'If $a_{k,n}:=$ the number of ways of partitioning $n$ distinct objects into $k$ odd parts, what is $F_k(x)=\sum_{n\geq 0} a_{n,k}\frac{x^n}{n!}=?$'
If I understand correctly, $a_{k,n}$ is the $n$th coefficient of the generating function $\left(\frac{e^{x}-e^{-x}}{2}\right)^k=F_k(x)$. This seems very straightforward too me. However, I know it's wrong because this questions is followed up with:
'If $b_n:=$ the number of ways of partitioning $n$ distinct objects into odd parts, then show $\sum_{n\geq 0} b_n \frac{x^n}{n!}=e^{\sinh(x)}$.'
So my solution for $F(x)$ has to be wrong since $e^{\sinh(x)}\neq\frac{e^{x}-e^{-x}}{2}$. I'm not sure where I went wrong?
You are almost correct - you need to divide by $k!$. $(\frac{e^x - e^{-x}}{2})^k$ gives the number of ways of partitioning $n$ labeled objects into $k$ ordered parts of odd size, but you want unordered parts. Then $$\sum_n b_n \frac{x^n}{n!} = \sum_{k=0}^\infty F_k(x) = \cdots$$ Do you see how to finish?