Questions tagged [combinatorial-identities]

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A = B (but not quite); 3-d arrays with multiple recurrences

Many years ago, I discovered the remarkable array (apparently originally discovered by Ramanujan) 1 1 3 2 10 15 6 40 105 105 24 196 700 1260 945 ...
Peter Shor's user avatar
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18 votes
0 answers
745 views

Two curious series for $1/\pi$

On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ ...
Zhi-Wei Sun's user avatar
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12 votes
0 answers
497 views

$q$-analogue of the multinomial theorem?

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
10 votes
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490 views

New series for $\zeta(5)$ involving second-order harmonic numbers

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
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10 votes
0 answers
342 views

A bijective proof for the odd companion to Shapiro's Catalan convolution

Shapiro's Catalan convolution is the following formula (where $C_n$ is the $n$th Catalan number): $$ \sum_{k=0}^{n}{C_{2k}C_{2(n-k)}}=4^nC_n. $$ In other words, letting $C(z)=\sum_{n=0}^{\infty}{C_nz^...
Alexander Burstein's user avatar
9 votes
0 answers
191 views

For $q$-analogues of a known curious identity

In 2002 I published the folllowing curious combinatorial identity: $$(x+m+1)\sum_{i=0}^m(-1)^i\binom{x+y+i}{m-i}\binom{y+2i}i-\sum_{i=0}^m\binom{x+i}{m-i}(-4)^i=(x-m)\binom xm.$$ My original proof is ...
Zhi-Wei Sun's user avatar
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8 votes
0 answers
315 views

A hypergeometric series for $\sqrt3\pi$ with converging rate $1/9$

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
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6 votes
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221 views

A curious series for $L(2,(\frac{-3}{\cdot}))$

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
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6 votes
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282 views

A new series for $\sqrt3/\pi$?

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
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5 votes
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338 views

Sum over permutations involving sign

The problem is to evaluate the following sum over all permutations $\sigma\in S_{d}$ of $\{1,2,...,d\}$: $\displaystyle\sum_{\sigma\in S_{d}}\text{sgn}(\sigma)\displaystyle\frac{1}{\prod_{i=1}^{d}(\...
Abhishek Halder's user avatar
5 votes
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344 views

When does a triangle of numbers have a zero row sum?

Suppose we have a triangle of numbers defined by the recurrence relation $$\left| n \atop k \right| = f(n,k) \left| n-1 \atop k \right| +g(n,k) \left| n-1 \atop k-1 \right| + [n=k=0],$$ for some ...
Mike Spivey's user avatar
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4 votes
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"Convolving" a general Catalan with classical Catalan

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
4 votes
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200 views

Extract this constant term

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
4 votes
0 answers
214 views

Curious double sum identity

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
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4 votes
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For a combinatorial proof of a symmetric identity

In my paper Supercongruences involving dual sequences [Finite Fields Appl. 46(2017), 179-216], I gave a new symmetric identity which states that if $x+y=-1$ then $$\sum_{k=0}^n(-1)^k\binom xk^2\binom{...
Zhi-Wei Sun's user avatar
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4 votes
0 answers
312 views

Generalization of Tamarkin’s ARO 1993, final round, problem 10/8: part II

Let us use the notations of my previous question about Tamarkin's problem. Let $\ell\in\left\lbrace 0,1,...,p\right\rbrace$. An element $f\in \mathbb Z^{\mathbb Z}$ is said to be $\ell$-average-...
darij grinberg's user avatar
3 votes
0 answers
172 views

Equality of determinants: a direct justification request

Let $I_k$ denote the enumeration of involutions among permutations in $\mathfrak{S}_k$. Here is yet another cute finding for which I ask a: Question. Is there a direct proof (or interpretation or ...
T. Amdeberhan's user avatar
2 votes
0 answers
194 views

For human proofs of two novel combinatorial identities

For $n=0,1,2,\ldots$, let us define the polynomial $$S_n(x):=\sum_{k=0}^n\binom{x/2}k\binom{(x-1)/2}k\binom{-(x+1)/2}{n-k}\binom{-(x+2)/2}{n-k}.$$ Such polynomials occur in some series for $1/\pi$ ...
Zhi-Wei Sun's user avatar
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2 votes
0 answers
347 views

Reflection formula for the Hurwitz zeta function and odd zeta values

A reflection formula for the Hurwitz zeta function, which does not seem to be well known, uses half of the polynomials generated by $\frac{1}{-1+\sqrt{t-1}\cot(\sqrt{t-1}u)}$. (Look at the sections "...
Wolfgang's user avatar
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2 votes
0 answers
153 views

system of complex number equations

Let $a_1,a_2,a_3,a_4\in \mathbb{C}$ be distinct such that $$a_1^3+a_2^3+a_3^3+a_4^3=0$$ $$(1+|a_1|^2)a_1^2+(1+|a_2|^2)a_2^2+(1+|a_3|^2)a_3^2+(1+|a_4|^2)a_4^2=0$$ $$(1+2|a_1|^2+2|a_1|^4)a_1+(1+2|a_2|^...
Ma Na's user avatar
  • 309
2 votes
0 answers
186 views

Multiplying three factorials with three binomials in polynomial identity

I have checked the following identity (1) below for $n\leq 40$ with a computer. Let $(n)_k$ denote the falling factorial $n(n-1)\ldots (n-k+1)$, let $Z_n=\sum_{k=0}^n (n)_k x^{n-k}$, and finally let $...
Ewan Delanoy's user avatar
2 votes
1 answer
667 views

Transfinite sums related to a sequence

Given a sequence $S$ indexed by the finite ordinals, a limit ordinal $\alpha$, and $k \in \mathbb{N}$, define $S_{\alpha+k}$(the extension of $S$ to $\alpha+k$) to be the sum over the products of all $...
Michael Burge's user avatar
1 vote
0 answers
512 views

Two conjectural identities involving $\zeta(3)$ and the golden ratio $\phi$

Let $\phi$ be the famous golden ratio $\frac{\sqrt5+1}2$, and let $\zeta(3)=\sum_{n=1}^\infty\frac1{n^3}.$ In 2014, in the paper Zhi-Wei Sun, New series for some special values of $L$-functions, ...
Zhi-Wei Sun's user avatar
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1 vote
0 answers
113 views

In search of multiple expressions for a sequence

The sequence $a_n=\sum_{k=0}^n\binom{n}k^24^k$ is listed on OEIS along with a couple of combinatorial interpretations. What interested me at the moment is the plethora of binomial single-sums for the ...
T. Amdeberhan's user avatar
0 votes
0 answers
225 views

Looking for a combinatorial proof of an identity

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
0 votes
0 answers
256 views

An alternating sum involving a product of binomial coefficients

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
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