A245976 Decimal expansion of the number whose continued fraction is given by A245920 (limit-reverse of an infinite Fibonacci word).

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

Offset: 1

Author

Clark Kimberling and Peter J. C. Moses, Aug 08 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 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

Examples

			[2,1,2,1,2,2,1,2,1,2,...] =  2.72994419476785022907837430700599816738...
		

Crossrefs

Cf. A245920, A245975.

Programs

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 *)