All Questions
29 questions
32
votes
3
answers
2k
views
Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$
In my research, I found this identity and as I experienced, it's surely right. But I can't give a proof for it.
Could someone help me?
This is the identity:
let $a$ and $b$ be two positive integers; ...
- 321
27
votes
2
answers
1k
views
Some binomial coefficient determinants
It is well known that for $n>0$
$$d(n)=\det\left(\binom{2i+2j+1}{i+j}\right)_{i,j=0}^{n-1}=1.$$
Computer experiments suggest that more generally
$$d(n,k)=\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,...
- 5,622
26
votes
3
answers
4k
views
Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$
I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by ...
18
votes
3
answers
860
views
$\prod_k(x\pm k)$ in binomial basis?
Let $x$ be an indeterminate and $n$ a non-negative integer.
Question. The following seems to be true. Is it?
$$x\prod_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^...
- 43.2k
16
votes
2
answers
750
views
Sum of multinomals = sum of binomials: why?
I stumbled on the following identity, which has been checked numerically.
Question. Is this true? If so, any proof?
$$\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j}
=\sum_{j=0}...
- 43.2k
13
votes
1
answer
468
views
Four new series for $\pi$ and related identities involving harmonic numbers
Recently, I discovered the following four new (conjectural) series for $\pi$:
\begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\...
- 15.6k
13
votes
1
answer
385
views
Some more binomial coefficient determinants
The setup is similar to this question, but generalizes the size of the Hankel matrix. We'll define
$$d(n,k,r):=\det\left(\binom{2i+2j+k+r}{i+j}\right)_{i,j=0}^{kn-1}.$$
Edit: Thanks to Johann ...
- 13.4k
12
votes
3
answers
1k
views
A "quantum" identity: in search of a proof -Part II
As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on this ...
- 43.2k
12
votes
2
answers
1k
views
An interesting identity: in search of a proof -Part I
I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS.
Question. Can you show that
$$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-...
- 43.2k
9
votes
3
answers
2k
views
Combinatorial identities
I have computational evidence that
$$\sum_{k=0}^n \binom{4n+1}{k} \cdot \binom{3n-k}{2n}= 2^{2n+1}\cdot \binom{2n-1}{n}$$
but I cannot prove it. I tried by induction, but it seems hard. Does anyone ...
- 91
9
votes
0
answers
192
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 ...
- 15.6k
8
votes
1
answer
914
views
A special binomial identity in need of a proof
I've encountered a curious identity as a codicil in some work. Is there a proof or reference?
$$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$
- 43.2k
7
votes
1
answer
286
views
A reference for a sum found in Gould's Combinatorial Identities book
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 ...
- 135
6
votes
1
answer
2k
views
Double sum involving binomial coefficients
I came across a sum of binomial coefficients while trying to solve a problem involving $SU(2)$ group integrals. I am not able to solve it, nor I found a similar identity in the literature. I would ...
- 377
6
votes
1
answer
310
views
On the sum $\sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k}$
Let $n$ be a non-negative integer.
Does there always exist a polynomial $P_n(a,b)$ such that for all integers $a > b \geq n/2$ we have
$$
\sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k} = \...
- 3,927
4
votes
3
answers
509
views
How to calculate: $\sum\limits_{k=0}^{n-m} \frac{1}{n-k} {n-m \choose k}$
How to calculate:
$$\sum _{k=0}^{n-m} \frac{1}{n-k} {n-m \choose k}.$$
- 327
4
votes
2
answers
495
views
Showing this formula counts these things
I'm writing an article, and I got stuck trying to prove that some numbers are positive. I have a relatively good intuition for guessing what an expression is counting, but in this case I'm not being ...
- 1,889
4
votes
0
answers
225
views
For a combinatorial proof of a symmetric identity
In my paper Supercongruences involving dual sequences [Finite Fields Appl. 46(2017), 179-216], I gave a new symmetric identity which states that if $x+y=-1$ then
$$\sum_{k=0}^n(-1)^k\binom xk^2\binom{...
- 15.6k
3
votes
2
answers
244
views
Is there a combinatorial reason for variable-independence of this binomial-coefficient identity?
Consider the following identity
$$\sum_{n=0}^{R-t}\binom{n+\ell}n\binom{R-\ell-n}{R-t-n}=\binom{R+1}{t+1}.\tag1$$
It is relatively easy to give an algebraic or mechanical proof of (1). But, I like to ...
- 43.2k
3
votes
1
answer
466
views
How to prove this combinatorial identity?
If $n \in \mathbb N \setminus \{0\}$ and $x,y,z \in \mathbb R$ such that $x+y+z=n-1$, show that
$$\dfrac{(-4)^n}{\binom{2x}{n}}\sum_{r+s=n,r,s\in Z}\dfrac{\binom{y}{r}\binom{y-a}{r}\binom{z}{s}\binom{...
- 4,280
2
votes
1
answer
476
views
Alternating sign binomial identity [closed]
I recently noticed that for a triple of integers $k \geq 2$, $k \geq m \geq t \geq 1$, the following identity seems to hold
$\sum_{j=0}^{m-t} (-1)^{m-t-j}{k \choose j}{m-1-j \choose t-1}={k-t \choose ...
- 21
2
votes
0
answers
196
views
For human proofs of two novel combinatorial identities
For $n=0,1,2,\ldots$, let us define the polynomial
$$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}.$$
Such polynomials occur in some series for $1/\pi$ ...
- 15.6k
2
votes
0
answers
138
views
A combinatorial sum involving ratios of binomials [closed]
Can anyone suggest how to prove the following (for $k \le n$):
$$\sum \limits_{s=0}^N \frac{\binom{n}{k} \binom{N-n}{s-k} }{\binom{N}{s}} = \frac{N+1}{n+1}$$
I am assuming it to be true, and ...
- 21
1
vote
2
answers
1k
views
An identity involving a sum of binomial coefficients
I am moving through a classic paper (On Average Height of Planted Plane Trees by Knuth, de Bruijn and Rice, 1972), and I would like to trade a weaker result for simpler mathematical tools, because my ...
1
vote
0
answers
116
views
In search of multiple expressions for a sequence
The sequence $a_n=\sum_{k=0}^n\binom{n}k^24^k$ is listed on OEIS along with a couple of combinatorial interpretations. What interested me at the moment is the plethora of binomial single-sums for the ...
- 43.2k
0
votes
3
answers
694
views
How to calculate$ \sum \limits_{k=0}^{m-n} {m-k-1 \choose n-1} {k+n \choose n}$?
How to calculate $$\sum\limits_{k=0}^{m-n} {m-k-1 \choose n-1} {k+n \choose n}.$$
- 327
0
votes
1
answer
403
views
Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?
In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation*}%...
- 1,091
0
votes
0
answers
244
views
Looking for a combinatorial proof of an identity
I've come up with an interesting combinatorial identity (thanks to P. Belmans who precomputed the numbers and pointed out to me that they correspond to OEIS A002697):
$$
\sum_{i=0}^{n-1}\binom{n+1-i}{...
- 1,810
0
votes
0
answers
302
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}...
- 109