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Fix natural numbers $n,m\in\mathbb{N}$. Given a partition $\lambda\vdash d$ with at most $n$ rows (and at most $m$ columns), we can define a partition $\lambda^\ast=(\lambda^\ast_1,\ldots,\lambda^\ast_n)$ by setting $\lambda^\ast_i:=m-\lambda_{n+1-i}$. Graphically, it is obtained by taking the complement of the Young diagram of $\lambda$ inside the $n\times m$ $-$ box $(m^n)$. Now, I'd be curious to learn about any combinatorial results involving this construction - but I don't know what it is called. If someone could tell me, that would be great - this way I'd not have such a hard time searching.

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A partition of $mn$ with at most $n$ rows and the most $m$ columns IS the $n\times m$ box, no? – Vladimir Dotsenko May 24 '12 at 7:27
How silly. That was supposed to be some $d$, not $nm$. I fixed it. – Jesko Hüttenhain May 24 '12 at 7:31
I've heard both "reverse partition" and "complementary partition" used to describe this. – Gjergji Zaimi May 24 '12 at 8:10
That looks good, thanks a bunch. – Jesko Hüttenhain May 24 '12 at 11:49
up vote 1 down vote accepted

The "right" answer was given by Gjergji Zaimi in his comment above: Both "reverse partition" and "complementary partition" are terms that seem to be in use.

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