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Newman
Dividing up the unrestricted partitions ★★
Begin with the generating function for unrestricted partitions:
(1+x+x^2+...)(1+x^2+x^4+...)(1+x^3+x^6+...)...
Now change some of the plus signs to minus signs. The resulting series will have coefficients congruent, mod 2, to the coefficients of the generating series for unrestricted partitions. I conjecture that the signs may be chosen such that all the coefficients of the series are either 1, -1, or zero.
Keywords: congruence properties; partition
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