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%I #10 Sep 01 2014 05:00:02
%S 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,
%T 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,
%U 2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,2
%N 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).
%C 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 ...
%C 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.
%C Apparently a(n) = A082389(n+2) = A059426(n+1). - _R. J. Mathar_, Sep 01 2014
%H Clark Kimberling, <a href="/A245977/b245977.txt">Table of n, a(n) for n = 0..300</a>
%e S = the infinite Fibonacci word A014675, with B = (s(2), s(3)); that is, (m,k) = (2,3)
%e S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
%e B'(0) = (2,2)
%e B'(1) = (2,2,1)
%e B'(2) = (2,2,1,2)
%e B'(3) = (2,2,1,2,1)
%e B'(4) = (2,2,1,2,1,2)
%e B'(5) = (2,2,1,2,1,2,2)
%e 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 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 *)
%Y Cf. A245978, A245979, A245920, A014675.
%K nonn
%O 0,1
%A _Clark Kimberling_ and _Peter J. C. Moses_, Aug 10 2014
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