Questions tagged [binomial-coefficients]
For questions that explicitly reference the binomial coefficients, Pascal's Triangle, and Binomial identities.
76 questions
121
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15
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Sum of 'the first k' binomial coefficients for fixed $N$
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 ...
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votes
8
answers
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Lower bound for sum of binomial coefficients?
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 $\...
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votes
4
answers
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Estimating a partial sum of weighted binomial coefficients
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 $0 < \alpha \leq 1/2$:
$$ \sum_{k=0}^{\alpha n}\...
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46
votes
5
answers
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Integer-valued factorial ratios
This historical question recalls
Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation
Chebyshev used the factorial ratio sequence
$$
u_n=\frac{(30n)!n!}{(15n)!(10n)!(6n)...
- 13.5k
23
votes
1
answer
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Are there good bounds on binomial coefficients?
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 ...
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answer
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What is known about sums of the form $\sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $?
A fair bit is known about rational zeta series. This includes identities like $$ \sum_{n=2}^{\infty} [\zeta(n) -1] = 1 . $$
Many more identities can be found in articles by e.g. Borwein and Adamchik &...
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votes
4
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When do binomial coefficients sum to a power of 2?
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 ...
- 5,227
43
votes
2
answers
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Alternating sum of square roots of binomial coefficients
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 ...
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26
votes
3
answers
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Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$
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 ...
4
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2
answers
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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,
$a_\...
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votes
3
answers
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A binomial generalization of the FLT: Bombieri's Napkin Problem
This is an extract from Apéry's biography
(which some of the people have already enjoyed in
this answer).
During a mathematician's dinner in
Kingston, Canada, in 1979, the
conversation turned ...
- 13.5k
18
votes
3
answers
860
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$\prod_k(x\pm k)$ in binomial basis?
Let $x$ be an indeterminate and $n$ a non-negative integer.
Question. The following seems to be true. Is it?
$$x\prod_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^...
- 43.2k
9
votes
1
answer
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Eigenvalues of a matrix with binomial entries
I am trying to determine the eigenvalues and eigenvectors of the following matrix:
$$M_{ij} = 4^{-j}\binom{2j}{i}$$
where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k&...
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votes
1
answer
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What is the degree of a symmetric boolean function?
(previous title " Zero sum of binomials coefficients - a stronger version ")
This is a stronger version of another question.
Is there an $N\in \mathbb N$ and a sequence of non-constant functions $ \...
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votes
3
answers
6k
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A limit involving binomial coefficients: $\lim_{n\to\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac12$?
Experimentation suggests the limit
$$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac{1}{2}\ .$$
Does somebody have an idea for (a start of) a proof?
Added: There seem to ...
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votes
3
answers
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Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$
In my research, I found this identity and as I experienced, it's surely right. But I can't give a proof for it.
Could someone help me?
This is the identity:
let $a$ and $b$ be two positive integers; ...
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votes
2
answers
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Some binomial coefficient determinants
It is well known that for $n>0$
$$d(n)=\det\left(\binom{2i+2j+1}{i+j}\right)_{i,j=0}^{n-1}=1.$$
Computer experiments suggest that more generally
$$d(n,k)=\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,...
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1
answer
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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 ...
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votes
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answer
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We have $\binom{62}{26}^2+\binom{62}{27}^2=\binom{62}{28}^2$. How many other Pythagorean triples are contained in a single row of Pascal's triangle?
At MSE I asked, "Does any row of Pascal's triangle contain a Pythagorean triple?" The answer is yes; the example $\binom{62}{26}^2+\binom{62}{27}^2=\binom{62}{28}^2$ was given. In that ...
- 3,527
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votes
2
answers
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Real rootedness of a polynomial
Let's consider $m$ and $n$ arbitrary positive integers, with $m\leq n$, and the polynomial given by:
$$ P_{m,n}(t) := \sum_{j=0}^m \binom{m}{j}\binom{n}{j} t^j$$
I've found with Sage that for every $...
- 1,889
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votes
7
answers
6k
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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 $. ...
- 351
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votes
1
answer
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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!
$$\sum_{m=s}^{2s}\sum_{k=0}^{...
- 2,183
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votes
3
answers
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a sum with binomial coefficients
Let integers $n,k$ satisfy $0 \le k \le n$. We desire proof that
$$
{n\choose k} =
\sum {n\choose a}(-1)^a\;{-k\choose b}(-1)^b\;{-(n-k)\choose c}(-1)^c
\tag{$*$}$$
where the (finite) sum is over all ...
- 41.1k
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votes
2
answers
576
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Eigenvalues and eigenvectors of the matrix with entries $\dbinom{n+1}{2j-i}$ for $i, j = 1, 2, \ldots, n$
Let $n$ be a nonnegative integer, and let $B$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, ...
- 33.8k
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votes
2
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Binomial supercongruences: is there any reason for them?
One of the recent questions, in fact
the answer
to it, reminded me about the binomial sequence
$$
a_n=\sum_{k=0}^n{\binom{n}{k}}^2{\binom{n+k}{k}}^2,
\qquad n=0,1,2,\dots,
$$
of the Apéry ...
- 13.5k
16
votes
2
answers
618
views
3-adic valuation of a sum involving binomial coefficients
Let $$a(n) = \sum_{0 \leq k \leq n} {n \choose k}{{n+k} \choose k},$$ and define
$b(n) = \nu_3 \bigl(a(n)\bigr)$, where $\nu_3$ is the $3$-adic valuation. About twenty years ago or so, I discovered (...
- 3,028
14
votes
2
answers
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Integral of power of binomials equal to sum of power of binomials?
Inspired by this MO question about integrating binomial coefficients and the answers, I was wondering whether integrating powers of binomial coefficients also relates to the respective sums. And ...
