A245975 Decimal expansion of the number whose continued fraction is the (2,1)-version of the infinite Fibonacci word A014675.

2, 7, 0, 2, 9, 3, 8, 3, 5, 8, 0, 2, 2, 5, 1, 0, 2, 9, 4, 4, 4, 5, 0, 5, 0, 9, 7, 4, 6, 9, 3, 0, 0, 3, 7, 3, 4, 5, 3, 2, 7, 0, 3, 1, 5, 2, 9, 0, 9, 2, 3, 1, 2, 2, 1, 4, 0, 1, 4, 1, 2, 0, 0, 0, 3, 0, 7, 7, 4, 6, 9, 8, 3, 7, 2, 6, 6, 4, 8, 0, 2, 7, 0, 3, 5, 5

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,...); see Example.

Examples

			[2,1,2,2,1,2,1,2,2,...] = 2.702938358022510294445050974693003734532...
		

Crossrefs

Cf. A014675, A245976.

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] (* A245976 *)