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Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Further, let $h_u$ denote the [hook-length][1] 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 $$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$ [1]: https://en.wikipedia.org/wiki/Hook_length_formula#Example

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 $$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$

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

Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Further, let $h_u$ denote the [hook-length][1] 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 $$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$ [1]: https://en.wikipedia.org/wiki/Hook_length_formula#Example

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 not easy to see $d_2(n)=c_2(n)$.

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

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