Let $\mathcal{O}(n)$ and $\mathcal{D}(n)$ denote the set of all integer partitions of $n$ into odd parts and distinct parts, respectively. Let $o(n)=\#\mathcal{O}(n)$ and $d(n)=\#\mathcal{D}(n)$. Euler established that $o(n)=d(n)$.


Introduce the enumerations: $a(n)=$ number of parts in $\mathcal{O}(n)$ and $b(n)=$ number of parts in $\mathcal{D}(n)$. Then, George Beck conjectured that
$a(n)-b(n)=\alpha(n)$
where $\alpha(n)$ counts all partitions of $n$ with odd parts, except one even part (which may be repeated as many times). Soon enough, George Andrews proved this claim and added that $a(n)-b(n)=\beta(n)$ where $\beta(n)$ counts all partitions of $n$ with distinct parts, except one part repeated at least twice. I learned of these from Cristina Ballantine, who gave a seminar this week offering combinatorial proofs as well as further generalizations.


This prompted me to consider a further refinement. So, consider $\alpha\_u(n)$ and $\beta\_u(n)$ as enumerating all partitions of $n$ having odd parts with **one even part appearing $u$-times** and all partitions of $n$ having distinct parts with **the number $u$ being the repeating part**, respectively.


I would now ask:



> 
> **QUESTION 1.** Is it true that $\alpha\_u(n)=\beta\_u(n)$ for $u=1,2,3\dots$? **Note.** $u$ is bounded by $n$.
> 
> 
> 



> 
> **QUESTION 2.** Is this true too? We have the bivariate generating function
> $$\sum\_{n,u=0}^{\infty}\alpha\_u(n)z^uq^n=\sum\_{j=1}^{\infty}\frac{zq^{2j}}{1-zq^{2j}}\cdot\prod\_{k=1}^{\infty}\frac1{1-q^{2k-1}}
> =\sum\_{n,u=0}^{\infty}\beta\_u(n)z^uq^n.$$
> 
> 
> 



A refinment of Beck's conjecture
Given a partition $\lambda$ with all parts odd except for one even
part $2k$ appearing $u$ times, let $\mu$ be $\lambda$ with the parts
equal to $u$ removed. Apply your favorite bijection to transform $\mu$
into a partition $\nu$ with distinct parts. Then adjoin to $\nu$ $2k$
parts equal to $u$, obtaining a partition $\rho$ with distinct parts
except for the repeating part $u$. (Note that we may have $2k$ or $2k+1$ parts equal to $u$ in $\rho$, depending on whether $\nu$ already had a part equal to $u$.) The map $\lambda\mapsto \rho$ is a bijection, proving Question 1. It is straightforward to verify Question 2.


