Importance: Medium ✭✭
Author(s): David S.
Newman
Subject: Combinatorics
Recomm. for undergrads: no
Posted by: DavidSNewman
on: May 11th, 2010

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.

I've been thinking about this problem since about 1970. Emory Starke thought that it was a good problem, but not suitable for the Problems section of the AMM, because it was unsolved. George Andrews and Freeman Dyson also thought that it is a good problem, but neither had any ideas how to solve it.

I've found choices of sign which yield series with coefficients 1, -1, or 0 for all exponents about as high as 110 using computer searches. One thing which mitigates against finding a meaningful solution is that there is no known pattern for the number of unrestricted partitions modulo 2.

Bibliography

Andrews, George E., The Theory of Partitions, Cambridge University Press (1984)


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