# Generating function for $3$-core partitions

Let $$\lambda$$ be an integer partition: $$\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$$. Further, let $$h_u$$ denote the hook-length of the cell $$u$$.

We call $$\lambda$$ a $$t$$-core partition if none of its hooks $$h_u$$ equals $$t$$. Define $$c_t(n)$$ to be the number of partitions of $$n$$ that are $$t$$-core partitions. It's well-known that $$\sum_{n\geq0}c_t(n)\,q^n=\prod_{k=1}^{\infty}\frac{(1-q^{tk})^t}{1-q^k}.$$ For example, $$\sum_{n=0}^{\infty}c_2(n)\,q^n=\sum_{k=0}^{\infty}q^{\binom{k+1}2}$$.

Now, consider only those partitions of $$n$$ with distinct parts and let $$d_t(n)$$ be the number of such partitions that are $$t$$-cores. Then it is easy to see $$d_2(n)=c_2(n)$$.

QUESTION. Is this true? $$\sum_{n\geq0}d_3(n)\,q^n=\sum_{k\geq0}q^{k^2} +\sum_{k\geq1}q^{2\binom{k+1}2}.$$

Note that I have simplified the generating function from $$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$

The set of $$3$$-core partitions can be described explicitly.
Theorem The partition $$\lambda=\{\lambda_1,\lambda_2,\dots\}$$ is a $$3$$ core if and only if the sequence of differences $$\{\lambda_1-\lambda_2,\lambda_2-\lambda_3,\dots\}$$ is of the form $$\{2,2,\dots,2,1,0,1,0,\dots 1\}$$ or $$\{2,2,\dots,2,0,1,0,1,\dots 1\}$$.
Proof: It is easy to check by hand that a $$3$$-hook appears in the situations where a) some member of the sequence is $$\geq 3$$ b) two members of the sequence in a row are $$0$$'s c) there is a $$1$$ in the sequence that is not followed by $$0$$ d) there is a $$0$$ in the sequence that is not followed by a $$1$$. These correspond to the four possible shapes of a $$3$$ hook-strip in the boundary. If all of these patterns are avoided, then the partition has no hooks of length $$3$$.
Therefore the partitions with distinct parts that are $$3$$ cores have a difference sequence $$\{2,2,\dots,2\}$$ or $$\{2,2,\dots,2,1\}$$. The size of the partitions in the first case are given by $$2\binom{k+1}{2}$$ and the sizes in the second case are given by $$k^2$$, where $$k\geq 1$$, and this implies your generating function.
• The sequence is meant to be finite, right? And by "some member is $3$" you mean $\geq 3$? – darij grinberg Oct 14 '18 at 17:03
• +1, nice work! I had no idea that any $r$-cores with $r \geq 2$ could be characterized in such an explicit manner. – darij grinberg Oct 14 '18 at 17:09