Questions tagged [binomial-coefficients]
For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.
476 questions
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answer
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On sums of a prime and a central binomial coefficient
Let $\mathbb Z^+$ be the set of positive integers. In 1934, Romanoff proved that
$$\liminf_{x\to+\infty}\frac{|\{n\le x:\ 2n+1=p+2^k\ \text{for some prime}\ p\ \text{and}\ k\in\mathbb Z^+\}|}x>0.$$
...
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2
votes
1
answer
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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
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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
$$\sum_{k=0}^...
14
votes
6
answers
2k
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Another binomial identity
Via two calculations of the same quantity within a probability model,
Svante Janson and I observed a very indirect proof that for all $0 \le i \le n-2$, the identity
$$ n \sum_{j = i+1}^{n-1} \frac{...
- 486
6
votes
2
answers
822
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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
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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:
$$
\sum_{k,\ell,m}
(-1)^k
\frac{(k+\ell)!(a+b-k-\ell)!^2(a+b-m)!}{k!(a-k)!\ell!(b-\ell)!m!(c-m)!(a+b-k-\ell-...
- 23.7k
2
votes
1
answer
381
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Derive homogeneous recurrence from second order one
Let $a,b \in \mathbb{R}$ and sequece $\{f(n)\}_{n=1}^{\infty}$ is given by homogeneous second order recursive relation
$$
f(n):=af(n-1)-b^2f(n-2), \:\:\: n>2
$$
with two arbitrary starting values $...
0
votes
1
answer
106
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Closed forms or special function representations for certain binomial sums involving harmonic-like terms?
I’m studying the following family of polynomials defined for integers $n \geq 1$:
\begin{aligned}
A_n(x) &= \frac{x^n}{(n-1)!} \left[
\sum_{k=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor} \binom{n-...
9
votes
1
answer
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Nonnegativity of an alternating combinatorial sum
Let $u,a,b,n$ be nonnegative integers such that $n\le a+b$.
Define the quantity
$$
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
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Exact form of eigenvalues of pentadiagonal Toeplitz matrices
The tridiagonal Toeplitz matrices
$$\begin{pmatrix}
a & b & & \\
c & \ddots & \ddots \\
& \ddots & \ddots & b \\
& & c ...
3
votes
2
answers
504
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Fast converging series for $L(2,(\frac{d}{\cdot}))$
Dirichlet's $L$-function plays a central role in analytic number theory. For any integer $d\equiv0,1\pmod4$, let
$$L_d(2):=L\left(2,\left(\frac{d}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac dk)...
- 18k
11
votes
0
answers
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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 ...
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42
votes
1
answer
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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
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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 $a!b!$ ...
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