# Definition

Call a partition $$\lambda$$ of an even integer $$2n$$ "black-white balanced" if the following equivalent conditions are satisfied:

• In the usual (Ferrers-)Young diagram of $$\lambda$$, if you color alternating squares black and white in a checkerboard pattern, there are an equal number of black and white squares.
• The Young diagram of $$\lambda$$ can be covered by dominoes.
• The partition $$\lambda$$ has empty 2-core.
• If you sort the parts of $$\lambda$$ (as is usual), then the number of odd parts in even places is equal to the number of odd parts in odd places. (A part $$\lambda_i$$ is odd if the number $$\lambda_i$$ itself is odd; the place is odd or even based on the parity of the index $$i$$.)
• If you sort the parts of $$\lambda$$, then $$\sum_i (-1)^i(1-(-1)^{\lambda_i}) = 0$$.

The number of black-white balanced partitions of $$2n$$ is OEIS sequence A000712, which begins $$1$$, $$2$$, $$5$$, $$10$$, $$20$$, $$36$$, $$65$$, $$110$$. The name of this sequence is "Number of partitions of n into parts of 2 kinds". Alternatively, it is the convolution of the partition numbers with themselves. The description above appears on the linked page as "Also equals number of partitions of 2n in which the odd parts appear as many times in even as in odd positions."

# Question

In other words, there appears to be a bijection between black-white balanced partitions $$\lambda$$ of $$2n$$ and ordered pairs of partitions ($$\lambda_1$$, $$\lambda_2$$) of a combined $$n$$ (that is, if $$\lambda_1$$ is a partition of $$n_1$$ and $$\lambda_2$$ is a partition of $$n_2$$, then $$n=n_1+n_2$$).

Can anyone describe this bijection, or provide a reference for it?

## Remarks

I'm fairly sure I'm just a Google search away from finding a reference, but I just haven't succeeded yet. I found this question on Mathematics Stack Exchange and this identical question on MathOverflow asking about a bijection between strict (ie, distinct parts) black-white balanced partitions of $$2n$$ and unrestricted partitions of $$n$$ (not pairs thereof). The paper "Balanced partitions" by Sam Vandervelde also proves this result.

• I think "abaci" will provide the bijection you want. Jul 24, 2021 at 20:23
• @SamHopkins: I was unfamiliar with abaci until I read this related question mathoverflow.net/questions/147326/unique-domino-tiling , which casually mentions that the number of partitions of $2n$ with a unique domino tiling is twice the number of unrestricted partitions of $n$. It sketches a proof using abacus diagrams. Armed with this search term, I could find lots of interesting stuff to read, but still no reference for my desired result.
– aorq
Jul 24, 2021 at 20:34
• Have you looked at the so-called "2-quotient" of a partition, which is the counterpart to the 2-core? Jul 24, 2021 at 20:37
• @SamHopkins: No, I hadn't! If the 2-core is empty, then is the 2-quotient literally what I want?!
– aorq
Jul 24, 2021 at 20:50
• This is the case $p=2$ and $\mu=\emptyset$ of Exercise 7.59(e) of Enumerative Combinatorics, vol. 2. The proof is a disguised version of the abaci proof of James and Kerber. Some additional references are in the solution to the exercise. Jul 24, 2021 at 22:02

Richard Stanley and Sam Hopkins have answered this question in the comments. (Thanks!)

Professor Stanley mentions that this is a special case of exercise 7.59(e) in Enumerative Combinatorics, volume 2:

7.59(e) Let $$\mu$$ be a $$p$$-core. Let $$Y_{p,\mu}$$ be the set of all partitions whose $$p$$-core is $$\mu$$. Define $$\lambda \le \nu$$ in $$Y_{p,\mu}$$ if $$\lambda$$ can be obtained from $$\nu$$ by removing border strips of size $$p$$. Show that $$Y_{p,\mu} \cong Y^k$$, where $$Y$$ denotes Young's lattice. Deduce that if $$f_{\mu}(n)$$ is the number of partitions of $$n$$ with $$p$$-core $$\mu$$, then $$\sum_{n\ge 0} f_{\mu}(n)x^n = x^{\lvert \mu\rvert} \prod_{i\ge 1} (1-x^{pi})^{-p}.$$

(Presumably the $$k$$ in $$Y^k$$ is meant to be a $$p$$ as in $$Y^p$$.) The proof sketched by the exercise is very nice and uses an infinite binary sequence called the code of a partition $$\lambda$$. The bijection from this question can be obtained by considering every other bit from this binary sequence.

Professor Hopkins mentions that my desired bijection is known under the name "$$2$$-quotient". Specifically, every partition $$\lambda$$ has a $$2$$-core $$\alpha$$, which is a single partition, and a $$2$$-quotient $$(\beta_1, \beta_2)$$, which is an ordered pair of partitions. Two key facts are the following:

• $$\lvert \lambda\rvert = \lvert \alpha \rvert + 2\lvert \beta_0\rvert + 2\lvert \beta_1\rvert.$$
• The triple $$(\alpha; \beta_0, \beta_1)$$ uniquely determines $$\lambda$$.

In the case of this question, the $$2$$-core $$\alpha$$ is empty (by assumption), so the $$2$$-quotient $$(\beta_0, \beta_1)$$ is precisely the desired bijection. One way of describing the $$2$$-quotient is by coloring the Young diagram of $$\lambda$$ in a checkerboard pattern. Then, $$\beta_0$$ (say) is obtained by considering the subdiagram of $$\lambda$$ restricted to those rows ending in a black square and those columns ending in a white square. Oppositely, $$\beta_1$$ is obtained by restricting to those rows ending in a white square and those columns ending in a black square.

One reference for this is James and Kerber, "The Representation Theory of the Symmetric Group," page 83, Theorem 2.7.30.