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 $1$.