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: ( +/- means plus or minus )
(1 +/- x +/- x^2 +/- x^3 +/- ...)(1 +/- x^2 +/- x^4 +/- x^6 +/- ...) (1 +/- x^3 +/- x^6 +/- x^9 +/- ...) ... Multiply everything out to give:
1 + a(1) x + a(2) x^2 + a(3) x^3 +...\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$n$, is it always possible to choose the signs such that a(n)$a(n)$ is equal to +1$+1$ or 0$0$ or -1$-1$?
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$+1$, $0$ or -1$-1$.
I'm convinced that the answer to this question is yes.
