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Let $\lambda$ be a partition. Suppose that $\lambda$ is both $2$- and $3$-decomposable, in the sense that $\lambda$ admits a total decomposition by both $2$-rim hooks (aka dominos) and $3$-rim hooks. Equivalently, assume that the $2$-core and $3$-core of $\lambda$ is zero. Then what can be said about the $6$-core of $\lambda$, if anything? Has this problem been studied (i.e. some sort of fundamental theorem of arithmetic for tableaux)? Must the $6$-core either be a $2\times 3$ or a $3\times 2$ rectangle? The problem is easily reduced to the classification of $6$-cores whose $2$-and $3$-cores are empty; I had difficulty writing down such a core other than the above two examples.

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Let me comment a bit further on the fact that there are other 6-cores with trivial 2 and 3-core, in fact infinitely many of them. The fundamental paper of Garvan, Stanton, Kim, "Cranks and T-cores", gives a bijection between t-cores and integer tuples $(n_0,n_1,\dots,n_{t-1})$ which satisfy $\sum_{i=0}^{t-1}n_i=0$. Under this bijection the corresponding partition has size $$\frac{t}{2}(n_0^2+\cdots +n_{t-1}^2)+n_1+2n_2+\cdots+(t-1)n_{t-1}.$$ Now, this means that our set of 6-cores can be identified with 6-tuples $(n_0,n_1,\cdots,n_5)$ which satisfy $\sum_{i=0}^5 n_i=0$. In order to have an empty 2-core they must further satisfy $n_0+n_2+n_4=n_1+n_3+n_5=0$, and in order to have empty 3-core they must satisfy $n_0+n_3=n_1+n_4=n_2+n_5=0$. Therefore we are left with a sublattice of $\mathbb Z^6$ of dimension 2 generated by $(1,1,0,-1,-1,0)$ and $(0,1,1,0,-1,-1)$. These correspond to your two partitions, $2^3, 3^2$. Thus, we can form the generating function for 6-cores with empty 2 and 3 core: $$\sum_{m,n\in \mathbb Z^2}q^{12(m^2+mn+n^2)-6(m+n)}$$ This ends up being expressible as an infinite product, or a ratio of eta functions, if you will, as follows $$\frac{\prod_{k\geq 1}(1-q^{12k})^3(1-q^{18k})^2}{\prod_{k\geq 1}(1-q^{6k})^2(1-q^{36k})}.$$ Presumably generating functions of partitions that are $n$-cores but have empty $d$-core for all $d|n$ (I would be tempted to call these "pure n-cores") also have infinite product expansions, although I haven't seen this written anywhere.

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    $\begingroup$ Thank you very much; this is extremely useful for what I had in mind. $\endgroup$ – Philip Engel Mar 2 '17 at 17:58
  • $\begingroup$ Very nice answer. I think the (6,2,2,2) partition I mentioned below corresponds under this bijection to (1,0,-1,-1,0,1), one of the differences between the two generating 6-tuples you gave. (The conjugate, (4,4,1,1,1,1), corresponds to the other difference, (-1,0,1,1,0,-1), consistent with the paper's result on conjugation and these tuples.) Glad for a reason to work through some examples of their bijection. $\endgroup$ – Brian Hopkins Mar 4 '17 at 5:03
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I believe a partition of 12 answers your 3rd question in the negative: Consider (6,2,2,2). Each hook length is in the set $\{1,2,3,4,8,9\}$, so it is 6-core. One can check that the partition admits total decompositions into both 2- and 3-rim hooks.

No answers for your more interesting questions. Initially I thought of Jaclyn Anderson's 2002 Discrete Mathematics article "Partitions that are simultaneously $t_1$- and $t_2$-core" and the work it inspired, but your question is at the opposite extreme by asking about $t_1$- and $t_2$-decomposable partitions.

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