Define an integer sequence $$a_n=4 d_n^5 \sum_{r=0}^{n}\binom{n}{r}^4\binom{n+r}{r}^2 $$ where $d_n=lcm(1,2,3,...,n)$, $n\in\mathbb{N}$.
Prove that $a_n\zeta(5)\notin \mathbb{Z}$ for all $n\in\mathbb{N}$ where $\zeta(5)=\sum_{n=1}^{\infty}\frac{1}{n^5}$
Define $b_n=a_n\zeta(5)$ where $n\in\math...
Ahh if it's a research problem, then it makes sense why you're considering it. Maybe you can derive some results when trying to prove it. The reason why I thought that it's unsolvable is because it is very related to the question of proving the irrationality of $\zeta(5)$, and no one has done that yet.
@KamalSaleh So on dividing by $a_n$, we get $\zeta(5)=\frac{b_n}{a_n}$. So if $b_n\notin\mathbb{Z}$ then how do we get $\zeta(5)$ irrational? It might happen that $b_n\in\mathbb{Q}$ and then $\zeta(5)=\frac{b_n}{a_n}\in\mathbb{Q}$. Please correct me if I am wrong.
If $b_n\in\mathbb{Q}$ then $b_n=\frac{p_n}{q_n}$ for integers $p_n$ and $q_n$. Then $\zeta(5)=\frac{p_n}{q_na_n}$. Note however that $a_n$ and $b_n$ are specific sequences of rationals and so it isn't necessarily true that this problem is as hard as proving that $\zeta(5)$ is irrational.
I honestly have no idea because I'm not an expert in analytic number theory. I will say that, because this is a research problem, consider posting this on MathOverflow instead.
Discussion on question by Max: Prove …
Discussion on question by Max: Prove …