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This question is motivated by my earlier (unanswered) MO post.

The number of partitions into distinct parts is generated by $\sum_{n\geq0}Q(n)x^n=\prod_{k\geq1}(1+x^k)$. Focusing on parity of coefficients $Q(n) \,\,\text{mod}\,\, 2$, the series takes the form $\sum_{j\in\mathbb{Z}}x^{j(3j-1)/2}$.

In light of these matters, consider the modified product $\prod_{k\geq2}(1+x^k)$. It might be easy but I still like to ask:

QUESTION. Is this true? $$\prod_{k\geq2}(1+x^k) \mod 2=1+(x^2+x^3+x^4)+(x^7+x^8+x^9+x^{10}+x^{11})+\cdots$$ to say that consecutive terms appear with block-lengths: $1, 3, 5, 7, 9,\dots$; that is, odd numbers.

NOTE. $\prod_{k\geq2}(1+x^k)$ generates partitions into distinct parts with parts larger than $1$.

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    $\begingroup$ just multiply the right hand side of your hypothetical relation by $1+x$ and verify that you get $\sum x^{j(3j-1)/2}$ $\endgroup$ – Fedor Petrov Apr 20 at 19:54
  • $\begingroup$ @FedorPetrov that's the thought, but let's have the details verified. I don't expect this problem to be difficult but more discussion may come out of this. $\endgroup$ – T. Amdeberhan Apr 20 at 19:56
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We have $$\prod_{k\geq 2} (1+x^k) = \frac{\prod_{k\geq 1} (1+x^k)}{1+x} \equiv \sum_{j\geq 0} \frac{x^{j(3j+1)/2} + x^{(j+1)(3j+2)/2}}{1+x}$$ $$\equiv \sum_{j\geq 0} x^{j(3j+1)/2}\frac{1 - x^{2j+1}}{1-x}\equiv \sum_{j\geq 0} x^{j(3j+1)/2}(1+x+\dots+x^{2j}) \pmod{2}.$$

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