_2F_1(-\frac{1}{2}, \frac{1}{3}, \frac{4}{3},-8) = \frac{6}{5} - \frac{\chi}{2} $$ where $$ \chi = \frac{6\sqrt{\pi}}{5}\frac{\Gamma(\frac{1}{3})}{\Gamma(-\frac{1}{6})}...
scipio1465's user avatar
  • 9
5 votes
1 answer
288 views

In this interview, Ira Gessel mentions the following results: Result 1: Let $B_n$ denote the $n^{\text{th}}$ Bernoulli number. Define the series $$B(x) = \sum_{n=2}^{\infty} \frac{B_nx^{n-1}}{n(n-1)}.$...
Naysh's user avatar
  • 599
23 votes
5 answers
2k views

I came across the following identity in my research: $$ \sum_{m=0}^s \frac{(-1)^m (a+2m)}{m!(s-m)! (a+m)_{s+1}}=\delta_{s,0} $$ where $(a)_n= a(a+1)\cdots (a+n-1)$ is the Pochhammer symbol. One can ...
XYX's user avatar
  • 341
6 votes
2 answers
781 views

Consider the well-known sequence $C_k=\frac1{k+1}\binom{2k}k$ of Catalan numbers. I came across the below identity while working with certain generating functions. I thought it might be of interest to ...
T. Amdeberhan's user avatar
  • 43.9k
3 votes
1 answer
490 views

I arrived at this formula by inductive reasoning, but I don’t know how to prove it. For any natural numbers $m$ and $k=0,1,2,\ldots, m-1$, $B_i$ - Bernoulli numbers we have: $$\sum_{i=0}^k (-1)^{k-i}\...
juna's user avatar
  • 31
1 vote
1 answer
329 views

Recently I had a curious discovery. Namely, I have made the following conjectures. Conjecture 1. We have the identity $$\sum_{k=0}^\infty\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}=0.\label{1}...
Zhi-Wei Sun's user avatar
  • 18k
8 votes
0 answers
448 views

Recently, I found a (conjectural) new series for $\sqrt3\pi$: $$\sum_{k=1}^\infty\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}=\frac{\sqrt3\pi}{18}.\label{1}\tag{1}$$ The series converges fast ...
Zhi-Wei Sun's user avatar
  • 18k
13 votes
1 answer
584 views

Recently, I discovered the following four new (conjectural) series for $\pi$: \begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\...
Zhi-Wei Sun's user avatar
  • 18k
7 votes
0 answers
278 views

Let $$K:=L\left(2,\left(\frac{-3}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac k3)}{k^2}=\sum_{j=0}^\infty\left(\frac1{(3j+1)^2}-\frac1{(3j+2)^2}\right),$$ where $(\frac k3)$ is the Legendre ...
Zhi-Wei Sun's user avatar
  • 18k
7 votes
1 answer
663 views

Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{...
Zhi-Wei Sun's user avatar
  • 18k
6 votes
0 answers
338 views

Recently, I conjectured the following identity: $$\sum_{k=0}^\infty\frac{(66k^2+37k+4)\binom{2k}k\binom{3k}k\binom{4k}{2k}}{(2k+1)729^k}=\frac{27\sqrt3}{2\pi}.\tag{1}$$ This can be easily checked ...
Zhi-Wei Sun's user avatar
  • 18k
4 votes
1 answer
286 views

Stanley famously conjectured ("Some combinatorial properties of Jack symmetric functions" Adv. in Math. (77) 1989, doi:10.1016/0001-8708(89)90015-7, MR1014073, Zbl 0743.05072) that the Jack ...
Ryan Mickler's user avatar
  • 375
3 votes
3 answers
836 views

Let $\mathcal{P}(n)$ be the set of all unrestricted partitions of $n$ while $\mathcal{O}(n)$ stand for the set of all partitions of $n$ into odd parts. We adopt the power notation for partitions $\...
T. Amdeberhan's user avatar
  • 43.9k
6 votes
1 answer
528 views

I need to prove the following statement. Let $ n, g, m, a ,t$ be integers. Prove that the following statement is true for all $ n \geq g(1+2m)+1 $, $ g\geq 2t $, $ m\geq t $, $ 0\leq a <t $, and $ ...
Arda Aydin's user avatar
  • 63
8 votes
3 answers
755 views

Recall that $$\mathrm{Li}_2(x):=\sum_{n=1}^\infty\frac{x^n}{n^2}.$$ I have found the following identity: \begin{equation}\begin{aligned}&\mathrm{Li}_2\left(\frac{-1-\sqrt{-7}}4\right)+\mathrm{Li}...
Zhi-Wei Sun's user avatar
  • 18k
4 votes
0 answers
116 views

Consider what is sometimes known as generalized Catalan sequence $$\mathcal{{\color{red}C}}_{a,b}:=\frac{2b+1}{a+b+1}\binom{2a}{a+b}.$$ Observe that $\mathcal{{\color{red}C}}_{n,0}$ reduces to the ...
T. Amdeberhan's user avatar
  • 43.9k
5 votes
0 answers
240 views

