

A245976


Decimal expansion of the number whose continued fraction is given by A245920 (limitreverse 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
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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 limitreverse, 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



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



KEYWORD



AUTHOR



STATUS

approved



