# Recently proposed problem by George Andrews on partitions in Mathstudent Journal (India)

Show that the number of parts having odd multiplicities in all partitions of $n$ is equal to difference between the number of odd parts in all partitions of $n$ and the number of even parts in all partitions of $n.$

Example: $n=5.$ The number of partitions $5',\quad4'1',\quad3'2',\quad3'11,\quad221',\quad2'1'11,\quad1'1111$ have $10$ parts with odd multiplicities (marked with $'$). On the other hand, the number of odd parts is $15$, $$5,1,3,3,1,1,1,1,1,1,1,1,1,1,1$$ and the number of even parts is $5$, $$4,2,2,2,2.$$ I gave up solving the problem because in this problem parts are taken in all partitions not any single partition. I tried thinking some bijective proof or analytic proof using generating function but gave up as it seems to me a very different problem . Does anyone have any idea? The problem is very nice and it is recently been proposed by George Andrews in Mathstudent Journal (India).

• A bijection sounds right. Hint: look at conjugate partitions. – Peter Taylor Jan 27 '17 at 11:18
• not getting how conjugate partition will help here – Subhash Chand Bhoria Jan 27 '17 at 11:49
• @SubhashChandBhoria: Very nice problem. Thanks for posting it. It was great fun to work on it. – Markus Scheuer Apr 22 '17 at 13:29

## 1 Answer

This answer is based upon generating functions. We start by considering the generating function of all partitions of $n$ \begin{align*} \prod_{m=1}^\infty\frac{1}{1-z^m}=\prod_{m=1}^\infty\frac{1}{(1-z^{2m})(1-z^{2m-1})}\tag{1} \end{align*}

From the RHS of (1) we derive $A(z,t)$, a generating function of all partitions with marked odd parts and $B(z,t)$, a generating function with marked even parts. \begin{align*} A(z,t)&=\prod_{m=1}^\infty\frac{1}{(1-z^{2m})(1-tz^{2m-1})}\\ &=1+tz+(t^2+1)z^2+(t^3+2t)z^3+(t^4+2t^2+2)z^4\\ &\qquad+(t^5+2t^3+4t)z^5+\cdots\\ B(z,t)&=\prod_{m=1}^\infty\frac{1}{(1-tz^{2m})(1-z^{2m-1})}\\ &=1+z+(t+1)z^2+(t+2)z^3+(t^2+2t+2)z^4\\ &\qquad+(t^2+3t+3)z^5+\cdots \end{align*}

Similarly we obtain a generating function $C(z,t)$ with marked odd multiplicities by separating terms with odd and even index \begin{align*} C(z,t)&=\prod_{m=1}^{\infty}\left(\frac{1}{1-z^{2m}}+\frac{tz^m}{1-z^{2m}}\right)\\ &=\prod_{m=1}^{\infty}\frac{1+tz^m}{1-z^{2m}}\\ &=1+tz+(t+1)z^2+(t^2+2t)z^3+(t^2+2t+2)z^4\\ &\qquad+(3t^2+4t)z^5+\cdots \end{align*}

In terms of generating functions the equality of the difference between the number of odd and even partitions of a number $n$ and the number of odd multiplicities of the partitions of $n$ translates to the following

Claim: \begin{align*} \left.\frac{\partial}{\partial t}A(z,t)\right|_{t=1}-\left.\frac{\partial}{\partial t}B(z,t)\right|_{t=1} =\left.\frac{\partial}{\partial t}C(z,t)\right|_{t=1} \end{align*}

Before we show the claim we do a plausibility check and look at partitions of small $n=1,\ldots,5$.

\begin{array}{c|ccc|l} n &[z^n]\left.\frac{\partial}{\partial t}A(z,t)\right|_{t=1} &[z^n]\left.\frac{\partial}{\partial t}B(z,t)\right|_{t=1} &[z^n]\left.\frac{\partial}{\partial t}C(z,t)\right|_{t=1} &partitions\\ 1&1&0&1&1\\ 2&2&1&1&2,1^2\\ 3&5&1&4&3,21,1^3\\ 4&8&4&4&4,31,2^2,21^2,1^4\\ 5&15&5&10&5,41,32,31^2,2^21,21^3,1^5\\ \hline \end{array}

Here we see the coefficients of the power series for $n=1,\ldots,5$ and observe the difference of odd and even parts coincides with the number of odd multiplicities. :-)

The right-most column lists the partitions of $n$ with the compact notation e.g. $21^3$ is a short-hand for $2+1+1+1$.

