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%I #20 Nov 04 2020 05:49:49
%S 3,9,16,22,27,32,38,45,51,56,62,67,73,80,86,91,96,102,109,115,120,126,
%T 133,139,144,150,157,163,168,173,179,186,192,197,203,208,214,221,227,
%U 232,237,243,250,256,261,267,274,280,285,290,296,303,309,314,320,327
%N Indices of c in the sequence closed under a -> abc, b -> ab, c -> b . Start with a.
%C Indices of 3 in A214640.
%H Alois P. Heinz, <a href="/A214644/b214644.txt">Table of n, a(n) for n = 1..1000</a>
%e Start: a
%e Rules:
%e a --> abc
%e b --> ab
%e c --> b
%e -------------
%e 0: (#=1)
%e a
%e 1: (#=3)
%e abc
%e 2: (#=6)
%e abcabb
%e 3: (#=13)
%e abcabbabcabab
%e 4: (#=29)
%e abcabbabcabababcabbabcababcab
%e 5: (#=64)
%e abcabbabcabababcabbabcababcababcabbabcabababcabbabcababcabbabcab
%e 3 9 16 22 27 ... (positions of 'c')
%p S:= proc(n) option remember; `if`(n=0, [1], subs(x=[1, 2, 3][],
%p y=[1, 2][], z=2, subs(1=x, 2=y, 3=z, S(n-1))))
%p end:
%p a:= proc(n) option remember; local i, k; i:=1;
%p for k from 1+`if`(n=1, 0, a(n-1)) do
%p while nops(S(i))<k do i:=i+1 od;
%p if S(i)[k] = 3 then return k fi
%p od
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jul 28 2012
%t SubstitutionSystem[{"a" -> {"a", "b", "c"}, "b" -> {"a", "b"}, "c" -> {"b"}}, {"a"}, 8] // Last // Position[#, "c"]& // Flatten (* _Jean-François Alcover_, Nov 04 2020 *)
%Y Cf. A214640, A214641, A214642.
%K nonn,easy
%O 1,1
%A _Philippe Deléham_, Jul 24 2012