%I A133768
%S A133768 2,9,0,12,9,0,12,0,8,12,6,6,12,0,0,11,4,0,23,0,9,10,0,8,23,4,13,11,0,0,
%T A133768 11,0,16,0,3,7,11,4,7,9,3,7,10,0,0,20,1,10,0,8,6,0,4,8,11,5,13,9,2,0,11,
%U A133768 3,0,10,9,9,11,0,16,0,4,5,12,4,7,11,2,0,9,15,0,11,2,0,1,5,5,12,2,13,11
%N A133768 An example of a sequence composed of two twelve tone substitutions and 
               two binary sequences: c(n)=bn1(n)*a(n)+bn2(n)*b(n) The resulting 
               sequence is a substitution on 24 tones and zero: 25 tones total.
%C A133768 Suppose that you have two known sequences of finite length on a limited 
               alphabet: a[n], b[n] Such that a[n]<>b[n] ( doesn't equal) and you 
               have two binary sequences bn1[n] and bn2[n] such that you can construct: 
               c[n]=bn1[n]*a[n]+bn2[n]*b[n] Then there are four possibilities for 
               the sums: c[n]=0 c[n]=a[n] c[n]=b[n] c[n]=a[n]+b[n] Since a[n]<>b[n], 
               these are distinguishable, so that using these two sequences a simple 
               code of two binary sequences can be coded on a single sequence with 
               a limited alphabet. My 12 tone sequences would limit the total alphabet 
               to 24 characters. By skipping any case where a[n]=b[n], the use of 
               any general set of {a,b} sequences is possible. It doesn't matter 
               if they repeat in one sequence of the other, just that they aren't 
               the same. The information is in the binary sequence and this can 
               be taken as a coding of any number of bits in pairs. It can work 
               on lower substitutions as well, but maybe not as well since it is 
               more likely in a level 3 substitution that a[n]=b[n] than in a 12th 
               level substitution. Something like a Fibonacci sequence would involve 
               too large of numbers, so substitutions are more natural. In this 
               sequence I haven't worried about a(n)=b(n) and just constructed a 
               general sequence of this type.
%F A133768 a(n)=A133270(n); b(n)=A133269(n); bn1(n)=Mod[A004001(n),2]; bn2[n]=Mod[A005229(n),
               2]; c(n)=bn1(n)*a(n)+bn2(n)*b(n)
%t A133768 Clear[a, Conway, Mallows] (* sequence A004001*) Conway[1] = Conway[2] 
               = 1; Conway[n_Integer?Positive] := Conway[n] = Conway[Conway[n - 
               1]] + Conway[n - Conway[n - 1]] (* sequence A005229*) Mallows[n_Integer?Positive] 
               := Mallows[n] = Mallows[Mallows[n - 2]] + Mallows[n - Mallows[n - 
               2]] Mallows[0] = Mallows[1] = Mallows[2] = 1; (* minor A133270*) 
               Clear[s, p] s[i_] = {i, If[i + 3 > 12, i - 7, i + 3], If[i + 7 > 
               12, i - 5, i + 7], If[i + 10 > 12, i - 2, i + 10]}; t[a_] := Flatten[s 
               /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; P1 = p[4]; 
               (*Major A133269*) Clear[s, p] s[i_] = {i, If[i + 4 > 12, i - 8, i 
               + 4], If[i + 7 > 12, i - 5, i + 7], If[i + 11 > 12, i - 1, i + 11]}; 
               t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n 
               - 1]]; P2 = p[4]; aout = Table[Mod[Conway[n], 2]*P1[[n]] + Mod[Mallows[n], 
               2]*P2[[n]], { n, 1, Min[Length[P1], Length[P2]]}]
%Y A133768 Cf. A133270, A133269, A004001, A005229.
%Y A133768 Sequence in context: A140239 A021348 A020817 this_sequence A085333 A128892 
               A105548
%Y A133768 Adjacent sequences: A133765 A133766 A133767 this_sequence A133769 A133770 
               A133771
%K A133768 nonn,uned
%O A133768 1,1
%A A133768 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 02 2008