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A075100
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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.
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2
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OFFSET
| 1,3
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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 (njas(AT)research.att.com), Dec 25 2006. Comment from Benoit Jubin, Jan 24 2009: I think the confusion comes from the fact that N. J. A. Sloane (njas(AT)research.att.com) 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.
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EXAMPLE
| a(3) = 3 because we need only xx, xy, yy to generate each of xxx, xxy, xyx, yxx, xyy, yxy, yyx, yyy.
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CROSSREFS
| Cf. A075099, A003313, A124677.
Sequence in context: A023896 A128488 A117781 * A066861 A191192 A139063
Adjacent sequences: A075097 A075098 A075099 * A075101 A075102 A075103
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KEYWORD
| hard,more,nonn,obsc
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AUTHOR
| Colin Mallows (colinm(AT)research.avayalabs.com), Aug 31 2002
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