# Tagged Questions

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### Binomial triplets

Solutions to the equation $$\dbinom{a}{n}+\dbinom{b}{n}=\dbinom{c}{n}$$ I will refer to as 'Binomial triplets of order $n$'. These triplets describe simplicial $n$-polytopic numbers that can be ...
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### Binomial Congruence

How can we show that $\dbinom{pm}{pn}\equiv\dbinom{m}{n}\pmod {p^3}$ for positive integers m and n and p a prime greater than 5? I can do it for mod p^2 but Im stuck here.
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### For what $n$ is it true that $(1+\sum_{k=0}^{\infty}x^{2^k})^n+(\sum_{k=0}^{\infty}x^{2^k})^n\equiv1\mod2$

Let $A:=\sum_{k=0}^{\infty}x^{2^k}$. For what $n$ is it true that $(A+1)^n+A^n\equiv1\mod2$ (here we are basically working in $\mathbb{F}_2$.) The answer is all powers of 2, and it's fairly simple ...
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### Given $p, m$, how many $r, k$ exist such that $\sum_{i=0}^k{m+i \choose p} = {m+r \choose p}$?

I know that ${m+1 \choose p+1} = {m \choose p} + {m \choose p+1}$, does this identity extend further out? My guess is that there exist certain $k$ such that there exists $r > k$ where the title ...
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### Bertrand's postulate proof

Regarding http://michaelnielsen.org/polymath1/index.php?title=Bertrand%27s_postulate I think the last inequality should be $4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}}$. But even when the RHS is decreased from ...
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### how compute $\gcd\Bigl(\binom{n}{1},\binom{n}{2},\binom{n}{3},…,\binom{n}{n-1}\Bigr)?$

how to prove $$\gcd\Biggl(\binom{n}{1},\binom{n}{2},\binom{n}{3},...,\binom{n}{n-1}\Biggr)=\begin{cases} p, & \text{n=p^m ;p is prime} \\ 1, & \text{o.w} \\ \end{cases}$$ thanks in ...
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### Stirling's Approximation

A sharp Stirling's approximation form states that $$n! \sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}.$$ Use that form to show that: $$\binom{2m}{m} = \Theta\left(\frac{2^{2m}}{\sqrt{m}}\right).$$
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### Feature of the Pascal's triangle (OEIS A007318)?

If rows of Pascal's triangle (OEIS's A007318) after their content concatenation {1-1, 1-2-1, 1-3-3-1, 1-4-6-4-1, 1-5-10-10-5-1, 1-6-15-20-15-6-1, 1-7-21-35-35-21-7-1 and so on } be considered as ...
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### prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared. My thought process: The numerator is the product of the first n even ...