[seqfan] Re: Lyndon words

[Heinz, Alois]

Third way: reverse and then convert to decimal:

0, 1, 2, 4, 6, 8, 12, 14, 16, 24, 20, 28, 26, 30, 32, 48, 40, 56, 52, 
44, 60, 58, 62, 64, 96, 80, 112, 72, 104, 88, 120, 100, ...

Fourth way: shift left until leftmost bit is 1 (exception for 0) and 
then convert to decimal:

0, 1, 2, 4, 6, 8, 12, 14, 16, 24, 20, 28, 22, 30, 32, 48, 40, 56, 44, 
52, 60, 46, 62, 64, 96, 80, 112, 72, 88, 104, 120, 76, ...

All not in OEIS.

Best  regards, Alois

Am 09.03.2014 01:21, schrieb Wouter Meeussen:
> In http://oeis.org/A001037  under ‘Example’ and in http://en.wikipedia.org/wiki/Lyndon_word under ‘Enumeration’ we find the binary Lyndon words themselves (not their counts) as
> 0,1,01,001,011,0001,0011,0111,00001,00011,00101,00111,01011,01111,...
> Since each starts with a zero bit, I can see two ways of encoding these as decimal numbers:
> either prepend a “1” bit
> {2,3,5,9,11,17,19,23,33,35,37,39,43,47,65,67,69,71,75,77,79,87,95,...}
> or switch  1’s and 0’s .
> {1,0,2,6,4,14,12,8,30,28,26,24,20,16,62,60,58,56,52,50,48,40,32,...}
> The first seems more natural, and produces an increasing sequence.
> For such sequence, split in runs by bit-count, the number of Lyndon words in each run is A001037.
>
> Would you also prefer the first option, and, if so, how should I name it?
>
> Wouter.