What can be said of a $6$-core Young diagram whose $2$-and $3$-cores are empty

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.