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How to prove or disprove this statement:

For all $c<z<0<s$, there exists $0<k\leq i$, $0\leq j<s+i$, such that all conditions hold simultaneously:

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

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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Mar 1 '12 at 2:09
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    $\begingroup$ I sort of appreciate the quirky humor involved in the current title, but I also think that I have a strange sense of humor. $\endgroup$ – davidlowryduda Mar 1 '12 at 8:24
  • $\begingroup$ It's basically the same algorithm as in my answer here, but in 1 dimension: mathoverflow.net/questions/88659/… $\endgroup$ – user26004 Mar 1 '12 at 8:25
  • $\begingroup$ Every unintresting question is intresting to somebody who was looking for unintresting questions. $\endgroup$ – Arjang Mar 1 '12 at 11:06
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    $\begingroup$ @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. $\endgroup$ – Willie Wong Mar 2 '12 at 9:03

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