Let $f$ be a function from the positive integers to the real numbers (or some ring...). Let $$(\star) \quad F(n) = \sum_{n_1 \leq \cdots \leq n_j\atop n_1 + \cdots + n_j = n} f(n_1) \cdots f(n_j), $$ for each positive integer $n$, where the sum runs over all the partitions of $n$, i.e., $n_1 \leq \cdots \leq n_j$ and $n_1 + \cdots + n_j = n$.

Computing $F(n)$ directly from $(\star)$ is computationally expensive since it require to sum $p(n) \sim e^{\pi\sqrt{2n/3}} / (4n\sqrt{3})$ addends, as $n \to \infty$, where $p(n)$ is the partition function.

My question is: Assume that we have computed $F(1), \dots, F(n-1)$, there is some (recursive) formula to find $F(n)$ in a way faster than $(\star)$?

I think the answer is affermative and that $F(n)$ can be expressed as a sum involving $F(k)$, $k < n$ and $f(h)$, $h \leq n$ with fewer addends than $(\star)$, but I am unable to figure out it. Thank you in advance for any suggestion.