2
$\begingroup$

How to prove or disprove this statement:

For all $c, there exists $0, $0\leq j, such that all conditions hold simultaneously:

  1. $z=is+i(i-1)/2-j-k(s+i+2j)-k(k-1)/2$,
  2. $z and $0
  3. for all $0\leq m and $0, a) holds

a) $ms+m(m-1)/2\neq is+i(i-1)/2-j-n(s+i+2j)-n(n-1)/2$

All variables are integers. I have tried computationally to find a counterexample but without luck.


This would imply the existence of several Self-avoiding walk on $\mathbb{Z}$ . (but not the converse.)

  • 7
    Seems like an extremely uninteresting question, which no one would ever want to think about. Maybe if we knew how you stumbled upon all those formulas, we'd find it worth a thought.2012-03-01
  • 3
    I sort of appreciate the quirky humor involved in the current title, but I also think that I have a strange sense of humor.2012-03-01
  • 0
    It's basically the same algorithm as in my answer here, but in 1 dimension: http://mathoverflow.net/questions/88659/traversing-the-infinite-square-grid2012-03-01
  • 0
    Every unintresting question is intresting to somebody who was looking for unintresting questions.2012-03-01
  • 0
    This might be a silly question, but are automatic theorem provers anywhere close to proving statements like these?2012-03-01
  • 6
    @mmm: please stop with these ridiculous edits. You are needlessly bumping this question up to the front page, and those edits can be seen as vandalism.2012-03-02
  • 0
    Since the "Self-avoiding walk" problem (the motivation for the OP) has been solved on mathoverflow without any ugly equations, I think we may safely declare this question completely unintereseting.2012-03-04

0 Answers 0