Questions tagged [binomial-coefficients]
For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.
476 questions
3
votes
1
answer
226
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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
$$\liminf_{x\to+\infty}\frac{|\{n\le x:\ 2n+1=p+2^k\ \text{for some prime}\ p\ \text{and}\ k\in\mathbb Z^+\}|}x>0.$$
...
2
votes
1
answer
161
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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 $\...
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
$$ n \sum_{j = i+1}^{n-1} \frac{...
6
votes
2
answers
822
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identity involving products of binomial coefficients
I need the following identity to prove something (regarding functions of two variables):
$$\sum_{k=0}^{n-r} (-1)^k \binom{n-k}{k} \binom{n-2k}{n-k-r} = 1$$
Is this well-known? Is there a quick proof?
3
votes
1
answer
655
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Proving that a certain factorial double sum collapses to a double-factorial series
While solving a few sums, I came across the following double sum
$$\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}^...
11
votes
2
answers
739
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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:
$$
\sum_{k,\ell,m}
(-1)^k
\frac{(k+\ell)!(a+b-k-\ell)!^2(a+b-m)!}{k!(a-k)!\ell!(b-\ell)!m!(c-m)!(a+b-k-\ell-...
52
votes
2
answers
2k
views
Gap in binomial coefficients
Let $a,b$ be positive integers. Because binomial coefficients are integers, we know that $a!b!$ divides $(a+b)!$. For particular $a$ and $b$ there may be a gap $g$ with a tighter result, so $a!b!$ ...
42
votes
1
answer
2k
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A conjecture concerning odd binomial coefficients
Let $n$ be a positive integer with binary expansion $2^{a_1}+\cdots +
2^{a_s}$. For $S\subseteq [s]=\{1,2,\dots,s\}$, let $k_S= \sum_{i\in S}
2^{a_i}$. Thus by a fundamental result of Lucas, ${n\...
3
votes
2
answers
504
views
Fast converging series for $L(2,(\frac{d}{\cdot}))$
Dirichlet's $L$-function plays a central role in analytic number theory. For any integer $d\equiv0,1\pmod4$, let
$$L_d(2):=L\left(2,\left(\frac{d}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac dk)...
9
votes
1
answer
1k
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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
$$
L(u,a,b,n):=
(u+a+b-n)!\times\sum_{i,k,\ell}\
\frac{(-1)^k\ \ (u+a+b-i)!\ (k+\ell)!\ (a+b-k-\ell)!\ (u+a+b-k-\ell)!}...
16
votes
2
answers
966
views
An identity involving binomial coefficients
How to prove that
$$\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)!},$$
where $n$ and $i$ are integers such $1\le i\...
5
votes
1
answer
511
views
An analogue of the sum of binomial coefficients
Let $a$ and $p$ be positive integers, and consider the polynomial $$(1+x+\cdots+x^{p-1})^a = \sum_{i=0}^{a(p-1)} a_ix^i.$$ I'm looking for an asymptotic estimate of $\sum_{i=0}^{b} a_i$. If $p=2$, ...
2
votes
1
answer
381
views
Derive homogeneous recurrence from second order one
Let $a,b \in \mathbb{R}$ and sequece $\{f(n)\}_{n=1}^{\infty}$ is given by homogeneous second order recursive relation
$$
f(n):=af(n-1)-b^2f(n-2), \:\:\: n>2
$$
with two arbitrary starting values $...
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, ...
23
votes
3
answers
3k
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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 ...