A245977 Limit-reverse of the infinite Fibonacci word A014675 = (s(0),s(1),...) = (2,1,2,2,1,2,1,2, ...) using initial block (s(2),s(3)) = (2,2).

2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2

Offset: 0

Author

Clark Kimberling and Peter J. C. Moses, Aug 10 2014

Keywords

• nonn

Comments

Suppose, as in A245920, that S = (s(0),s(1),s(2),...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S. (It is assumed that A014675 is such a sequence.) Let B = B(m,k) = (s(m-k),s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1),s(m(1)-k),...,s(m(1))). Let m(2) be the least i > m(1) such that (s(i-k-1),s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2),s(m(2)-k-1),...,s(m(2))). Continuing in this manner gives a sequence of blocks B(m(n),k+n). Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.

...

The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245921 and A245978.

Apparently a(n) = A082389(n+2) = A059426(n+1). - R. J. Mathar, Sep 01 2014

Examples

			S = the infinite Fibonacci word A014675, with B = (s(2), s(3)); that is, (m,k) = (2,3)
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2,2)
B'(1) = (2,2,1)
B'(2) = (2,2,1,2)
B'(3) = (2,2,1,2,1)
B'(4) = (2,2,1,2,1,2)
B'(5) = (2,2,1,2,1,2,2)
S* = (2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,...), with index sequence (3,8,11,16,21,29,...)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A245978, A245979, A245920, A014675.

Programs

Mathematica

z = 100; 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}]]; ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[3]], s[[4]]}]; (* Initial block is (s(3),s(4)) [OR (s(2),s(3)) if using offset 0] *); cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]] (* A245977 *)