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46 votes
5 answers
6k views

Integer-valued factorial ratios

This historical question recalls Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation Chebyshev used the factorial ratio sequence $$ u_n=\frac{(30n)!n!}{(15n)!(10n)!(6n)...
Wadim Zudilin's user avatar
4 votes
2 answers
2k views

Estimate on sum of squares of multinomial coefficients

I am interested in approximating the sum of the squares of the multinomial coefficients, i.e. $a_\ell^p := \sum_{k_0+\ldots+k_p = \ell} (\frac{\ell!}{k_0! \ldots k_p!})^2$ or more general, $a_\...
Liss's user avatar
  • 145
27 votes
1 answer
2k views

Solutions to $\binom{n}{5} = 2 \binom{m}{5}$

In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says: On National Public Radio, the Weekend Edition program posed the following probability problem: Given a certain number of ...
Nick Matteo's user avatar
21 votes
7 answers
6k views

Upper limit on the central binomial coefficient

What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ? I just tried to proceed a bit, like this: $$ n! > n^{\frac{n}{2}} $$ for all $ n>2 $. ...
RandomStudent's user avatar
20 votes
1 answer
2k views

How to prove that the following double sum is always an integer?

I have verified the following double sum is always an integer for $s$ up to $1000$ via Maple. But I can not prove it. Proofs, hints, or references are all welcome. Thanks! $$\sum_{m=s}^{2s}\sum_{k=0}^{...
Chitsai Liu's user avatar
  • 2,183
16 votes
2 answers
618 views

3-adic valuation of a sum involving binomial coefficients

Let $$a(n) = \sum_{0 \leq k \leq n} {n \choose k}{{n+k} \choose k},$$ and define $b(n) = \nu_3 \bigl(a(n)\bigr)$, where $\nu_3$ is the $3$-adic valuation. About twenty years ago or so, I discovered (...
Jeffrey Shallit's user avatar
13 votes
2 answers
2k views

Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$

I found the following formula in a book without any proof: $$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$ This does not seem to follow immediately from the basic ...
mathlove's user avatar
  • 4,757
11 votes
1 answer
681 views

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 ...
Neil Strickland's user avatar
10 votes
4 answers
1k views

Binomial coefficient in Andrews' partition book

First of all, I think MathOverflow is a very great community to discuss math, either basic or advanced, and I'm glad to participate here. It's my first post, so I'm sorry if i did anything wrong, and ...
Guilherme's user avatar
  • 103
3 votes
0 answers
151 views

Combinatorial interpretation of inverse modulo $2$ binomial transform of A284005

My question is related to the following: Sum with products turned into subsequences We have an identity $$a(n, -1) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}(-1)^{\operatorname{wt}(n)-\...
Notamathematician's user avatar
3 votes
0 answers
144 views

Flat polynomials with factors of big height

Let $p(x)$ be a polynomial of degree $n$ with all coefficients in $\{-1,0,1\}$ (such polynomials are sometimes called flat). I am wondering how big the coefficients of a factor of $p$ can be. Call ...
Wolfgang's user avatar
  • 13.4k
1 vote
1 answer
3k views

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}, \ m=1,2,....$

Consider a sum $$\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j$$ which returns an odd power $n^{2m+1}$ of $n$, for $\ m=0,1,2,...$ given fixed $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$. The ...
Petro Kolosov's user avatar