The large-parts formula for p(n)

TL;DR

This paper introduces a novel formula for the partition function p(n), expressing it in terms of the two largest parts of all partitions, providing a new combinatorial perspective on partition enumeration.

Contribution

The paper presents a new formula for p(n) based on the sum over all partitions involving their two largest parts, offering a fresh approach to partition function calculations.

Findings

The sum of a specific function over all partitions equals 2p(n) - 1.

The formula relates the partition function to the largest parts of partitions.

Provides a combinatorial interpretation of p(n) in terms of partition parts.

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 $2p(n) - 1$.