A245977 -id:A245977 - cp's OEIS frontend

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).

3, 8, 11, 16, 21, 29, 37, 42, 50, 63, 71, 84, 92, 105, 118, 126, 139, 152, 173, 194, 207, 228, 249, 262, 283, 296, 317, 338, 351, 372, 406, 427, 461, 482, 516, 550, 571, 605, 626, 660, 694, 715, 749, 783, 804, 838, 859, 893, 927, 948, 982, 1016, 1071, 1126

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 10 2014

nonn

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*.
...
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.

			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,...)

Cf. A014675, A245977, A245979, A245921, A014675.

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 *)
q1 = Differences[q]  (* A245979 *)

A245979 First differences of A245978.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling and Peter J. C. Moses, Aug 10 2014

nonn

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*.
...
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.

			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,...)

Cf. A245978, A245977, A245979, A245921, A014675, A000045.

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 *)
q1 = Differences[q]  (* A245979 *)

cp's OEIS Frontend

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).

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