

A152214


The digits of Pi from BBP modulo 4 as symbols {0,1,2,3} interpreted as groups of symbols {a,b,c} in a 4^3==64 symbol set.


0



59, 25, 38, 2, 19, 37, 52, 33, 46, 48, 59, 37, 25, 25, 32, 8, 26, 1, 10, 34, 55, 48, 6, 15, 47, 63, 14, 50, 32, 21, 14, 53, 29, 17, 62, 49, 14, 5, 2, 22, 38, 19, 60, 55, 16, 52, 52, 11, 56, 41, 42, 40, 18, 20, 42, 57, 47, 11, 21, 35, 29, 40, 54, 18, 41, 61, 8, 43, 36, 23, 6, 4, 14, 8
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OFFSET

0,1


COMMENTS

This sequence analogous to the 4 base DNA coding problem into length 3 codons. It interprets the digits of Pi as if they were a DNA coding sequence. There are 24 =4! ways that you can assign bases to the numbers/ symbols. This sequence is the answer to the question of how a normal "noise" with an equal appearing digits set could also be a code of information of an higher type.


LINKS



FORMULA

BBP Pi digits modulo 4: a(n)=Floor[Mod[(4/(8*n + 1)  2/(8*n + 4)  1/(8*n + 5)  1/(8*n + 6))*16^n, 4]].


MATHEMATICA

Clear[a0, b0, c0, f, a, b, c]; f[n_] = Floor[Mod[(4/(8*n + 1)  2/(8*n + 4)  1/(8*n + 5)  1/(8*n + 6))*16^n, 4]]; a0 = Table[{f[n], f[n + 1], f[n + 2]}, {n, 0, 300, 3}]; b0 = Flatten[Table[{a, b, c}, {a, 0, 3}, {b, 0, 3}, {c, 0, 3}], 2]; Length[b0]; c0 = Delete[Union[Flatten[Table[If[a0[[n]] == b0[[m]], {n, m}, {}], {n, 1, Length[a0]}, {m, 1, Length[b0]}], 1]], 1]; d0 = Table[c0[[n]][[2]], {n, 1, Length[c0]}]


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



