Search: id:A133769
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%I A133769
%S A133769 9,23,0,9,23,0,12,0,11,7,20,4,9,0,0,9,10,0,13,0,7,9,0,8,20,12,23,7,0,0,
%T A133769 11,0,3,0,9,2,4,11,7,9,8,3,6,0,0,10,9,15,0,17,4,0,12,4,7,11,6,5,12,0,11,
%U A133769 4,0,11,15,9,10,0,13,0,5,5,12,2,8,11,12,0,8,15,0,23,6,0,8,3,6,11,1,10
%N A133769 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. 
               Here a(n)=b(n) have been eleminated from the 12 tone substitutions.
%C A133769 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 worried about a(n)<>b(n) and constructed sequence of this 
               type.
%F A133769 a(n)=A133270(n); b(n)=A133269(n); a(n)<>b(n); bn1(n)=Mod[A004001(n),2]; 
               bn2[n]=Mod[A005229(n),2]; c(n)=bn1(n)*a(n)+bn2(n)*b(n)
%t A133769 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]; P11 = Flatten[Table[If[P1[[ n]] - P2[[n]] == 0, 
               {}, P1[[n]]], {n, 1, Length[P1]}]]; P22 = Flatten[Table[If[P1[[n]] 
               - P2[[ n]] == 0, {}, P2[[n]]], {n, 1, Length[P2]}]]; aout = Table[Mod[Conway[n], 
               2]*P11[[n]] + Mod[Mallows[n], 2]*P22[[n]], {n, 1, Min[Length[P11], 
               Length[P22]]}]
%Y A133769 Cf. A133270, A133269, A004001, A005229.
%Y A133769 Sequence in context: A123833 A084023 A156342 this_sequence A165484 A009235 
               A031023
%Y A133769 Adjacent sequences: A133766 A133767 A133768 this_sequence A133770 A133771 
               A133772
%K A133769 nonn,uned
%O A133769 1,1
%A A133769 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 02 2008

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