A246129 Decimal expansion of the number whose continued fraction is given by A246127 (limiting block extension of an infinite Fibonacci word).

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...

Links

Table of n, a(n) for n=1..86.

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

Cf. A246127, A246128, A014675, A245975.

Sequence in context: A333304 A248896 A102402 * A359123 A124498 A197334

Adjacent sequences: A246126 A246127 A246128 * A246130 A246131 A246132

Keyword

nonn,cons

Author

Clark Kimberling and Peter J. C. Moses, Aug 15 2014

Status

approved