Open Problem Garden - Dividing up the unrestricted partitions - Comments

a question on the numerical work (re: Dividing up the unrestricted partitions)

Benjamin Young — Tue, 18 May 2010 21:37:21 +0200

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 (re: Dividing up the unrestricted partitions)

Benjamin Young — Tue, 18 May 2010 21:33:19 +0200

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?

Dividing up the unrestricted partitions

DavidSNewman — Tue, 11 May 2010 18:43:00 +0200

Author(s): David S.; Newman

Subject: Combinatorics

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.