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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
0, 0, 3, 4, 10, 11 (list; graph; refs; listen; history; text; internal format)



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.


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


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


Cf. A075099, A003313, A124677.

Sequence in context: A128488 A226303 A117781 * A288660 A212440 A066861

Adjacent sequences:  A075097 A075098 A075099 * A075101 A075102 A075103




Colin Mallows, Aug 31 2002



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Last modified November 22 15:27 EST 2017. Contains 295089 sequences.