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A245979
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First differences of A245978.
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3
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5, 3, 5, 5, 8, 8, 5, 8, 13, 8, 13, 8, 13, 13, 8, 13, 13, 21, 21, 13, 21, 21, 13, 21, 13, 21, 21, 13, 21, 34, 21, 34, 21, 34, 34, 21, 34, 21, 34, 34, 21, 34, 34, 21, 34, 21, 34, 34, 21, 34, 34, 55, 55, 34, 55, 55, 34, 55, 34, 55, 55, 34, 55, 55, 34, 55, 34
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OFFSET
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1,1
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COMMENTS
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It appears that every term is a Fibonacci number, as in A000045. The sequence A245979 arises from A245977 and A245978 in the following manner.
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*.
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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. Then A245979 is the difference sequence of A245978, as in the Example.
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LINKS
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EXAMPLE
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S = the infinite Fibonacci word A014675, with B = (s(2),s(3)); that is, (m,k) = (2,3)
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2,2)
B'(1) = (2,2,1)
B'(2) = (2,2,1,2)
B'(3) = (2,2,1,2,1)
B'(4) = (2,2,1,2,1,2)
B'(5) = (2,2,1,2,1,2,2)
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,...), of which the difference sequence is (5,3,5,5,8,8,5,8,13,8,...)
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MATHEMATICA
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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] *) 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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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