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Start with the product for unrestricted partitions: $(1+x+x^2+...)(1+x^2+x^4+...)(1+x^3+x^6+...)...$

Now replace some of the plus signs with minus signs and expand the product into a series. Is it possible that the coefficients of the resulting series can all be in the set {+1,0,-1}?

Here's an example for which the coefficients of x$^n$ are from the required set for n<=9.

(1+x-x$^2$-x$^3$-x$^4$-x$^5$+x$^6$+x$^7$+x$^8$+x$^9$) \ (1+x$^2$-x$^4$-x$^6$-x$^8$)
(1+x$^3$-x$^6$-x$^9$)
(1+x$^4$-x$^8$)(1+x$^5$)( 1+x$^6$)(1+x$^7$)(1+x$^8$)(1+x$^9$)

=1+x+x$^3$-x$^4$-x$^5$+x$^6$-x$^7$+...

This can be extended to a product which satisfies the requirements for n<65.

I've found examples where the coefficients are good for n exceeding 100, but not as large as 110.

Over the years I've asked this question of several mathematicians. Two of them ventured their opinions. Freeman Dyson thought it couldn't be done. George Andrews thought it couldn't be done in any meaningful way. (These are not exact quotes, just the sense of their answers.)

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  • $\begingroup$ Must all the parentheses begin with +1? $\endgroup$
    – Konstantinos Gaitanas
    Commented Oct 30, 2013 at 10:22
  • $\begingroup$ It would suffice to find a product $\prod_{k\rm{\ odd}}(1+\varepsilon_k x^k)$ ($\varepsilon_k=\pm 1$) whose Taylor expansion has coefficients in $\{-1,0,1\}$. Is there any obvious reason why that is impossible? $\endgroup$
    – fedja
    Commented Oct 30, 2013 at 12:13
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    $\begingroup$ @Konstantinos: I don't think it makes a difference. Just negate the ones that don't. $\endgroup$
    – Brendan McKay
    Commented Oct 30, 2013 at 12:17
  • $\begingroup$ @Fedja I don't see why that suffices, although I'm not saying that it is hard. However your product would include $\epsilon_1\epsilon_7+\epsilon_3\epsilon_5)x^8$ the coefficient being even must be zero so...$\epsilon_1\epsilon_3\epsilon_5\epsilon_7=-1.$ I could be wrong, but I think that similar considerations easily force $\epsilon_3=\epsilon_5=\cdots=\epsilon_{25}=-\epsilon_1$ but then $\epsilon_{27}=\epsilon_1$ and then one ends up with $2x^{28}.$ $\endgroup$
    – Aaron Meyerowitz
    Commented Oct 31, 2013 at 6:42
  • $\begingroup$ @fedja You are correct that a product of the sort that you describe would suffice to answer my question. I searched for such products. If I remember correctly there were none with coefficients of the required sort that went beyond x^19 or thereabouts. This was a relatively easy computation. But the search space for what I am looking for is much larger. $\endgroup$
    – David S. Newman
    Commented Nov 10, 2013 at 1:54

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