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 A245976 Decimal expansion of the number whose continued fraction is given by A245920 (limit-reverse of an infinite Fibonacci word). 2
 2, 7, 2, 9, 9, 4, 4, 1, 9, 4, 7, 6, 7, 8, 5, 0, 2, 2, 9, 0, 7, 8, 3, 7, 4, 3, 0, 7, 0, 0, 5, 9, 9, 8, 1, 6, 7, 3, 8, 1, 8, 8, 7, 0, 1, 6, 4, 0, 5, 2, 5, 8, 0, 2, 0, 4, 9, 2, 7, 5, 4, 1, 0, 1, 9, 9, 6, 3, 3, 6, 2, 4, 3, 4, 5, 7, 7, 8, 6, 7, 1, 3, 1, 1, 6, 8 (list; constant; graph; refs; listen; history; text; internal format)
 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 limit-reverse, A245920, is the sequence (2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...), which is the continued fraction for 2.729944... For the (0,1)-version of the infinite Fibonacci word 0100101001001... (A003849), the decimal expansion is the same except for the first digit. That is 0.729944194... . - Gandhar Joshi, Mar 28 2024 LINKS Table of n, a(n) for n=1..86. EXAMPLE [2,1,2,1,2,2,1,2,1,2,...] = 2.72994419476785022907837430700599816738... MATHEMATICA z = 300; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 2]]] &[seqtofind]; x = GoldenRatio; s = Differences[Table[Floor[n*x], {n, 1, z^2}]]; (* A014675 *) x1 = N[FromContinuedFraction[s], 100] r1 = RealDigits[x1, 10] (* A245975 *) ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]] (* A245920 *) x2 = N[FromContinuedFraction[rcf], z] r2 = RealDigits[x2, 10] (* this sequence *) CROSSREFS Cf. A245920, A245975. Sequence in context: A278419 A197133 A178206 * A245216 A223149 A222853 Adjacent sequences: A245973 A245974 A245975 * A245977 A245978 A245979 KEYWORD nonn,cons AUTHOR Clark Kimberling and Peter J. C. Moses, Aug 08 2014 STATUS approved

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Last modified May 20 11:42 EDT 2024. Contains 372712 sequences. (Running on oeis4.)