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A245921
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Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.
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9
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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,...)
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,obsc
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AUTHOR
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STATUS
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approved
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