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%I #13 Sep 04 2014 09:26:31
%S 1,2,1,3,2,3,1,3,1,2,1,3,2,3,1,1,3,1,1,2,1,3,2,3,1,1,1,2,1,3,2,3,1,3,
%T 1,2,1,3,2,3,1,1,3,1,1,2,1,3,2,3,1,1,2,1,3,2,3,1,1,1,2,1,3,2,3,1,1,2,
%U 1,3,2,3,1,3,1,2,1,3,2,3,1,1,3,1,1,2,1,3,2,3,1,1,2,1,3,2,3,1,1,2,1,3,2,3,1
%N Substitution 1->{1, 2, 1, 3, 2, 3, 1}, 2->{3}, 3->{1}, starting with 1.
%C Original name was: 223 Pisot cubic substitution with characteristic polynomial: x^3-3*x^2-2*x-2.
%e Start: 1
%e Rules:
%e 1 --> 1213231
%e 2 --> 3
%e 3 --> 1
%e -------------
%e 0: (#=1)
%e 1
%e 1: (#=7)
%e 1213231
%e 2: (#=25)
%e 1213231312132311311213231
%e 3: (#=91)
%e 1213231312132311311213231112132313121323113112132311213231112132311213231312132311311213231
%e 4: (#=337)
%e 12132313121323113112132311121323131213231131121323112132 ...
%t s[1] = {1, 2, 1, 3, 2, 3, 1}; s[2] = {3}; s[3] = {1}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]] aa = p[4]
%o (Python)
%o from itertools import chain
%o A107337 = [1]
%o for _ in range(4):
%o ....A107337 = list(chain.from_iterable(([1,2,1,3,2,3,1] if d == 1 else [3] if d == 2 else [1] for d in A107337)))
%o # _Chai Wah Wu_, Sep 03 2014
%K nonn,easy
%O 0,2
%A _Roger L. Bagula_, May 22 2005
%E Edited by _Joerg Arndt_, Jun 26 2011
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