Given a Laurent polynomial $F$ in the variables $\mathbf{t}=(t_1,\dots,t_n)$, let $CT_{\vec{\mathbf{t}}}\,F$ denote its constant term. For example, $CT_{t_1,t_2}((8t_1-\frac1{3t_1t_2})(5t_1t_2+t_2^2+\...
T. Amdeberhan's user avatar
  • 43.9k
1 vote
0 answers
300 views

I've come up with an interesting combinatorial identity (thanks to P. Belmans who precomputed the numbers and pointed out to me that they correspond to OEIS A002697): $$ \sum_{i=0}^{n-1}\binom{n+1-i}{...
Anton Fonarev's user avatar
  • 1,840
3 votes
1 answer
256 views

Given an integer partition $\lambda$, introduce the following quantities: \begin{align*} c(\lambda)&=\sum_{i\geq1}\left\lceil\frac{\lambda_i}2\right\rceil, \qquad c_o(\lambda)=\sum_{i\geq1}\left\...
T. Amdeberhan's user avatar
  • 43.9k
10 votes
2 answers
2k views

$$C_{n} = \sum_{i=1}^n (-1)^{i-1} \binom{n-i+1}{i} C_{n-i}$$ Are there any good combinatorial proofs or algebraic proofs of this?
banana's user avatar
  • 111
0 votes
0 answers
366 views

I encountered the sum below, where $c_{1}$, $c_{2}$, $c_{3}$, $c_{4}$ and $d$ are some given positive constants. Does anyone have an idea how to simplify it? $$ \sum\limits_{k=1}^{d} \frac{(-1)^{k-1}k}...
sdd's user avatar
  • 109
7 votes
1 answer
350 views

On p. 49 in Gould's book Combinatorial Identities, the author states that the sum $$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{2n}{2k}^{-1}$$ "... arises naturally in a statistical problem; it ...
Sela Fried's user avatar
  • 135
2 votes
1 answer
306 views

Let $n\geq m$ be non negative integers, and consider a list of $(n+m+1)$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $\det\left[\...
Athena's user avatar
  • 275
10 votes
0 answers
589 views

In 1997 T. Amdeberhan and D. Zeilberger proved that $$\sum_{k=1}^\infty\frac{(-1)^k(205k^2-160k+32)}{k^5\binom{2k}k^5}=-2\zeta(3).\tag{1}$$ In 2008 J. Guillera obtained that $$\sum_{k=1}^\infty\frac{(...
Zhi-Wei Sun's user avatar
  • 18k
10 votes
1 answer
648 views

The classical rational Ramanujan-type series for $1/\pi$ have the following four forms: \begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1} \\\sum_{k=0}^\...
Zhi-Wei Sun's user avatar
  • 18k
4 votes
0 answers
279 views

The following identity seems to hold for $a>1$ : $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{ \frac{a^m}{m}\left( \frac{a^m}{m}+ \frac{a^n}{n} \right) } = \frac{a^2}{2(a-1)^4}$$ I've tested ...
Jim Bryan's user avatar
  • 6,072
1 vote
1 answer
276 views

Can you prove (preferably combinatorially) the following identity for the total number of perfect matchings of the complete graph $K_{2n}$, where the edges in the matching are ordered, i.e., $\binom{...
sdd's user avatar
  • 109
8 votes
1 answer
597 views

In an old paper of Glaisher, I find the following formulas: $$\dfrac{\sin(\pi x)}{\pi x}=1-\dfrac{x^2}{1^2}-\dfrac{x^2(1^2-x^2)}{(1.2)^2}-\dfrac{x^2(1^2-x^2)(2^2-x^2)}{(1.2.3)^2}-\cdots$$ $$\cos(\pi x/...
Henri Cohen's user avatar
  • 14k
15 votes
1 answer
1k views

The $q$-binomial theorem states that $$ \prod_{k=0}^{n-1}(1+q^kt) = \sum_{k=0}^n q^{\binom k2}{n\brack k}_q t^k. $$ This identity is a $q$-analogue of the binomial theorem $$ (1+t)^n = \sum_{k=0}^n \...
Amritanshu Prasad's user avatar
  • 5,894
2 votes
1 answer
261 views

Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ with $[0]_q!:=1$ and the Gaussian polynomials $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,\cdot\,[n-k]_q!}$. Adopt the convention that $\binom{n}k_q=0$ whenever $k&...
T. Amdeberhan's user avatar
  • 43.9k
3 votes
2 answers
473 views

How to prove the following identity? Let $r = (r_1, r_2, \ldots, r_d)$ and $c = (c_1, c_2, \ldots, c_d)$ be sequences of natural numbers such that $s = r_1 + r_2 + \cdots + r_d = c_1 + c_2 + \ldots + ...
MMM's user avatar
  • 325

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