Questions tagged [binomial-coefficients]
For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.
476 questions
3
votes
1
answer
244
views
On sums of a prime and a central binomial coefficient
Let $\mathbb Z^+$ be the set of positive integers. In 1934, Romanoff proved that
$$\liminf_{x\to+\infty}\frac{|\{n\le x:\ 2n+1=p+2^k\ \text{for some prime}\ p\ \text{and}\ k\in\mathbb Z^+\}|}x>0.$$
...
- 18k
2
votes
1
answer
161
views
For which $x, c$ does $\sum_{n=0}^{\infty}\binom{x}{n}\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k} |k - c|^{\alpha}$ converge?
Consider the Newton series
$$ \sum_{n=0}^{\infty}\binom{x}{n}\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{\alpha} $$
for $x, c\in\mathbb{R}$ and $\alpha\in (0, 1)$. For given values of $c$ and $\...
- 171
3
votes
1
answer
655
views
Proving that a certain factorial double sum collapses to a double-factorial series
While solving a few sums, I came across the following double sum
$$\sum_{k=0}^{n} (-1)^k\frac{(n+k)!}{2^kk!(n-k)!}x^{n-k}\sum_{j=0}^{n+k} \frac{x^j}{j!}$$
which is expected to evaluate to
$$\sum_{k=0}^...
14
votes
6
answers
2k
views
Another binomial identity
Via two calculations of the same quantity within a probability model,
Svante Janson and I observed a very indirect proof that for all $0 \le i \le n-2$, the identity
$$ n \sum_{j = i+1}^{n-1} \frac{...
- 486
6
votes
2
answers
822
views
identity involving products of binomial coefficients
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?
11
votes
2
answers
739
views
Reducing a triple combinatorial sum to a single sum
I conjecture the following identity is true for $a,b,c$ nonnegative integers with $a$ even:
$$
\sum_{k,\ell,m}
(-1)^k
\frac{(k+\ell)!(a+b-k-\ell)!^2(a+b-m)!}{k!(a-k)!\ell!(b-\ell)!m!(c-m)!(a+b-k-\ell-...
- 23.7k
2
votes
1
answer
381
views
Derive homogeneous recurrence from second order one
Let $a,b \in \mathbb{R}$ and sequece $\{f(n)\}_{n=1}^{\infty}$ is given by homogeneous second order recursive relation
$$
f(n):=af(n-1)-b^2f(n-2), \:\:\: n>2
$$
with two arbitrary starting values $...
0
votes
1
answer
106
views
Closed forms or special function representations for certain binomial sums involving harmonic-like terms?
I’m studying the following family of polynomials defined for integers $n \geq 1$:
\begin{aligned}
A_n(x) &= \frac{x^n}{(n-1)!} \left[
\sum_{k=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor} \binom{n-...
9
votes
1
answer
1k
views
Nonnegativity of an alternating combinatorial sum
Let $u,a,b,n$ be nonnegative integers such that $n\le a+b$.
Define the quantity
$$
L(u,a,b,n):=
(u+a+b-n)!\times\sum_{i,k,\ell}\
\frac{(-1)^k\ \ (u+a+b-i)!\ (k+\ell)!\ (a+b-k-\ell)!\ (u+a+b-k-\ell)!}...
- 23.7k
1
vote
1
answer
264
views
Exact form of eigenvalues of pentadiagonal Toeplitz matrices
The tridiagonal Toeplitz matrices
$$\begin{pmatrix}
a & b & & \\
c & \ddots & \ddots \\
& \ddots & \ddots & b \\
& & c ...
3
votes
2
answers
504
views
Fast converging series for $L(2,(\frac{d}{\cdot}))$
Dirichlet's $L$-function plays a central role in analytic number theory. For any integer $d\equiv0,1\pmod4$, let
$$L_d(2):=L\left(2,\left(\frac{d}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac dk)...
- 18k
11
votes
0
answers
622
views
A $q$-analogue of a conjecture (now proved) on odd binomial coefficients
At A conjecture concerning odd binomial coefficients I conjectured the following, which was proved by Fedor
Petrov and Fedor Ushakov. Let the positive integer $n$ have binary expansion
$2^{a_1}+\cdots ...
