# Questions tagged [binomial-coefficients]

For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.

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### Are there any numerically-plausible perfect binary code parameters besides (90,2)? [duplicate]

(Formerly on Math StackExchange here, without much progress.) In order for a perfect binary code on $n$ symbols to correct $k$ errors, we need the sum $${n\choose 0}+{n\choose 1}+\ldots+{n\choose k}$$ ...
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### how to prove identity for nth derivative of $(\text{arctanh}(x))^j$?

this question asked on MSE I worked on integral problem and I got that $$\int_0^1 \frac{x^n}{\ln \left(\frac{1-x}{1+x} \right) } dx=-\frac{2}{(n+1)!}\sum_{j=1}^{n+1}F(n,j) \eta'(-j)$$ where $\eta(x)$ ...
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### An integer sequence related to Pascal’s triangle

We need someone expert in binomial coefficients (subject 11B65) to recognize the integer sequence generated by an iterative formula we have encountered while working on a project about Pascal’s ...
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### About the exact origin of a binomial congruence

Given a prime $p$ and an integer $0 \leq k \leq p-1$, a famous congruence on binomial coefficients states: $$\binom{p-1}{k} \equiv (-1)^k \pmod{p}$$ It is generally taught as a consequence of Pascal’s ...
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### Are (55, 165, 495, 1485) and (286, 1716, 10296, 61776) the only geometric sequences of length 4 among non-trivial binomials?

Let's define non-trivial binomial coefficients as values of $\binom{n}{k}$, where $n$ and $k$ are positive integers such that $2 \le k \le \frac{n}{2}$. (Therefore, $6$ is the smallest non-trivial ...
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### Solve $\binom{n}{k}=m$ for $(n,k)$

For an integer $m>0$, put $X(m)=\{(n,k):4\leq 2k\leq n \text{ and } \binom{n}{k}=m\}$. Is there an efficient method to calculate $X(m)$? Is there a uniform upper bound for $|X(m)|$? By ...
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### Question in a paper by Erdős on divisibility properties of central binomial coefficient

In Erdős, Graham, Ruzsa, and Straus - On the prime factors of $\binom{2n}n$, at the beginning of the proof of theorem 1, they consider the case where $\log p$ and $\log q$ are commensurable numbers (...
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### 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,\...
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### Why does this combinatorial sum vanish?

We define the coefficients $c_{k,k-i}$ of ${n \choose k}$ by the following easy expansion: \begin{align*} & {n \choose k} = \frac{1}{k!} n(n-1) \dots (n-k+1) = \frac{1}{k!} \prod\limits_{t=...
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### sum of binomial coefficient approximation by geometric series

I follow a subject almost like this link: Sum of 'the first k' binomial coefficients for fixed $N$ $$f(N,k) = \sum^{k}_{i=0} \binom{N}{i} .$$ Michael Lugo suggest a way with geometric series ...
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### Sign-reversing involution for $q$-binomial coefficient at $q=-1$

Consider the q-binomial coefficient $\binom{n}{k}_q$. One combinatorial way to define it is as follows. Let $W_{n,k}$ be the set of binary words of length $n$ with $(n-k)$ 0's and $k$ 1's. An ...
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### Solving a Catalan-like recursion of polynomials, related to the KdV energies

I am working on a PDE problem. The goal is to connect the higher order energies of the Gross-Pitaevskii equation to those of the Korteweg-de-Vries equation. As these higher order energies are ...
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### 3D generalization of Gaussian q-binomial coefficient

It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions. Is there a closed ...
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### Integer solutions of system of inequalities

I am trying to solve a problem in combinatorics and I came up with the following system of inequalities: $0\leq x<y<z\leq n$ and $x+y<n$ and I am trying to count the number of integer ...
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### 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 ...
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### Maybe with the multinomial theorem

I'm looking for some idea to prove \begin{equation*} \sum_{\substack{ \left( s_{0},\ldots ,s_{r-1}\right) \in \left\{ \mathbb{N}% \cup \left\{ 0\right\} \right\} ^{r}, \\ \sum_{a=0}^{r-1}\left( a+1\...
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### Why is this alternating sum involving Catalan numbers $\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0$ for all $t$?

I need the result that for all $t$, $$\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0,$$ where $C_{t-i-1}$ is the $(t-i-1)$-th Catalan number. I've checked for $t$ up to ...
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### A congruence for a product of binomial coefficients?

For every prime $p\geq 5$ one seems to have the congruence $$(-1)^{(p-1)/2}\prod_{k=0}^{p-1}{p-1\choose k}\equiv 1-p+\frac{3}{2}p^2-\frac{7}{6}p^3\pmod{p^4}\ .$$ (I have checked all primes up to $5000$...
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### Greatest common divisors of some binomial coefficients

This is cross-posted from math.stackexchange. While making some computation, I stumbled upon a curious relation among some binomial coefficients. Consider the sequence of binomial coefficients $a(k,n)$...
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### (Conceptual) proof and/or interpretation of a $q$-binomial identity

There is a $q$-binomial identity that I encountered in one paper I am reading (https://arxiv.org/abs/1910.06193) which probably admits a very simple proof that I do not see: for two nonnegative ...
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### Series for $\frac{\log m}{\pi}$ with summands involving harmonic numbers

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}^\...
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### Binomial coefficient congruence modulo $p^n$

I am interested in the following congruence $$\binom{ap^n}{bp^n}\equiv \binom{a}{b}\pmod{p^n}$$ I am aware that by some reference in a book the above it should actually hold modulo $p^{3n}$; the ...
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### What is the exact value of the series $\sum_{k=0}^\infty \binom{2k}k^4/256^k$?
By Stirling's formula $n!\sim\sqrt{2\pi n}(n/e)^n$, we have $$\binom{2k}k\sim\frac{4^k}{\sqrt{k\pi}}$$ and hence $$\frac{\binom{2k}k^4}{256^k}\sim\frac1{k^2\pi^2}.\tag{1}$$ So the series \sum_{k=0}^\...