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%I #2 Mar 30 2012 17:34:22
%S 11,7,3,2,12,8,4,3,3,4,12,8,8,3,4,12,12,9,5,1,9,1,5,12,3,4,12,8,1,5,
%T 12,9,1,5,12,9,12,9,5,1,9,1,5,12,3,4,12,8,3,4,12,8,1,5,12,9,9,1,5,12,
%U 12,9,5,1,2,11,7,3,2,2,3,11,7,7,2,3,11,7,2,3,11,2,11,7,3,2,2,3,11,7,2,3,11,7
%N A ten-out-of-twelve fractal substitution sequence based on 7th like chords: 1->{1,5,9,12}->{A,Db,F,Ab}, 3->{3,7,11,2}->{B,E, G,Bb}, 4->{4,8,12,3}->{C,Eb,Ab,B}.
%C This sequence uses nine basic tones with {2}->Bb as a terminator. If the triad chords are used only it has less of a blue sound and only has nine tones. It also goes from 4!=24 permutations possible to 3!=6: that significantly lowers the possible out comes for automated output. Substitutions with the permutations increase the variability of this type of sequence.
%F Substitution function that skip {6,10} and use {2} as a terminator:
%F s[1] = {12, 9, 5, 1};
%F s[2] = {2};
%F s[3] = {11, 7, 3, 2};
%F s[4] = {12, 8, 4, 3};
%F s[5] = Permutations[s[1]][[12]];
%F s[6] = {6};
%F s[7] = Permutations[s[3]][[12]]; s[8] = Permutations[s[4]][[12]];
%F s[9] = Permutations[s[1]][[23]];
%F s[10] = {10};
%F s[11] = Permutations[s[3]][[23]];
%F s[12] = Permutations[s[4]][[23]];
%F Permutations {12,23} are used for spread, but any two combinations of 24:Binomial[24,2]=46 can be used if the first substitutions are the reference state.
%t Clear[s]
%t s[1] = {12, 9, 5, 1};
%t s[2] = {2};
%t s[3] = {11, 7, 3, 2};
%t s[4] = {12, 8, 4, 3};
%t s[5] = Permutations[s[1]][[12]];
%t s[6] = {6};
%t s[7] = Permutations[s[3]][[12]]; s[8] = Permutations[s[4]][[12]];
%t s[9] = Permutations[s[1]][[23]];
%t s[10] = {10};
%t s[11] = Permutations[s[3]][[23]];
%t s[12] = Permutations[s[4]][[23]];
%t t[a_] := Flatten[s /@ a]; p[0] = {1, 3, 4}; p[1] = t[p[0]];
%t p[n_] := t[p[n - 1]];
%t p[3]
%Y Cf. A132150, A132160.
%K nonn,uned
%O 1,1
%A _Roger L. Bagula_, Nov 08 2007, Nov 09 2007
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