We obtain using logarithmic differentiation \begin{align*} \left.\frac{\partial}{\partial t}A(z,t)\right|_{t=1} &=\left.\frac{\partial}{\partial t}\prod_{m=1}^\infty\frac{1}{(1-z^{2m})(1-tz^{2m-1})}\right|_{t=1}\\ &=\left.\prod_{m=1}^\infty\frac{1}{(1-z^{2m})(1-tz^{2m-1})}\sum_{k=1}^\infty\frac{z^{2k-1}}{1-z^{2k-1}}\right|_{t=1}\\ &=\prod_{m=1}^\infty\frac{1}{(1-z^{2m})(1-z^{2m-1})}\sum_{k=1}^\infty\frac{z^{2k-1}}{1-z^{2k-1}}\\ &=\prod_{m=1}^\infty\frac{1}{1-z^{m}}\color{blue}{\sum_{k=1}^\infty\frac{z^{2k-1}}{1-z^{2k-1}}}\\ \end{align*} and similarly \begin{align*} \left.\frac{\partial}{\partial t}B(z,t)\right|_{t=1} &=\left.\frac{\partial}{\partial t}\prod_{m=1}^\infty\frac{1}{(1-tz^{2m})(1-z^{2m-1})}\right|_{t=1}\\ &=\prod_{m=1}^\infty\frac{1}{1-z^{m}}\color{blue}{\sum_{k=1}^\infty\frac{z^{2k}}{1-z^{2k}}}\\ \end{align*} Next we get \begin{align*} \left.\frac{\partial}{\partial t}C(z,t)\right|_{t=1} &=\left.\frac{\partial}{\partial t}\prod_{m=1}^\infty\frac{1+tz^m}{1-z^{2m}}\right|_{t=1}\\ &=\left.\prod_{m=1}^\infty\frac{1+tz^m}{1-z^{2m}}\sum_{k=1}^\infty\frac{z^k}{1+tz^k}\right|_{t=1}\\ &=\prod_{m=1}^\infty\frac{1+z^m}{1-z^{2m}}\sum_{k=1}^\infty\frac{z^k}{1+z^k}\\ &=\prod_{m=1}^\infty\frac{1}{1-z^{m}}\color{blue}{\sum_{k=1}^\infty\frac{z^k}{1+z^k}} \end{align*}

We observe the claim is valid if the following holds: \begin{align*} \color{blue}{\sum_{k=1}^\infty\frac{z^{2k-1}}{1-z^{2k-1}} -\sum_{k=1}^\infty\frac{z^{2k}}{1-z^{2k}} =\sum_{k=1}^\infty\frac{z^k}{1+z^k}}\tag{2} \end{align*}

We start with the left-hand side of (2) and get \begin{align*} \color{blue}{\sum_{k=1}^\infty}&\color{blue}{\frac{z^{2k-1}}{1-z^{2k-1}} -\sum_{k=1}^\infty\frac{z^{2k}}{1-z^{2k}}}\\ &=\sum_{k=1}^\infty\left(\frac{z^{k}}{1-z^{k}}-\frac{z^{2k}}{1-z^{2k}}\right) -\sum_{k=1}^\infty\frac{z^{2k}}{1-z^{2k}}\\ &=\sum_{k=1}^\infty\left(\frac{z^{k}}{1-z^{k}}-\frac{2z^{2k}}{1-z^{2k}}\right)\\ &=\sum_{k=1}^\infty\frac{z^k}{1-z^k}\left(1-\frac{2z^k}{1+z^k}\right)\\ &=\sum_{k=1}^\infty\frac{z^k}{1-z^k}\cdot\frac{1-z^k}{1+z^k}\\ &\color{blue}{=\sum_{k=1}^\infty\frac{z^k}{1+z^k}} \end{align*} and the claim follows.

• very nice solution sir – Subhash Chand Bhoria May 27 '17 at 7:23
• @SubhashChandBhoria: Thanks, my pleasure! :-) – Markus Scheuer May 27 '17 at 7:33