The integer-valued sequence $u_n=\binom{2n}{n}$ satisfies the recurrence relation $(n+1)u_{n+1}=2(2n+1)u_n$.
My question : for which values $a,b,c,d\in{\mathbb Z}$ is there a sequence $(u_n)$ (not eventually constant) satisfying the recurrence relation $(an+b)u_{n+1}=(cn+d)u_n$ ?
My thoughts : Legendre's formula shows that $\nu_p(n!) \leq \frac{n}{p-1}$ for large $n$ when $p$ is prime, so this shows that if $a=0$ and $c=1$, $b$ cannot have an odd prime divisor, so $b$ must be a power of $2$. Not sure how the general case can be handled.
If you have $$(a\,n+b)\,u_{n+1}=(c\,n+d)\,u_n$$ the general term is given in terms of Pochhammer symbols $$u_n=u_0\,\frac db\,\left(\frac{c}{a}\right)^{n-1}\, \frac{ \left(\frac{c+d}{c}\right)_{n-1}}{ \left(\frac{a+b}{a}\right)_{n-1}}$$ which, for large values of $n$, is $$u_n \sim u_0\,\frac db\,\,\frac{\Gamma \left(\frac{a+b}{a}\right)}{\Gamma \left(\frac{c+d}{c}\right)}\,\left(\frac{c}{a}\right)^{n-1}\,(n-1)^{\left(\frac{d}{c}-\frac{b}{a}\right)}$$ If you want a better approximation, muliply the above by $$\exp{\left(\frac{(a d-b c) (ac+ad +b c)}{2 a^2 c^2 (n-1)} \right) }$$