- 53.6k
42
votes
1
answer
2k
views
A conjecture concerning odd binomial coefficients
Let $n$ be a positive integer with binary expansion $2^{a_1}+\cdots +
2^{a_s}$. For $S\subseteq [s]=\{1,2,\dots,s\}$, let $k_S= \sum_{i\in S}
2^{a_i}$. Thus by a fundamental result of Lucas, ${n\...
- 53.6k
1
vote
0
answers
257
views
New irrational series for $1/\pi$ involving Apéry numbers
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:
$...
- 18k
52
votes
2
answers
2k
views
Gap in binomial coefficients
Let $a,b$ be positive integers. Because binomial coefficients are integers, we know that $a!b!$ divides $(a+b)!$. For particular $a$ and $b$ there may be a gap $g$ with a tighter result, so $a!b!$ ...
- 4,589
5
votes
1
answer
511
views
An analogue of the sum of binomial coefficients
Let $a$ and $p$ be positive integers, and consider the polynomial $$(1+x+\cdots+x^{p-1})^a = \sum_{i=0}^{a(p-1)} a_ix^i.$$ I'm looking for an asymptotic estimate of $\sum_{i=0}^{b} a_i$. If $p=2$, ...
0
votes
1
answer
187
views
How can I prove following equation (containing a sum of binomial coefficients)?
I'm trying to prove the following equation:
$$\sum_{r=0}^i (-1)^{i-r} \binom{k-r}{i-r}\binom{v-k-1+r}{r}\binom{k-1}{r} = (1 - \frac{i}{k(1-\frac{k}{v})}) \binom{k}{k-i}\binom{v-k}{i}.$$
I already ...
- 3
16
votes
2
answers
966
views
An identity involving binomial coefficients
How to prove that
$$\sum _{j=0}^{n-i} \frac{(-1)^j \binom{n-i}{j}}
{(i+j) (i+j+1) (n+i+j+1)\binom{n+i+j}{n-i}}
=\frac{4 (2 i-1)!\, (2 n-2 i+1)!}{(2n+2)!},$$
where $n$ and $i$ are integers such $1\le i\...
- 143k
16
votes
1
answer
1k
views
Diagonalizing Pascal's triangle
Let $D_n$ be the $n \times n$ diagonal matrix with entries $1, 2, \dots, n$.
Let $P_n$ be the $n \times n$ upper triangular matrix whose entry $a_{i,i+j}$ is given by $\binom{i+j}{i-1}$. For instance, ...
- 10.1k
0
votes
1
answer
144
views
Binomial coefficients identity of the form $\sum_{k=0}^{n} (-1)^{k+n} ... $ [closed]
Is there a binomial coefficient identity with alternating sums as
$$...= \sum_{k=0}^{n} (-1)^{k+n} ... $$ presented? With ... I mean something connected to binomial coefficients.
I searched many books ...
user514000
2
votes
0
answers
163
views
On $a_n(x)=\sum_{i,j=0}^n \binom{n}i^2\binom{n}j^2\binom{i+j}ix^{i+j}$ (II)
As in Question 491655, we define the polynomial
$$a_n(x):=\sum_{i,j=0}^n\binom ni^2\binom nj^2\binom{i+j}ix^{i+j}$$
for every $n=0,1,2,\ldots$. Motivated by Antoine Labelle's approach to Question ...
- 18k
2
votes
0
answers
238
views
On $a_n(x)=\sum_{i,j=0}^n\binom ni^2\binom nj^2\binom{i+j}ix^{i+j}$ (I)
Recall that the Apéry numbers are given by
$$A_n=\sum_{k=0}^n\binom nk^2\binom {n+k}k^2\ \ (n\in\mathbb N=\{0,1,2,\ldots\}).$$
In a 2012 JNT paper I conjectured that for any odd prime $p$ we have
$$\...
- 18k
0
votes
2
answers
300
views
Identity involving binomials and sum of nested binomials
Let $t,d$ be positive integers. Is there a simple way to prove
$$\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}$$
(Added)
I was actually ...
