There are the following analogs of the famous identity $$ \prod_{n\geqslant1}(1-q^n)=\sum_{n\in\mathbb Z}(-1)^nq^{\frac{3n^2-n}2}. $$

Let $v_2(n)$ denote the 2-adic valuation of $n$, that is, the number determined by $n=2^{v_2(n)}(2k+1)$. Then \begin{align*} \prod_{n\geqslant1}(1+q^n)^{1-v_2(n)}&=\sum_{n\geqslant1}q^{\frac{n^2-n}2},\\ \prod_{n\geqslant1}(1+q^n)^{-v_2(n)}&=\sum_{n\in\mathbb Z}(-1)^nq^{3n^2-n},\\ \prod_{n\geqslant1}(1+q^n)^{-1-v_2(n)}&=\sum_{n\in\mathbb Z}(-1)^nq^{\frac{3n^2-n}2},\\ \prod_{n\geqslant1}(1+q^n)^{-2-v_2(n)}&=\sum_{n\in\mathbb Z}(-1)^nq^{n^2}. \end{align*} I am not so much interested in proofs - I have almost finished proofs, but they are ad hoc and different in each of the cases. What I want to know is whether there is a general explanation for such identities, and whether there are more identities of this kind.

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    $\begingroup$ Basically you are multiplying with varying powers of $q_{24}^{-1}\eta(2\tau)/\eta(\tau)$ where $q_n=\exp(2\pi\mathrm{i}\tau/n)$. Therefore, once you have proven one of these identities, you can deduce the others by means of the known representations of Jacobi thetanulls as eta quotients, which are consequences of the Jacobi triple product identity. Is this what you have done? $\endgroup$ – ccorn Jul 30 '15 at 12:53
  • $\begingroup$ @ccorn Yes you are right, but it did not occur to me. I just stumbled on the last one by asking myself which powers of $1+q^n$ would yield $\vartheta_4(0)$, and having obtained $-2-v_2(n)$ asked myself which other $k-v_2(n)$ would give something recognizable. $\endgroup$ – მამუკა ჯიბლაძე Jul 30 '15 at 19:13

Let $q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$ and $q=q_1$, so we can regard expressions with $q$ as functions of $\tau\in\mathbb{H}$. This is only needed to draw a connection to the Dedekind eta function $\eta(\tau)$. It is not needed when focussing on formal $q$-power series only.

The following proof is based on binary representation of nonnegative integers, which gives rise to the factorization $$\frac{1}{1-q} = \sum_{m=0}^\infty q^m = \prod_{k=0}^\infty\left(1+q^{2^k}\right)$$ and consequently $$\begin{align} \sum_{n\in\mathbb Z}(-1)^n q^{n(3n-1)/2} = q_{24}^{-1}\eta(\tau) &= \prod_{n=1}^\infty(1-q^n) = \prod_{j=1}^\infty\prod_{m=0}^\infty\left(1-q^{2^m(2j-1)}\right) \\ &= \prod_{j=1}^\infty\prod_{m=0}^\infty\prod_{k=0}^\infty \left(1+q^{2^{k+m}(2j-1)}\right)^{-1} \\ &= \prod_{n=1}^\infty(1+q^{n})^{-1-\nu_2(n)} \end{align}$$ The other identities follow from $$q_{24}^{-1}\frac{\eta(2\tau)}{\eta(\tau)} = \prod_{n=1}^\infty\frac{1-q^{2n}}{1-q^n} = \prod_{n=1}^\infty(1+q^n)$$ and from the known corollaries of Jacobi's triple product identity: $$\begin{align} \sum_{n=1}^\infty q^{n(n-1)/2} = \frac{1}{2}q_8^{-1}\Theta_{10}(0;\tau) &= q_8^{-1}\frac{\eta^2(2\tau)}{\eta(\tau)} \\ \sum_{n\in\mathbb{Z}}(-1)^n q^{n^2} = \Theta_{01}(0;2\tau) &= \frac{\eta^2(\tau)}{\eta(2\tau)} \end{align}$$ Another corollary covers partition numbers $P(n)$: $$\sum_{n=0}^\infty P(n)\,q^n = \prod_{n=1}^\infty\frac{1}{1-q^n} = \frac{q_{24}}{\eta(\tau)} = \prod_{n=1}^\infty(1+q^{n})^{1+\nu_2(n)}$$ Here is another twist, using base-$3$ representations of natural numbers: $$\begin{align} \frac{1}{1-q} &= \sum_{m=0}^\infty q^m = \prod_{k=0}^\infty\left(1+q^{3^k}+q^{2\cdot3^k}\right) = \prod_{k=0}^\infty \left(1-\zeta_3 q^{3^k}\right)\left(1-\zeta_3^{-1}q^{3^k}\right) \\ \quad\text{where}\quad \zeta_n &= \exp\frac{2\pi\mathrm{i}}{n} \\ \therefore\quad q_{24}^{-1}\eta(\tau) &= \prod_{n=1}^\infty(1-q^n) = \prod_{j=0}^\infty\prod_{m=0}^\infty \left(1-q^{3^m(3j+1)}\right)\left(1-q^{3^m(3j+2)}\right) \\ &= \prod_{j=0}^\infty\prod_{m=0}^\infty\prod_{k=0}^\infty \left(1+q^{3^{k+m}(3j+1)}+q^{2\cdot3^{k+m}(3j+1)}\right)^{-1} \left(1+q^{3^{k+m}(3j+2)}+q^{2\cdot3^{k+m}(3j+2)}\right)^{-1} \\ &= \prod_{n=1}^\infty\left(1+q^n+q^{2n}\right)^{-1-\nu_3(n)} = \prod_{n=1}^\infty \left((1-\zeta_3 q^n)(1-\zeta_3^{-1}q^n)\right)^{-1-\nu_3(n)} \end{align}$$

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