The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ of $2$ such that $2^t$ divides $n$. The number of integer partitions of $n$, denoted by $p(n)$, has generating function $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}\frac1{1-x^k}.$$ However, I am finding an alternative: $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$

Question. Is this known? If so, any reference? If not, then any proof?

Caveat. I would be surprised if this is new.

The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ such that $2^t$ divides $n$. The number of integer partitions of $n$, denoted by $p(n)$, has generating function $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}\frac1{1-x^k}.$$ However, I am finding an alternative: $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$

Question. Is this known? If so, any reference? If not, then any proof?

Caveat. I would be surprised if this is new.

The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ of $2$ such that $2^t$ divides $n$. The number of integer partitions of $n$, denoted by $p(n)$, has generating function $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}\frac1{1-x^k}.$$ However, I am finding an alternative: $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$

Question. Is this known? If so, any reference? If not, then any proof?

Note. I would be surprised if this is new.

The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ of $2$ such that $2^t$ divides $n$. The number of integer partitions of $n$, denoted by $p(n)$, has generating function $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}\frac1{1-x^k}.$$ However, I am finding an alternative: $$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$

Question. Is this known? If so, any reference? If not, then any proof?

Caveat. I would be surprised if this is new.