# Dividing up the unrestricted partitions

 Importance: Medium ✭✭
 Author(s): David S. Newman
 Subject: Combinatorics
 Keywords: congruence properties partition
 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)

* indicates original appearance(s) of problem.

### a question on the numerical work

I'm also curious to know a little more about the experimental work that was done -- roughly how many ways were there to choose the signs to make things work up to degree 110? were there any choices which gave you lots of zeros?

### pentagonal number theorem

Euler's famous pentagonal number theorem is somewhat like this problem, except it deals with the generating function for partitions into distinct parts:

(1+x)(1+x^2)(1+x^3)...

If you change *all* of the + signs in the above into minus signs, then the statement of your conjecture holds; indeed there is an explicit formula for the terms of the generating function involving the pentagonal numbers, hence the name of the theorem. This theorem has several pretty and well-publicized proofs (see Chapter 1 of the introduction to "The Theory of Partitions" by George Andrews, or Chapter 14.5 of "Introduction to Analytic Number Theory" by Tom Apostol, or "Proofs from the book" by Aigner-Ziegler, or Wikipedia).

I would wager that this observation isn't terribly helpful, but still. Was this was the motivation of the problem?