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%I #12 Aug 21 2014 17:57:51
%S 0,2,5,7,15,20,28,36,41,54,75,96,109,130,143,164,185,198,219,240,253,
%T 274,308,329,363,397,418,452,473,507,541,562,596,617,651,685,706,740,
%U 774,795,829,850,884,918,973,1007,1062,1117,1151,1206,1261,1295,1350
%N Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.
%C 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.
%e S = infinite Fibonacci word A014675, B = (s(0)); that is, (m,k) = (0,0);
%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)
%e B'(1) = (2,1)
%e B'(2) = (2,1,2)
%e B'(3) = (2,1,2,1)
%e B'(4) = (2,1,2,1,2)
%e B'(5) = (2,1,2,1,2,2)
%e S* = (2,1,2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...),
%e with index sequence (0,2,5,7,15,...)
%t 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*)
%t 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 *)
%t q1 = Differences[q] (* A245922 *)
%Y Cf. A245920, A245922.
%K nonn,obsc
%O 0,2
%A _Clark Kimberling_ and _Peter J. C. Moses_, Aug 07 2014
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