The large-parts formula for p(n)

Abstract

A new formula for the partition function $p(n)$ is developed. We show that the number of partitions of $n$ can be expressed as the sum of a simple function of the two largest parts of all partitions. Specifically, if $a_1 + >... + a_k = n$ is a partition of $n$ with $a_1 \leq ... \leq a_k$ and $a_0 = 0$, then the sum of $\lfloor(a_k + a_{k-1}) / (a_{k-1} + 1)\rfloor$ over all partitions of $n$ is equal to $2p(n) - 1$.Comment: Four pages; 1 figur