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.

0, 0, 3, 4, 10, 11

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 one-letter 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: A359366 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