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Jun 6, 2016 at 22:49 comment added Gerhard Paseman An even better upper bound would be $(1+n-k)!p_k(n)$, where we count the number of partitions of $n$ with $k$ pieces by $p_k(n)$. The lower bound has a multiplicative term close to $(n/k)!^k$, and I imagine the harmonic mean of the two terms is a good approximating multiplicative term for $p_k(n)$ for the desired sum. Gerhard "Or Maybe The Geometric Mean" Paseman, 2016.06.06.
Jun 6, 2016 at 14:03 history answered Geoff Robinson CC BY-SA 3.0