%I #9 Jul 12 2015 20:34:45
%S 0,0,3,4,10,11
%N 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.
%C I believe a(2n) = a(n)+ 2^n. I think a(7) = 28.
%C _Benoit Jubin_ (Jan 24 2009) suggests replacing "monoid" in the definition by "semigroup".
%C 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.
%e a(3) = 3 because we need only xx, xy, yy to generate each of xxx, xxy, xyx, yxx, xyy, yxy, yyx, yyy.
%Y Cf. A075099, A003313, A124677.
%K hard,more,nonn,obsc
%O 1,3
%A _Colin Mallows_, Aug 31 2002