This question is motivated by [my earlier (unanswered) MO post](https://mathoverflow.net/questions/390446/being-even-or-odd-in-the-product-expansion-prod1xkxk1).


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 $1$.



Partity of partitions with distinct parts of parts $>1$
We have
$$\prod\_{k\geq 2} (1+x^k) = \frac{\prod\_{k\geq 1} (1+x^k)}{1+x} \equiv \sum\_{j\geq 0} \frac{x^{j(3j+1)/2} + x^{(j+1)(3j+2)/2}}{1+x}$$
$$\equiv \sum\_{j\geq 0} x^{j(3j+1)/2}\frac{1 - x^{2j+1}}{1-x}\equiv \sum\_{j\geq 0} x^{j(3j+1)/2}(1+x+\dots+x^{2j}) \pmod{2}.$$


