整数序列 $n^2 b_n-(n-1)^2 b_{n-2}=\left(11 n^2-11 n+3\right) b_{n-1}$

hbghlyj

数列$\{b_n\},b_0=1,b_1=3,$
$$n^2 b_n-(n-1)^2 b_{n-2}=\left(11 n^2-11 n+3\right) b_{n-1}$$
求证$b_n\in\mathbb Z$


Source:A proof that Euler missed . . . Apéry’s proof of the irrationality of $ζ(3)$
似乎不包含证明

Czhang271828

显然 $b_n=\sum_{k=0}^n\binom n k^2\binom{n+k}k$.

Czhang271828

文中还有个式子. 若 $n^3b_n+(n-1)^3b_{n-1}=(34n^2-51n^3+27n-5)b_{n-1}$, $b_0=1, b_1=5$, 则通项 $b_n=\sum_{k=0}^n\binom n k^2\binom{n+k}k^2$.