A246128 Index sequence for limit-block extending the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

0, 2, 7, 10, 15, 23, 31, 36, 44, 49, 57, 70, 78, 91, 104, 112, 125, 138, 159, 193, 214, 248, 282, 303, 337, 371, 392, 426, 447, 481, 515, 536, 570, 591, 625, 659, 680, 714, 748, 803, 892, 981, 1036, 1125, 1180, 1269, 1358, 1413, 1502, 1557, 1646, 1735, 1790

Offset: 0

Author

Clark Kimberling and Peter J. C. Moses, Aug 15 2014

Keywords

• nonn

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), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^.

...

The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246127.

Examples

			S = the infinite Fibonacci word A014675, with 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) = (2,2)
B'(2) = (2,2,1)
B'(3) = (2,2,1,2)
B'(4) = (2,2,1,2,1)
B'(5) = (2,2,1,2,1,2)
S^ = (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...),
with index sequence (0,2,7,10,15,...)
		

Crossrefs

Cf. A245921, A246127, A246129, A014675.

Programs

Mathematica

seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s = Differences[Table[Floor[n*GoldenRatio], {n, 10000}]]; t = {{2}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; t1 = Last[t] (*A246127*)
q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)