cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A245920 Limit-reverse of the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

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, 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, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2
Offset: 0

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

Keywords

Comments

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.
For numbers represented by taking S and S* as continued fractions, see A245975 and A245976. If S is taken to be the classical (0,1)-version of the infinite Fibonacci word, then S* is obtained from the present sequence by substituting 0 for 2 throughout, as in A241422.
The limit-reverse, S*, is analogous to a limiting block extension, S^, defined at A246127. The essential difference is that S^ is formed by extending each new block one term to the right, whereas S* is formed by extending each new block one term to the left (and then reversing).

Examples

			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,...)

Links

Crossrefs

Cf. A245921, A245922, A003849, A014675, A245975, A245976.

Programs

  • Mathematica
    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[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]]

A245921 Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

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
Offset: 0

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

Keywords

Comments

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.

Examples

			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,...)

Crossrefs

Cf. A245920, A245922.

Programs

  • Mathematica
    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 *)
q1 = Differences[q] (* A245922 *)

A245935 First differences of A245934; see Comments.

Original entry on oeis.org

2, 2, 3, 5, 7, 5, 7, 5, 5, 12, 12, 5, 12, 12, 12, 17, 12, 17, 12, 12, 17, 12, 17, 12, 17, 41, 29, 41, 29, 29, 41, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 29, 70, 70, 29, 70, 70, 70, 29, 70, 70, 29, 70, 70, 70
Offset: 1

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

Keywords

Comments

It appears that every term is a term of A002965. The sequence A245935 arises from A245933 and A245934, in which the limit-reverse of certain sequences is defined, as follows. Suppose 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 A006337 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*. (Since Beatty sequences are usually written with offset 1, the above definition is adapted accordingly, so that s(n) = A006337(n+1) for n >= 0.)
...
The sequence (m(i)), where m(0) = 1, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245934.

Examples

			S = A006337 (the Beatty sequence of sqrt(2)), re-indexed to start with s(0) = 1, with B = (s(0)); that is, (m,k) = (0,0)
S = (1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,...)
B'(0) = (1)
B'(1) = (2,1)
B'(2) = (1,2,1)
B'(3) = (1,2,1,1)
B'(4) = (1,2,1,1,2)
B'(5) = (1,2,1,1,2,1)
S* = (1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,...), with index sequence (1,3,5,8,13,20,25,32,37,...), with difference sequence (2,2,3,5,7,5,7,5,5,12,12,...).

Crossrefs

Cf. A245934, A245933, A245935, A245922.

Programs

  • Mathematica
    z = 140; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = Sqrt[2]; s =  Differences[Table[Floor[n*x], {n, 1, z^2}]];  ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; q = Accumulate[Join[{1}, Table[p[n], {n, 0, z}]]]  (* A245934 *)
q1 = Differences[q]  (* A245935 *)
Showing 1-3 of 3 results.

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