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A245978 Index sequence for limit-reversing 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).

%I A245978 #9 Jun 14 2025 21:32:21
%S A245978 3,8,11,16,21,29,37,42,50,63,71,84,92,105,118,126,139,152,173,194,207,
%T A245978 228,249,262,283,296,317,338,351,372,406,427,461,482,516,550,571,605,
%U A245978 626,660,694,715,749,783,804,838,859,893,927,948,982,1016,1071,1126
%N A245978 Index sequence for limit-reversing 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).
%C A245978 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*.
%C A245978 ...
%C A245978 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.
%e A245978 S = the infinite Fibonacci word A014675, with B = (s(2), s(3)); that is, (m,k) = (2,3)
%e A245978 S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
%e A245978 B'(0) = (2,2)
%e A245978 B'(1) = (2,2,1)
%e A245978 B'(2) = (2,2,1,2)
%e A245978 B'(3) = (2,2,1,2,1)
%e A245978 B'(4) = (2,2,1,2,1,2)
%e A245978 B'(5) = (2,2,1,2,1,2,2)
%e A245978 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,...)
%t A245978 z = 140; 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] *)
%t A245978 cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; q = Join[{3}, Rest[Accumulate[Join[{1}, Table[p[n], {n, 0, z}]]]]] (* A245978 *)
%t A245978 q1 = Differences[q]  (* A245979 *)
%Y A245978 Cf. A014675, A245977, A245979, A245921, A014675.
%K A245978 nonn
%O A245978 0,1
%A A245978 _Clark Kimberling_ and _Peter J. C. Moses_, Aug 10 2014