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

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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
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