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

Author

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

Keywords

• nonn
• cons

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...

Examples

			[2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...] =  2.3663046946532726566824972058...
		

Crossrefs

Cf. A246127, A246128, A014675, A245975.

Programs

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 *)