Another question related to the generating function for unrestricted partitions

Question

This question is similar to another that I asked, but should be, I think, very much easier.

Start with the generating function for unrestricted partitions and replace some of the plus signs with minus signs to get:

\begin{align} &(1\pm x\pm x^2\pm x^3\pm\cdots)(1\pm x^2\pm x^4\pm x^6\pm\cdots)(1\pm x^3\pm x^6\pm x^9\pm\cdots)\cdots \\ = & 1 + a(1) x + a(2) x^2 + a(3) x^3 +\cdots \end{align}

For a given positive integer, $n$, is it always possible to choose the signs such that $a(n)$ is equal to $+1$ or $0$ or $-1$?

My previous question on the same topic asked if it is possible to choose the signs such that every coefficient in the series is $+1$, $0$ or $-1$.

I'm convinced that the answer to this question is yes.

Answer 1

It is possible. Select positive signs for all factors but the first. The product of all these factors is then the generating function for partitions without 1's. Since this sequence is sub-doubling, G Pasemans' answer to question Will this greedy algorithm always work? shows that we can select the signs in the first factor so as to make the coefficient $a(n)$ be $1$, $-1$ or $0$.