# Another question related to the generating function for unrestricted partitions

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.

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 $0$.