Let $n$ an integer sufficiently large.
I'm looking for a "big" set $S$ of prime number $p$ satisfying $10.n \geq p > \sqrt{18n} $ and $p| C_{9.n}^{k-n}C_{8n+k}^{8.n}$ for all integer $k$ satifying $ n \leq k \leq 10n$ and $k \geq p $
that means I'm looking a heavy condition $(C)$ over the $p$ prime $10n \geq p > \sqrt{18n} $ giving me for all $p$ prime satisfying $(C)$ we have for all integers $k$ such that $\max(n,p) \leq k \leq 10n$ one has
$$v_p(C_{9n}^{k-n}C_{8n+k}^{8n})=\Big[\frac{8n+k}{p}\Big]-\Big[\frac{k}{p}\Big]-\Big[\frac{8n}{p}\Big]+\Big[\frac{9n}{p}\Big]-\Big[\frac{k-n}{p}\Big]-\Big[\frac{10n-k}{p}\Big] \geq 1$$
where $[x]$ is the integer part of a real number $x$.
I think $(C) $ over the prime number satisfying $10n \geq p \geq \sqrt{18n}$ is:
$$p \in \displaystyle {\cup}_{h=1}^8 ]\displaystyle \frac{8n}{h},\displaystyle \frac{9n}{h}[ $$
I have proved that for all $p $ satifying $ 10.n \geq p > \sqrt{18n} $ and in $]8n,9n[,$ one has $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$. There are many others prime number satisfying $v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$, maybe in $\big(\frac{8n}{2},\frac{9n}{2}\big), \big(\frac{8n}{3},\frac{9n}{3}\big)...$.
Numerically, I have found many $p$ $10n \geq p > \sqrt{18n}$ satisfying $ \forall k $ such that $10n \geq k \geq p$ we have $ v_p(C_{9n}^{k-n}C_{8n+k}^{8n})\ge 1$ and $10n \geq p > \sqrt{18n}$.
Any help is appreciated.
