Interpreting the Truncated Pentagonal Number Theorem using Partition Pairs

Louis W Kolitsch,Michael Burnette

doi:10.37236/4917

Abstract

In 2012 Andrews and Merca gave a new expansion for partial sums of Euler's pentagonal number series and expressed \[\sum_{j=0}^{k-1}(-1)^j(p(n-j(3j+1)/2)-p(n-j(3j+5)/2-1))=(-1)^{k-1}M_k(n)\] where $M_k(n)$ is the number of partitions of $n$ where $k$ is the least integer that does not occur as a part and there are more parts greater than $k$ than there are less than $k$. We will show that $M_k(n)=C_k(n)$ where $C_k(n)$ is the number of partition pairs $(S, U)$ where $S$ is a partition with parts greater than $k$, $U$ is a partition with $k-1$ distinct parts all of which are greater than the smallest part in $S$, and the sum of the parts in $S \cup U$ is $n$. We use partition pairs to determine what is counted by three similar expressions involving linear combinations of pentagonal numbers. Most of the results will be presented analytically and combinatorially.

Highlights

Euler’s pentagonal number theorem gives an easy recurrence for the number of partitions of n, denoted by p(n)
If we define Bk(n) for k 0 to be the number of partition pairs (S, T ) where S is a partition with parts greater than k, T is a partition with k distinct parts all of which are greater than the smallest part in S, and the sum of the parts in S ∪ T is n, the generating function for Bk(n) is given by
If we define Ck(n) for k > 0 to be the number of partition pairs (S, U ) where S is a partition with parts greater than k, U is a partition with k − 1 distinct parts all of which are greater than the smallest part in S, and the sum of the parts in S ∪ U is n we have the following theorem

Summary

Introduction

Euler’s pentagonal number theorem gives an easy recurrence for the number of partitions of n, denoted by p(n). The electronic journal of combinatorics 22(2) (2015), #P2.55 where p(k) = 0 if k < 0. An interesting question is to determine how far off from p(n) we are if we truncate this recurrence sum before we reach n − j(3j − 1)/2 < 0 or n − j(3j + 1)/2 < 0. In [1] Andrews and Merca answered this question when we stop the recurrence sum after an odd number of terms. In [3] Kolitsch gave an answer when we stop the recurrence sum with p(n − 1) + p(n − 2).

A Generating Function Proof

A Combinatorial Look at Our Results

Concluding Remarks