A245921 Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

0, 2, 5, 7, 15, 20, 28, 36, 41, 54, 75, 96, 109, 130, 143, 164, 185, 198, 219, 240, 253, 274, 308, 329, 363, 397, 418, 452, 473, 507, 541, 562, 596, 617, 651, 685, 706, 740, 774, 795, 829, 850, 884, 918, 973, 1007, 1062, 1117, 1151, 1206, 1261, 1295, 1350

Offset: 0

Author

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

Keywords

• nonn
• obsc

Comments

Suppose 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.

Examples

			S = infinite Fibonacci word A014675, B = (s(0)); that is, (m,k) = (0,0);
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2)
B'(1) = (2,1)
B'(2) = (2,1,2)
B'(3) = (2,1,2,1)
B'(4) = (2,1,2,1,2)
B'(5) = (2,1,2,1,2,2)
S* = (2,1,2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...),
with index sequence (0,2,5,7,15,...)
		

Crossrefs

Cf. A245920, A245922.

Programs

Mathematica

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind] (*finds the position of the SECOND appearance of seqtofind. Example: seqPosition2[{1,2,3,4,2,3},{2}] = 5*)
A014675 = Nest[Flatten[# /. {1 -> 2, 2 -> {2, 1}}] &, {1}, 25]; ans = Join[{A014675[[p[0] = pos = seqPosition2[A014675, #] - 1]]}, #] &[{A014675[[1]]}]; cfs = Table[A014675 = Drop[A014675, pos - 1]; ans = Join[{A014675[[p[n] = pos = seqPosition2[A014675, #] - 1]]}, #] &[ans], {n, z}]; q = -1+Accumulate[Join[{1}, Table[p[n], {n, 0, z}]]] (* A245921 *)
q1 = Differences[q] (* A245922 *)