Konigsberg wrote: Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that (a) $f(1)=1$ (b) $f(n+2)+(n^2+4n+3)f(n)=(2n+5)f(n+1)$ for all $n \in \mathbb{N}$. (c) $f(n)$ divides $f(m)$ if $m>n$. $f(3)=7f(2)-8$ and so $f(2)>1$ and $f(2)|8$ and so $f(2)\in\{2,4,8\}$ If $f(2)=2$, simple induction implies $\boxed{f(n)=n!}$ which indeed is a solution If $f(2)=4$, simple induction implies $\boxed{f(n)=\frac{(n+2)!}6}$ which indeed is a solution If $f(2)=8$, we get $f(3)=48$ and $f(4)=312$ which is not a solution since $f(3)\not|f(4)$