- 2,109
0
votes
1
answer
183
views
Identity with Newton's binomial symbol
Let's $a>0$ and $b>0$ be some, fixed real numbers. Let us consider sequence $S_m$, $m=0,1,\ldots$ defined as follow:
$$
S_m = \sum_{n=0}^{\left\lfloor \frac{m-1}{2} \right\rfloor}
\;\sum_{k=0}^{...
- 21
3
votes
0
answers
202
views
On central trinomial coefficents
For $n\in\mathbb N=\{0,1,2,\ldots\}$, the central binomial coefficient $\binom{2n}n$ is the coefficient of $x^n$ in the expansion of $(x^2+2x+1)^n$. Similarly, the central trinomial coefficient $T_n$ ...
- 18k
3
votes
1
answer
115
views
Binomial Sum Transformed by a piecewise linear function
I want to understand a quantity that looks like this: for fixed, non-negative integers $k$ and $m$ so that $k\geq 2m$,
$$
\sum_{n=0}^k f(n)\binom{k}{n} x^n
$$
where $f(n)$ is a piecewise linear ...
- 45
2
votes
0
answers
331
views
Does $\lim\limits_{n\to\infty} S(n,x)$ exist for $x\geqslant 1$? What is its asymptotic expression?
Consider the sum:
$$
S(n,x) = \sum_{i=1}^n \binom{n}{i} \frac{(-1)^i}{i} \left(-1 + \sum_{m=0}^i \frac{x^{m-i}}{m!}\right)
$$
Numerical evidence suggests rapid convergence for $x \geqslant 1$. The ...
- 273
3
votes
1
answer
396
views
Integrality of a product involving binomial coefficients
In some computations I've been doing, the following numbers appeared:
$$\binom{2n}{n}\binom{n}{k}\binom{k}{j}\frac{2n+1}{j+k+1}$$
where $n\ge k\ge j\ge0$. It seems that these are always integers, ...
- 3,300
17
votes
3
answers
970
views
A formula for Apéry numbers
In my research, I came across the following formula for Apéry numbers:
$$ a_n = \sum_{i=0}^n \sum_{j=0}^n {n \choose i}^2 {n \choose j}^2{i+j \choose i}.$$
This formula does not seem to appear in the ...
- 4,097
0
votes
0
answers
99
views
Estimate the number of special strings and the asymptotic ratio
Suppose a ternary string $x=x_1x_2\cdots x_N\in\{0,1,2\}^N$. Counting the number of non-identical substrings (consecutive, length $L\geq2$) such that its coordinate-sum do not equal to the length, i.e....
- 19
1
vote
3
answers
395
views
Binomial coefficient C(2k,n-1) alternative formula equivalent to the Vandermonde identity?
I am working with a binomial sum that arises in some combinatorial arguments (and also appears in certain generating‐function manipulations). Specifically, I have this identity
$$
\sum_{j=0}^{n}
\...
- 13
13
votes
2
answers
459
views
Duality between binomial coefficients and Eulerian numbers
Apart from the binomial formula $\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}$, the binomial coefficients satisfy another recursive formula:
$$
\binom{n+1}{m}=(m+1)\binom{n}{m}-(n-m)\binom{n}{m-1}.
$$
...
- 1,350
1
vote
2
answers
433
views
Prove by induction a formula involving binomial coefficients
Define $$f(k) = - \frac{\Gamma(\frac{k-1}{2}) \pi^{3/2}}{\Gamma(\frac{k}{2}) 2^{\frac{k-1}{2}}} \sum_{m=0}^{\lfloor\frac{k-3}{4}\rfloor} \binom{\frac{k-1}{2}}{2m+1} \binom{2m}{m} 4^{-m},$$
And let $$M(...
- 439
4
votes
0
answers
251
views
New series identities motivated by Question 488044
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 $(1)$....
- 18k
10
votes
1
answer
602
views
How to prove the identity $\sum_{k=0}^\infty(22k^2-92k+11)\binom{4k}k/16^k=-5$?
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.
...
- 18k
2
votes
0
answers
258
views
A bunch of hypergeometric series with converging rate $-1/27$
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
$$\...
- 18k
2
votes
0
answers
376
views
A new series for $\pi^2$
In 2023, I discovered the identity
$$\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}$$
which was later confirmed by K. C. Au in ...
