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Dec 2, 2014 at 20:14 comment added Wolfgang @user38477 I just saw your comments (thanks to Darij). You are right. But I am still hoping to find the pattern(s) of square-initial Nyldon words... They seem to be quite restricted. Maybe all $swsswsw^r$ with words $s>w$ (and $r\ge1$) are such patterns, e.g. s=10, w=0, r=1 yield your $10010100100$. Are there square-initial Nyldon words not fitting this pattern?
Nov 29, 2014 at 16:51 comment added user38477 Also, $10010100100$ is a Nyldon word, and none of its conjugate starts with $1w11w1$ (it is $11$-free)
Nov 29, 2014 at 16:28 comment added user38477 No problem :) I comment here for @wolfgang (I still cannot comment elsewhere...). Is 10110100 a S-word ? Because it is a Nyldon word, and not of the form $kwkkwkw$
Nov 28, 2014 at 21:26 comment added darij grinberg OK, now this means a lot of reading-up for me... once the FPSAC deadline is over. Sorry for the slowness.
Nov 28, 2014 at 19:44 history edited user38477 CC BY-SA 3.0
added 2013 characters in body
Nov 22, 2014 at 9:23 comment added user38477 $\pi$ is a mapping between Lyndon/Nyldon words and binary tree whose leaves are labelled by letter of the alphabet. For example, for Lyndon words, $\pi(00101001101111) = [[[0[01]][01]][0[[[01]1][[[[01]1]1]1]]]]$, since $00101001101111= 00101.001101111$ and (recursively) $\pi(00101)= [[0[01]][01]]$ and $\pi(001101111)= [0[[[01]1][[[[01]1]1]1]]]$. Now if we take the image of the set of Lyndon words, we get a Hall set, and all the factorization properties follow. We can do the same thing for Nyldon words. Eg $\pi(10111100)= [[[[[10]1]1]1][[10]0]]$. But now we have to show that it is a hall set.
Nov 22, 2014 at 3:38 comment added darij grinberg Thanks a lot, but this feels like it's going to take me a while to grok. First and foremost, what is $\pi$ ? And does this all add up to a (conjectural, yet) intrinsic definition of a Nyldon word?
Nov 22, 2014 at 0:25 comment added user38477 Note: I checked this for binary alphabet up to n=17 and for ternary alphabet up to n=11.
Nov 21, 2014 at 23:43 comment added user38477 I think this comment is usefull to understand that I do not have yet an "answer" (i.e. a "proof"), it is just a remark to move forward on the problem. But since I do not have "50 points", I cannot "comment", so I have to put is as an "answer", and I ask for indulgence to not vote "down". And I read my word from left to right. (I edited the answer to use the lexicographic order.)
Nov 21, 2014 at 23:31 comment added David Hill Also, can you define you ordering more explicitly. Do you read words left-to-right or right-to-left?
Nov 21, 2014 at 23:30 comment added David Hill I don't think your comment above is helpful. I think it was appropriate to delete it.
Nov 21, 2014 at 23:05 history edited user38477 CC BY-SA 3.0
change > by <_lex
Nov 21, 2014 at 23:01 comment added user38477 Btw, "thanks" Joonas for the editing. I put here my first paragraph he deleted: "I cannot comment, since I am a new user and I do not have enough "points". So I put my "comment" here even it is not a proof. As a first feedback, this Mathoverflow rule really sucks and are far from academic standards...)"
Nov 21, 2014 at 22:59 comment added user38477 My order is the reverse lexicographic order... In this case $y=0 > v=1 > u= 10$. But yes, this is confusing... I will edit the comment.
Nov 21, 2014 at 22:03 history edited Joonas Ilmavirta CC BY-SA 3.0
If you wish to lament the way MO works, don't start your post with it. Leave it as a comment in the end, or after earning 5 reputation points, bring the issue up at meta.
Nov 21, 2014 at 21:58 comment added David Hill Isn't (3) already false for $w=101$, $u=10$ and $v=1$?
Nov 21, 2014 at 21:49 review First posts
Nov 21, 2014 at 22:03
Nov 21, 2014 at 21:49 history answered user38477 CC BY-SA 3.0