

A075100


Number of terms of length < n that are needed on the way to computing all words of length n in the free monoid with two generators.


2




OFFSET

1,3


COMMENTS

I believe a(2n) = a(n)+ 2^n. I think a(7) = 28.
Benoit Jubin (Jan 24 2009) suggests replacing "monoid" in the definition by "semigroup".
Shouldn't a(2) = 2 ? Shouldn't a(3) = 5, because we need x, y, xx, xy, yy ? I'm confused!  N. J. A. Sloane, Dec 25 2006. Comment from Benoit Jubin, Jan 24 2009: I think the confusion comes from the fact that N. J. A. Sloane counts the oneletter words, so obtains always the written value plus 2. I think N. J. A. Sloane's way of counting is preferable, so the terms should be changed accordingly.


LINKS

Table of n, a(n) for n=1..6.


EXAMPLE

a(3) = 3 because we need only xx, xy, yy to generate each of xxx, xxy, xyx, yxx, xyy, yxy, yyx, yyy.


CROSSREFS

Cf. A075099, A003313, A124677.
Sequence in context: A128488 A226303 A117781 * A288660 A212440 A066861
Adjacent sequences: A075097 A075098 A075099 * A075101 A075102 A075103


KEYWORD

hard,more,nonn,obsc


AUTHOR

Colin Mallows, Aug 31 2002


STATUS

approved