- 18k
-1
votes
1
answer
221
views
A series related to $1/\pi^2$
The following series is related to Example X of the paper 2312.14051v4:
$$\sum_{k=0}^{\infty}\frac{ (-1)^k (5532k^4 + 5464k^3 + 2173k^2 + 416k + 32)\binom{2k}{k}^9}{ (10k + 1)(10k + 3)\binom{10k}{5k} \...
- 1,248
6
votes
0
answers
417
views
Another series for $\pi^{10}$ with geometric converging rate
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.
...
- 18k
10
votes
2
answers
1k
views
Some series related to $\pi^4$ and $\zeta(3)$
Recently, I discovered the following identity
$$\sum_{k=0}^{\infty}\frac{\left(3 k +2\right) \left(7 k^{2}+9 k +3\right) 256^{k +1}}{\left(2 k +1\right)^{7} {\binom{2 k}{k}}^{7}}=(2\pi)^4.\tag{1}$$
My ...
- 1,248
2
votes
2
answers
583
views
A series involving $\log(2)$ and $\log(3)$
Recently, I discovered the following identity
$$\sum_{k=1}^{\infty}\frac{100+299 k}{\left(24 k^{2}+8 k \right) 576^{k}{\binom{3 k}{k}} }=\frac{1}{8}+3\log(2)-2\log(3). \tag{1}$$
Based on identity (3) ...
- 1,248
7
votes
1
answer
666
views
A series for $\log(2)$ with converging rate 1/22235661
Recently, I discovered the following identity
$$\sum_{k=1}^{\infty}\frac{ \binom{4k}{k} \binom{2k}{k}^2 P(k)}{224 \binom{3k}{k}^2 \binom{7k}{k} \binom{14k}{7k} 256^kQ(k)}=\frac{2329}{3360}-\ln \! \...
- 1,248
5
votes
0
answers
441
views
A series for $\pi^{10}$ with converging rate $1/64$
In the 2023 preprint [arXiv:2312.14051], K. C. Au used the WZ method to confirm the identity
$$\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}$$
conjectured by ...
- 18k
4
votes
0
answers
668
views
Coefficients in polynomial identity for odd powers
Recently, I've come back again to one of the prominent articles in area of polynomials, power sums etc.
That is Johann Faulhaber and sums of powers: https://arxiv.org/abs/math/9207222.
Indeed, a great ...
- 197
4
votes
1
answer
543
views
How to prove this combinatorial equality on $\sum_{k=0}^{n} \binom{k}{m}(-1)^{k}$?
I have been trying to prove the equality
$$\sum_{k=0}^{n} \binom{k}{m}(-1)^{k} = \left(\frac{-1}{2}\right)^m[\text{$n$ is even}] + \frac{(-1)^n}{2}\sum_{k=1}^{m}\binom{n+1}{k}\left(\frac{-1}{2}\right)^...
- 59
3
votes
0
answers
255
views
Doubling or addition formulas for the central binomial coefficients $\binom{2n}{n}$?
Let $B(n)$ denote the central binomial coefficient $B(n)=\binom{2n}{n}$.
We have
$$ B(n+1)=2(2n+1)B(n)/(n+1) \tag1\label1$$
With \eqref{1} we can compute near neighbors $B(n+i)$ given $n,B(n)$.
in $O(...
- 25.7k
4
votes
0
answers
199
views
Proving binomial identities
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-...
- 43.9k
15
votes
2
answers
912
views
Sign of the sum of alternating triple binomial coefficient
I encountered a research question and it can be turned equivalently into showing
$$\sum_{k=j}^n(-1)^{k+j}\binom{n}{k}\binom{n+k}{k}\Big(\frac{1}{10n+10}\Big)^k\binom{2k}{k+j}\ge0$$
holds for any $n\in\...
- 253
23
votes
3
answers
3k
views
Product of all binomial coefficients
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 ...
- 448
0
votes
1
answer
214
views
Closed form for $\sum\limits_{k=0}^{n} [\operatorname{wt}(k) = m]$ where $\operatorname{wt}(n)$ is the binary weight of $n$
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of $1$'s in binary expansion of $n$).
Let $a(n,m)$ be the family of integer sequences such that
$$
a(n,m) = \sum\limits_{k=0}^{n} [\operatorname{wt}(...
user231922