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Questions tagged [binomial-coefficients]

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

128 votes
17 answers
105k views

I am interested in the function $$f(N,k)=\sum_{i=0}^{k} {N \choose i}$$ for fixed $N$ and $0 \leq k \leq N $. Obviously it equals 1 for $k = 0$ and $2^{N}$ for $k = N$, but are there any other ...
mathy's user avatar
  • 1,358
23 votes
8 answers
14k views

Hi! I'm new here. It would be awesome if someone knows a good answer. Is there a good lower bound for the tail of sums of binomial coefficients? I'm particularly interested in the simplest case $\...
user13006's user avatar
  • 263
12 votes
4 answers
6k views

There is a well-known estimate for the sum of all binomial coefficients $\binom{n}{k}$ satisfying $k \leq \alpha n$ for some $\alpha$ satisfying $0 < \alpha \leq 1/2$: $$ \sum_{k=0}^{\alpha n}\...
bandini's user avatar
  • 491
48 votes
5 answers
6k views

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
23 votes
1 answer
7k views

Motivated by the central limit theorem, one expects that $$\binom{n}{k} \approx \frac{2^n}{\sqrt{\pi n/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right).$$ Computations suggest that the ratio of the two ...
Kevin O'Bryant's user avatar
56 votes
4 answers
5k views

Define the function $$S(N, n) = \sum_{k=0}^n \binom{N}{k}.$$ For what values of $N$ and $n$ does this function equal a power of 2? There are three classes of solutions: $n = 0$ or $n = N$, $N$ is odd ...
John D. Cook's user avatar
  • 5,317
43 votes
2 answers
7k views

Let $$ c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}. $$ It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum ...
Mark Wildon's user avatar
  • 11.9k
26 votes
3 answers
4k views

I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$. Some clues that might work (kindly provided by ...
Alexandra Seceleanu's user avatar
4 votes
2 answers
3k views

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
2 votes
0 answers
238 views

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 $$\...
Zhi-Wei Sun's user avatar
35 votes
3 answers
3k views

This is an extract from Apéry's biography (which some of the people have already enjoyed in this answer). During a mathematician's dinner in Kingston, Canada, in 1979, the conversation turned to ...
Wadim Zudilin's user avatar
18 votes
3 answers
909 views

Let $x$ be an indeterminate and $n$ a non-negative integer. Question. The following seems to be true. Is it? $$x\prod_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^{...
T. Amdeberhan's user avatar
10 votes
2 answers
1k views

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 ...
Deyi Chen's user avatar
  • 1,248
9 votes
1 answer
611 views

I am trying to determine the eigenvalues and eigenvectors of the following matrix: $$M_{ij} = 4^{-j}\binom{2j}{i}$$ where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k&...
valle's user avatar
  • 924
6 votes
1 answer
987 views

(previous title " Zero sum of binomials coefficients - a stronger version ") This is a stronger version of another question. Is there an $N\in \mathbb N$ and a sequence of non-constant functions $ \...
Shir's user avatar
  • 337

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