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I was surprised to learn of Durfee squares, which can be visually explained as the largest square contained within a partition's Ferrers diagram.

Moreover, partition identities have always amused me by how complicated they appear at first to prove. Out of curiosity, can Durfee squares be used to prove either of the following partition identities?

$$ \prod_{i\geq 1}(1-qx^i)^{-1}=\sum_{j\geq 0}\frac{x^{j^2}q^j}{(1-x)\cdots(1-x^j)(1-qx)\cdots(1-qx^j)} $$ or $$ \prod_{i\geq 1}(1+qx^{2i-1})=\sum_{j\geq 0}\frac{x^{j^2}q^j}{(1-x^2)(1-x^4)\cdots(1-x^{2j})}? $$

In fact, in Andrews and Eriksson's Integer Partitions, Exercise 101 on page 77 suggests that Durfee squares can be used to determine these formulas. How can this method be employed? Thanks.

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The idea is that the LHS of both identities count partitions based on number of rows (the coefficient of $q$) and number of squares (the coefficient of $x$) and the RHS counts those same partitions, but grouped together based on the size of their Durfee squares ($x^{j^2}$ refers to the number of squares in the Durfee square). When you remove a Durfee square from a partition, you get a pair of partitions, and... well, I'll let you fill in the details. –  Qiaochu Yuan Feb 9 '12 at 0:29
    
Thanks, I'll try to give it a shot. –  Buble Feb 9 '12 at 0:42

1 Answer 1

Andrews and Eriksson, Integer Partitions, has a chapter on Durfee squares, and in that chapter there are formulas that look a lot like the ones you are asking about, for example, $$\sum_{j=0}^{\infty}{q^{j^2}\over(1-q)^2(1-q^2)^2\cdots(1-q^j)^2}=\prod_{n=1}^{\infty}{1\over1-q^n}$$ and $$\sum_{j=0}^N\left[N\atop j\right]{z^jq^{j^2}\over(1-zq)(1-zq^2)\cdots(1-zq^j)}=\prod_{n=1}^N{1\over1-zq^n}$$ Perhaps what you want can be found in that chapter.

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Thank you Gerry. The second formula you write seems to be just the same as my first, with the $q$s and $x$s swapped. –  Buble Feb 7 '12 at 0:02

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