9
$\begingroup$

This is my latest attempt to simplify an old problem of mine so much that the simplified problem can actually be answered.

Starting with the generating function for unrestricted partitions:

$$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$

Change some of the plus signs in the leftmost expression in parentheses to minus signs. Is it possible that the resulting series has coefficients all of which are $1$, $-1$, or zero?

I believe that the answer is no, but I'm not convinced by my computer aided search.

$\endgroup$
  • $\begingroup$ Can you give any background? $\endgroup$ – Alexey Ustinov Jun 23 '17 at 3:24
  • $\begingroup$ @Alexey Ustinov. Ultimately, my motivation for this problem came from a comparison of the generating function for partitions into distinct parts (1+x)(1+x^2)(1+x^3)... with the series (1-x)(1-x^2)(1-x^3)...=1-x-x^2+x^5+x^7-x^12-... I wanted to find other instances where a change of sign led to some pretty result. $\endgroup$ – David S. Newman Jun 23 '17 at 12:04
  • $\begingroup$ @Alexey Ustinov I just noticed that when you edited the problem you left off the ellipsis in the term that begins 1+x^3+x^6+x^9. There should be a "+..." there. $\endgroup$ – David S. Newman Jun 24 '17 at 0:36
3
$\begingroup$

REVISED

No.

Let $$A=(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ and call a polynomial $f=1+\sum_1^da_ix^i$ feasible if all the $a_i \in \{{-1,1\}}$ and all the coefficients of $fA$ up to that of $x^d$ are in $\{{-1,0,1\}}.$ Finally, call $f$ maximal if it is feasible but neither of $f+x^{d+1}$ nor $f-x^{d+1}$ is feasible.

There are no feasible polynomials of degree 31. There are $40$ maximal polynomials.

The largest degree maximal polynomials are

${x}^{30}+{x}^{29}+{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}-{x}^{23 }-{x}^{22}-{x}^{21}-{x}^{20}+{x}^{19}+{x}^{18}+{x}^{17}-{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $

${x}^{30}+{x}^{29}-{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}+{x}^{23 }-{x}^{22}+{x}^{21}-{x}^{20}+{x}^{19}-{x}^{18}-{x}^{17}+{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $ and $-{x}^{25}-{x}^{24}-{x}^{23}+{x}^{22}+{x}^{21}+{x}^{20}+{x}^{19}-{x}^{ 18}+{x}^{17}-{x}^{16}-{x}^{15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x} ^{10}-{x}^{9}+{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}- x+1. $

the degrees of the maximal polynomials are

$30, 30, 25, 24, 24, 24, 24, 24, 24, 24, 22, 21, 21, 20, 19, 18, 18, 18, 18, 18, 18, 17, 17, 16, 15, 15, 14, 14, 14, 14, 14, 13, 10, 10, 9, 8, 8, 7, 7, 6.$

$\endgroup$
  • $\begingroup$ I'm glad to see that the branch that you've chosen here is in agreement with what I've calculated. It is impossible to continue this branch to 25. What you are indicating with the c's and d I do not understand. All that would be necessary to do is continue the same sort of calculations for all the other branches. I think I have accomplished this but it would be nice to have independent confirmation. Otherwise I'll just do the computations again by myself. $\endgroup$ – David S. Newman Jun 24 '17 at 21:34
  • $\begingroup$ I've now read the revised version of the answer by Aaron Meyerowitz and checked one of the largest degree maximal polynomials. This case of my old problem seems to be solved. Does anyone think they can solve the next case. By this I mean: s $\endgroup$ – David S. Newman Jun 25 '17 at 20:59
  • $\begingroup$ I've now read the revised version of the answer by Aaron Meyerowitz and checked one of the largest degree maximal polynomials. This case of my old problem seems to be solved. Does anyone think they can solve the next case? By this I mean: start with the generating function for unrestricted partitions as above and change signs in the first two expressions in parentheses of the product. Can the new series have coefficients all of which are 1, -1, and zero? This is much harder. $\endgroup$ – David S. Newman Jun 25 '17 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.