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A246129
Decimal expansion of the number whose continued fraction is given by A246127 (limiting block extension of an infinite Fibonacci word).
3
2, 3, 6, 6, 3, 0, 4, 6, 9, 4, 6, 5, 3, 2, 7, 2, 6, 5, 6, 6, 8, 2, 4, 9, 7, 2, 0, 5, 8, 6, 1, 4, 5, 6, 9, 1, 0, 0, 8, 1, 9, 9, 4, 8, 1, 0, 4, 0, 9, 5, 8, 9, 1, 0, 9, 3, 0, 5, 4, 1, 0, 2, 7, 1, 3, 8, 5, 3, 7, 7, 9, 1, 0, 1, 9, 1, 3, 5, 3, 1, 1, 3, 4, 6, 2, 6
OFFSET
1,1
COMMENTS
The (2,1)-version of the infinite Fibonacci word, A014675, as a sequence, is (2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...). Its limiting block extension, A246128, is the sequence (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...), which is the continued fraction for 2.366304...
EXAMPLE
[2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...] = 2.3663046946532726566824972058...
MATHEMATICA
seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[ Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 1]]]] &[seqtofind]; s = Differences[Table[Floor[n*GoldenRatio], {n, 10000}]]; t = {{2}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; t1 = Last[t] (*A246127*)
q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)
u = N[FromContinuedFraction[t1], 100]
r = RealDigits[u][[1]] (* A246129 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
STATUS
approved