This is my latest attempt to simplify an old problem of mine so much that the simplified problem can actually be answered.


Starting with the generating function for unrestricted partitions:


$$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$


Change some of the plus signs in the leftmost expression in parentheses to minus signs. Is it possible that the resulting series has coefficients all of which are $1$, $-1$, or zero?


I believe that the answer is no, but I'm not convinced by my computer aided search. 



A simplified version of an old problem about the generating function for unrestricted partitions
**REVISED**


No.


Let $$A=(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ and call a polynomial $f=1+\sum\_1^da\_ix^i$ *feasible* if all the $a\_i \in \{{-1,1\}}$ and all the coefficients of $fA$ up to that of $x^d$ are in $\{{-1,0,1\}}.$ Finally, call $f$ *maximal* if it is feasible but neither of $f+x^{d+1}$ nor $f-x^{d+1}$ is feasible.


There are no feasible polynomials of degree 31. There are $40$ maximal polynomials.


The largest degree maximal polynomials are


${x}^{30}+{x}^{29}+{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}-{x}^{23
}-{x}^{22}-{x}^{21}-{x}^{20}+{x}^{19}+{x}^{18}+{x}^{17}-{x}^{16}-{x}^{
15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{
7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1
$


${x}^{30}+{x}^{29}-{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}+{x}^{23
}-{x}^{22}+{x}^{21}-{x}^{20}+{x}^{19}-{x}^{18}-{x}^{17}+{x}^{16}-{x}^{
15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{
7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1
$
and 
$-{x}^{25}-{x}^{24}-{x}^{23}+{x}^{22}+{x}^{21}+{x}^{20}+{x}^{19}-{x}^{
18}+{x}^{17}-{x}^{16}-{x}^{15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}
^{10}-{x}^{9}+{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-
x+1.
$


the degrees of the maximal polynomials are


$30, 30, 25, 24, 24, 24, 24, 24, 24, 24, 22, 21, 21, 20, 19, 18, 18, 18, 18, 18, 18, 17, 17, 16, 15, 15, 14, 14, 14, 14, 14, 13, 10, 10, 9, 8, 8, 7, 7, 6.$


