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The following expression arose in a binomial coefficient summation :
$$f(n,k) = \frac{(2n + 1)!}{n!k!(n-k)!(2k + 1)} $$ where $k$ ranges from $0$ to $n$. This fraction is clearly an integer for $k = 0$ and $k = n$. It can also be proved to be integral in certain other cases including when $2k + 1$ is prime .

Question: Is $f(n,k)$ an integer for all values of $k$?

Thanks

Marcus M
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user2052
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  • You could take out a factor (2n+1)/(2k+1) and write the remaining as a trinomial coefficient. Then the only question is whether there is a superfluous factor 2k+1 either in 2n+1 or in the trinomial. – TMM Jul 01 '17 at 22:43
  • If I write $(2n + 1)! = (2n + 1)\frac{(2n)!}{n!}n!$ then $k!(n-k)! \mid n!$ and $n! \mid \frac{(2n)!}{n!}$. So if $(2k + 1) \mid (2n + 1)$ we would be golden. Otherwise it needs to divide some of the other factors. – Sera Gunn Jul 01 '17 at 23:04
  • Better yet, it seems that $g\left(n,k\right) := \dfrac{\left(2n+1\right)!}{n!k!\left(n-k\right)!\dbinom{2k+1}{k}}$ is an integer. This will yield the integrality of $f\left(n,k\right)$, since $f\left(n,k\right) / g\left(n,k\right) = \dfrac{1}{2k+1} \dbinom{2k+1}{k}$ is an integer (see https://math.stackexchange.com/questions/486179/arithmetical-proof-of-cfrac1ab-binomaba-is-an-integer-when-a-b-1?rq=1 ). – darij grinberg Jul 01 '17 at 23:52
  • Can we prove by induction. If f(n,k) is integer is f(n, k+1)? – fleablood Jul 02 '17 at 00:07
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    Better yet, it appears to me that $r\left(n,k,p\right) := \dfrac{\left(2n+p\right)! \left(k+p\right)!}{n! \left(n-k\right)! \left(2k+p\right)! p!}$ is an integer for all $0 \leq k \leq n$ and $0 \leq p$. Setting $p = 1$ would yield $g\left(n, k\right)$. (Checked using Sage for all $n \leq 50$ and $p \leq 50$.) – darij grinberg Jul 02 '17 at 00:16

1 Answers1

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If we consider darij's comment, it is enough to prove that $$ r(a, b, c) = \frac{(2a+2b+c)!(a+c)!}{(a+b)!b!(2a+c)!c!} $$ is an integer for all $a, b, c\geq 0$. (I just slightly changed the definition for convenience.) To prove this, we will prove the following lemma : $$ \lfloor 2x+2y+z\rfloor+\lfloor x+z\rfloor\geq \lfloor x+y\rfloor +\lfloor y\rfloor + \lfloor 2x+z \rfloor+\lfloor z \rfloor $$ for all $x, y, z\in \mathbb{R}$. If we can show this, our previous claim follows from the Lagrange's identity $$ \mathrm{ord}_{p}(n!)=\sum_{k=1}^{\infty}\bigg\lfloor\frac{n}{p^{k}}\bigg\rfloor. $$ By the way, since the equation in the lemma remains same if we change $x$ to $x+n$ for any $n\in \mathbb{Z}$ (and same for $y$ and $z$), we only need to check for $0\leq x, y, z<1$, which is an easy exercise.

Seewoo Lee
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