All Questions
96
questions
1
vote
1
answer
38
views
Summation of n-simplex numbers
Gauss proved that every positive integer is a sum of at most three triangular(2-simplex) numbers. I was thinking of an extension related to n-simplex.
Refer: https://upload.wikimedia.org/wikipedia/...
- 181k
3
votes
3
answers
704
views
Is it true that $\sum_{k=0}^m\binom{n-k}k$ outputs the $(n+1)$th Fibonacci number, where $m=\frac{n-1}2$ for odd $n$ and $m=\frac n2$ for even $n$?
Does $$\sum_{k=0}^m\binom{n-k}k=F_{n+1}$$ where $m=\left\{\begin{matrix}
\frac{n-1}{2}, \text{for odd} \,n\\
\frac n2, \text{for even} \,n
\end{matrix}\right.$ hold for all positive integers $n$?
...
- 17k
1
vote
2
answers
522
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Closed Formula Expression for Sum of Combinatorics
I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:
$n \choose 0 $+$ n \...
- 54.3k
5
votes
1
answer
205
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Conjecture: $\binom{n}{k } \mod m =0$ for all $k=1,2,3,\dots,n-1$ only when $m $ is a prime number and $n$ is a power of $m$
While playing with Pascal's triangle, I observed that $\binom{4}{k } \mod 2 =0$ for $k=1,2,3$,and $\binom{8}{k } \mod 2 =0$ for $k=1,2,3,4,5,6,7$ This made me curious about the values of $n>1$ and ...
- 6,565
1
vote
0
answers
323
views
A conjecture on representing $\sum\limits_{k=0} ^m (-1)^ka^{m-k}b^k$ as sum of powers of $(a+b)$.
UPATE: I asked this question on MO here.
I was solving problem 1.2.52 in "An introduction to the theory of numbers by by Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery"
Show that if ...
- 6,565
2
votes
1
answer
221
views
Generalization of binomial coefficients to both non-integer arguments
It is known that binomial coefficients can be generalized to the following: for $s\in\mathbb R$ and $k\in\mathbb N$,
\begin{equation*}
\binom{s}{k} := \prod_{i=0}^{k-1} \frac{s-i}{k-i} = \frac{s(s-1)...
- 40.1k
4
votes
0
answers
251
views
Sum of even binomial coefficients modulo $p$, without complex numbers
Let $p$ be a prime where $-1$ is not a quadratic residue, (no solutions to $m^2 = -1$ in $p$).
I want to find an easily computable expression for
$$\sum_{k=0}^n {n \choose 2k} (-x)^k$$
modulo $p$. ...
- 4,284
6
votes
3
answers
526
views
Proving a formula about binomial coefficients
I found the following formula in a book without any proof:
$$\sum_{j=k}^{\lfloor\frac n2\rfloor}{\binom{n}{2j}}{\binom{j}{k}}=\frac{n}{n-k}\cdot2^{n-2k-1}{\binom{n-k}{k}}$$
where $n$ is a natural ...
- 47.4k
11
votes
8
answers
17k
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Proof of binomial coefficient formula.
How can we prove that the number of ways choosing $k$ elements among $n$ is $\frac{n!}{k!(n-k)!} = \binom{n}{k}$ with $k\leq n$?
This is an accepted fact in every book but i couldn't find a ...
- 11.5k
1
vote
1
answer
213
views
Binomial Coefficient divisibility involving Multiples
Good afternoon. I'm looking for a proof, or a counterexample that, given $n,k,N\in\mathbb{Z}$, with $n>k>0$, $N\ge2$,
$$(N+1)|\binom{nN}{kN}$$
Just using the definition,
$$\binom{nN}{kN}=\frac{(...
- 1,254
0
votes
0
answers
100
views
Show that $2n\choose n$ divisible by primes $p,$ such that $n<p<2n$? [duplicate]
Suppose on the contrary that $2n \choose n$ is not divisible by $p\in (n,2n)$. There exis $k$ and $0\ne r\lt p$ such that
$(2n)\cdots (n+1)=kp \,n!+r \, n!$. The second term on RHS is not divisible by ...
- 54.3k
3
votes
3
answers
177
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Divisibility of Sum of Equally Spaced Binomial Coefficients
According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\...
-1
votes
1
answer
48
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Number of ones in the dyadic expansion of m [closed]
I was going through a paper where I stuck on a combinatorial argument as follows
I want help with the first assertion i.e proving the inequality $\alpha(m+l)\le\alpha(m)$.
As the author suggests it is ...
- 14.7k
1
vote
1
answer
83
views
prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors
prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors,so that all these factors are elements of $[k]$. Repetition of factors is allowed,
that is my attempt :
for ...
- 47.4k
0
votes
1
answer
61
views
$\sum\limits_{k=1}^{n-1}\binom{n-1}{k-1}\frac{a^{n-k}b^k}{k}=\frac{(a+b)^n-a^n-b^n}{n}$
In "New properties for a composition of some generating functions for primes", properties of generating functions are used to form a primality test, speaking specifically in example 1 the ...
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