Notation
- Let $[n]$ denote the set of integers from 1 up to $n$, inclusive.
- Let $A_n$ denote the average number of blocks over all partitions of $[n]$.
- Let $B_n$ denote the number of partitions of $[n]$ (the $n$th Bell number).
Question
For any given $n \geq 1$, how small can $m$ be (as a function of $n$) and still ensure $ A_{n-m} + 1 < A_n$ (or, equivalently, $ \frac{B_{n+1-m}}{B_{n-m}} < \frac{B_{n+1}}{B_n}-1)? $
Remarks
The lefthand side of the inequality comes from fixing a particular block of size $m$ and considering all partitions of $[n]$ containing that fixed block. The idea is that if a partition contains a single large block, the expected number of blocks must fall. For my purposes, smaller values of $m$ are superior, but I would be interested to see anything.
The following reasoning led me to believe $m = \lceil n / A_n \rceil$ would be big enough. The average size of a block in a partition of $[n]$ is at most $\lceil n / A_n \rceil$. Having a single larger-than-average block should reduce the average number of blocks appearing in such a partition. Exploring with Maple shows $\lceil n / A_n \rceil$ is not quite big enough to establish the inequality, but $\lceil n / A_n \rceil + 1$ works for $n \leq 500$.
I keep mentioning averages because I'm hoping there is a quick proof of the kind I attempted in the preceding paragraph. If that vantage point is not helpful, there is the alternative statement of the inequality in terms of Bell numbers. This expression can possibly be analyzed asymptotically (though the Moser-Wyman expansion fails here), but I was rather hoping to avoid it. Still, any suggestions in this direction are welcome.