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13 votes
1 answer
468 views

Four new series for $\pi$ and related identities involving harmonic numbers

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
  • 15.6k
12 votes
2 answers
1k views

An interesting identity: in search of a proof -Part I

I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS. Question. Can you show that $$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-...
T. Amdeberhan's user avatar
6 votes
2 answers
719 views

Recreation with Catalan

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
6 votes
1 answer
484 views

Three conjectural series for $\pi^2$ and related identities

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
  • 15.6k
6 votes
0 answers
235 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
  • 15.6k
6 votes
0 answers
298 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
  • 15.6k
4 votes
3 answers
322 views

An identity for product of central binomials

This "innocent-looking" identity came out of some calculation with determinants, and I like to inquire if one can provide a proof. Actually, different methods of proofs would be of valuable merit and ...
T. Amdeberhan's user avatar
2 votes
1 answer
253 views

In search of a binomial identity proof

The following has strong experimental evidence. Question. For $n\geq3k$, is this identity true? Proof? $$\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j}=\sum_{j=0}^{\lfloor\...
T. Amdeberhan's user avatar
1 vote
2 answers
211 views

Bilinear recurrence relation between even Bernoulli numbers

Throughout this question $n$ is a positive integer greater than 1. Consider the following well-known identity by Euler, $$\sum_{k=1}^{n-1} \binom{2n}{2k}B_{2k}B_{2n-2k}=-(2n+1)B_{2n}.$$ Rather ...
bryanjaeho's user avatar
1 vote
2 answers
820 views

Converting a recursive definition to an explicit one

Is there an explicit form for $a_x$ (whole numbers x) given that $a_x = \displaystyle\sum_{i=1}^{x-1} \binom{x-1}{i} a_i$? I've listed out the first few terms: for $x=0,1,2,3,4,5,6, 7$ we have $a_x ...
Hardik Shah's user avatar
1 vote
1 answer
124 views

A $1$-step convolution identity involving the Motzkin triangle

The Motzkin triangle $T(n,k)$ is built according to the rules: (1) $T(n,0)=1$; (2) $T(n,k)=0$ if $k<0$ or $k>n$; (3) $T(n,k)=T(n-1,k-2)+T(n-1,k-1)+T(n-1,k)$. After some numerical evidence I ...
T. Amdeberhan's user avatar
1 vote
0 answers
116 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