- 4,014
13
votes
5
answers
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Asymptotics of a Bernoulli-number-like function
Tony Lezard asked me the following question which seemed like it should not be too hard but which I did not immediately see how to answer. Define $f(n,k)$ recursively by $f(1,k) = 1$ and
$$f(n,k) = \...
- 82.7k
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votes
2
answers
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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 ...
- 4,757
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votes
5
answers
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Looking for a combinatorial proof for a Catalan identity
Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers.
QUESTION. Is there a combinatorial or conceptual justification for this identity?
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^...
- 43.2k
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votes
3
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Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers?
The Gauss-Legendre theorem on sums of three squares states that
$$\{x^2+y^2+z^2:\ x,y,z\in\mathbb Z\}=\mathbb N\setminus\{4^k(8m+7):\ k,m\in\mathbb N\},$$ where $\mathbb N=\{0,1,2,\ldots\}$.
It is ...
- 15.6k
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answer
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Total positivity of $q$-Pascal matrix?
A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$.
Is it true that the minors of the $q$-Pascal matrix ...
- 5,622
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2
answers
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An interesting identity: in search of a proof -Part I
I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS.
Question. Can you show that
$$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-...
- 43.2k
12
votes
5
answers
835
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A divisibility of q-binomial coefficients combinatorially
Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
- 8,866
11
votes
1
answer
681
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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 ...
- 56.9k
10
votes
3
answers
490
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Equality with binomials
I am reading a paper and I am trying to understand an equality which is given without proof:
$$\sum_{k=1}^s\binom{2s-k}{s}\frac{k}{2s-k}v^k(v-1)^{s-k}=v\sum_{k=0}^{s-1}\binom{2s}{k}\frac{s-k}{s}(v-1)^{...
- 333
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4
answers
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Binomial coefficient in Andrews' partition book
First of all, I think MathOverflow is a very great community to discuss math, either basic or advanced, and I'm glad to participate here. It's my first post, so I'm sorry if i did anything wrong, and ...
- 103
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votes
1
answer
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Series for $\frac{\log m}{\pi}$ with summands involving harmonic numbers
The classical rational Ramanujan-type series for $1/\pi$ have the following four forms:
\begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1}
\\\sum_{k=0}^\...
- 15.6k
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3
answers
1k
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Zero sum of binomial coefficients
Is there a function $p:\mathbb N\to \{ 1,-1 \} $ and a fixed $N\in \mathbb N$ such that for every $n \geq N$ we get:
$\sum _{i=0} ^{n} p(i)\binom {n}{i}=0$
?
Obviously $p(i)=(-1)^i$ works for $N=1$...
- 337
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votes
2
answers
791
views
Asymptotic difference between a function and its "binomial average"
(I posted this question on Math.SE a few weeks ago. I got a few comments, but nothing definite, and so I thought I would try MO.)
The origin of this question is the identity
$$\sum_{k=0}^n \binom{n}{...
- 3,283
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votes
1
answer
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Bounding the probability that two binomials are equal
Note: This question was migrated from this earlier post, where it initially appeared. Following suggestions, I moved this into its own question.
Let $B_{n,p}$ denote the usual binomial random ...
- 2,720
8
votes
2
answers
1k
views
Interpolating a sum of binomial coefficients using a sin function
While studying a problem about orthogonal polynomials I encountered the following
expressions
\begin{equation}
f(n)=\sum_{k=0}^{n}(-1)^k\binom{n+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\...
7
votes
0
answers
332
views
Do generalizations of the identity $\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) = 1 $ exist?
On p. 263 of Borwein's paper entitled “Computational Strategies for the Riemann zeta function”, the following identity is stated: $$\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) =1 . \qquad \...
- 4,829
7
votes
9
answers
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Binomial coefficient identity
It seems to be nontrivial (to me) to show that the following identity holds:
$$ \binom {m+n}{n} \sum_{k=0}^m \binom {m}{k} \frac {n(-1)^k}{n+k} = 1. $$
This quantity is related to the volume of the ...
- 245
6
votes
3
answers
745
views
Is there a closed formula for the generating function of some trinomial coefficients?
We learn in calculus how to obtain a sum of binomial coefficients $\frac{(2d)!}{(d!)^2}$ in terms of a generating function
$\sum_{d \geq 0} \frac{(2d)!}{(d!)^2} x^d$
by the Taylor series of $(1-4x)^...
- 61
6
votes
0
answers
214
views
Looking for a combinatorial proof for an identity involving $q$-Catalan triangles
Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
- 43.2k
5
votes
1
answer
685
views
Summing ratio of ratio of partial sums of binomial coefficients
I would like to approximate the following when $n \gg k$.
$\sum_{y = k + 1}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m} (y - 1)}{\sum_{m = 0}^k {y - 1 \choose m}}.$
The formula can be re-written ...
- 83
5
votes
2
answers
379
views
Asymptotic rate for $\sum\binom{n}k^{-1}$
This MO question prompted me to ask:
What is the second order asymptotic growth/decay rate for the sum
$$\sum_{k=0}^n\frac1{\binom{n}k}$$
as $n\rightarrow\infty$?
- 43.2k
5
votes
2
answers
566
views
Is there a simple proof of the following binomial Identity (part 2)?
This is a related question to the one I posted on MO earlier:
Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$?
It arose in the ...
- 2,337
5
votes
4
answers
917
views
Limit of a sum with binomial coefficients
Let $$A_k = \frac{\sum_{i=1}^ki{2k-i-1 \choose i-1}{i-1 \choose k-i}}{k{2k-1\choose k}}$$
$$B_k = \frac{\sum_{i=1}^ki{2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$
$$C_k = \frac{\sum_{i=1}^k(...
- 155