-----Original Message----- Von: Antti Karttunen [mailto:karttu@megabaud.fi] Gesendet am: 13 December, 2002 15:08 An: njas@research.att.com Cc: Pfoertner, Hugo; seqfan@ext.jussieu.fr; math-fun@mailman.xmission.com Betreff: Re: Lacings continued <sequence definition snipped>

Dear Neil,

Because you have counted 132546 as a solution in A078602, I reckon that

your condition

"and cannot pass though three consecutive and adjacent eyelets that are in a line" means that "cannot pass _in order_ three consecutive and adjacent eyelets on the same side."

If we want to count "real world" lacings things like the "132546" config should be excluded from being counted. I would like another condition that applies to all lacings I have seen:

From each eyelet there is at least one path to the opposite side. This condition should guarantee that there is a closing force on the eylet and not just a force in tangential direction as it is the case with eyelets 3 and 4 in the 1322546 config.

I think here it is a question of simply counting of permutations with some forbidden subsequences. I.e. in the case a(3) = 21, we have permutations of [1..6], but with the first and the last element fixed as 1 and 6, so actually counting the permutations of [2,3,4,5], but discarding the cases where we would have a either an increasing [i,i+1,i+2] or decreasing subsequence [i+2,i+1,i] (where i+2 is either <= n or i > n) e.g. from 4! = 24 permutations of [1,2,3,4,5,6] with 1 and 6 fixed we have to further discard the three cases [1,2,3,4,5,6] (violates the condition on both sides), [1,2,3,5,4,6] (on the left side) and [1,3,2,4,5,6] (on the right side), so we are left with 24-3 = 21 solutions.

Reasoning on these lines should easily (?) lead to a formula for A078602. At least it's now very easy to write a program in a declarative language like Prolog or Haskell to count the solutions.

Then regarding Hugo's point about whether the lace enters each hole (Btw, some of his pictures resemble my son's attempts at this difficult art...) from the outside or inside, doesn't this just add an extra factor of 2^(2n) to the count?

That would be the case, if all combinations were allowed. I want to exclude inside/outside changes if the lace connects two eyeholes on the same side. This complicates things.

Regarding which lace goes underneath or over which other, it seems more complex. I don't know whether thinking in terms of braids actually clarifies or confuses this issue:

http://mathworld.wolfram.com/BraidGroup.html and http://www.brown.edu/Students/OHJC/topology/index.html and http://math.ucr.edu/home/baez/braids.html

(at least one could borrow some notation from there?)

I'll try to learn something from the links.

Terveempänä,

Antti Karttunen

Best Regards, Hugo Pfoertner