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 1-5 of 5 results.

A246140 Limiting block extension of A006337 (difference sequence of the Beatty sequence for sqrt(2)) with first term as initial block.

Original entry on oeis.org

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

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 17 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 A006337 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128. If the sequence S is given with offset 1, then the role played by s(0) in the above definitions is played by s(1) instead, as in the case of A246140 and A246141.
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. 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 = A006337, with B = (s(1)); that is, (m,k) = (1,0)
S = (1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2,...)
B'(0) = (1)
B'(1) = (1,2)
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, 2, 1, 1, 2, 1,...),
with index sequence (1,3,6,8,15,...)

Links

Crossrefs

Cf. A246141, A246127, A246142, A246144, A246146, A006337.

Programs

  • Mathematica
    seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s =  Differences[Table[Floor[n*Sqrt[2]], {n, 10000}]]; Take[s, 60]
t = {{1}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # +Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (* A246140 *)
Accumulate[Table[p[k], {k, 0, n - 1}]] (* A246141 *)

A246145 Index sequence for limit-block extending A000002 (Kolakoski sequence) with first term as initial block.

Original entry on oeis.org

1, 4, 13, 16, 51, 78, 97, 124, 178, 247, 322, 402, 475, 578, 623, 746, 842, 1030, 1111, 1173, 1454, 1481, 2071, 2385, 2686, 4395, 5402, 5587, 5932, 6150, 6622, 6767, 7038, 7311, 7461, 10404, 10674, 12797, 18358, 20169, 20575, 21667, 23244, 25101, 26224
Offset: 1

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 17 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 A000002 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128. If the sequence S is given with offset 1, then the role played by s(0) in the above definitions is played by s(1) instead, as in the case of A246144 and A246145.
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. 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 = A000002, with B = (s(1)); that is, (m,k) = (1,0)
S = (1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,...)
B'(0) = (1)
B'(1) = (1,1)
B'(2) = (1,1,2)
B'(3) = (1,1,2,2)
B'(4) = (1,1,2,2,1)
B'(5) = (1,1,2,2,1,2)
S^ = (1,1,2,2,1,2,1,1,2,1,2,2,1,1,2,1,1,...),
with index sequence (1,4,13,16,51,78,97,124,178,247,322,...)

Crossrefs

Cf. A246140, A246142, A246144, A246146, A000002.

Programs

  • Mathematica
    seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; n = 30; s = Prepend[Nest[Flatten[Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, n], 1]; (* A246144 *)
Take[s, 30]
t = {{1}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (* A246144*)
Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246145*)

A246147 Index sequence for limit-block extending A010060 (Thue-Morse sequence) with first term as initial block.

Original entry on oeis.org

0, 3, 6, 12, 20, 30, 36, 68, 92, 116, 132, 156, 180, 228, 260, 308, 356, 420, 452, 516, 564, 612, 676, 708, 756, 804, 836, 900, 948, 996, 1076, 1188, 1268, 1316, 1460, 1572, 1716, 1764, 1844, 1956, 2100, 2212, 2292, 2340, 2484, 2740, 2868, 3060, 3252, 3380
Offset: 0

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 17 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 A010060 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246147.
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. 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 = A010060, with B = (s(0)); that is, (m,k) = (0,0)
S = (0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,...)
B'(0) = (0)
B'(1) = (0,1)
B'(2) = (0,1,1)
B'(3) = (0,1,1,0)
B'(4) = (0,1,1,0,0)
B'(5) = (0,1,1,0,0,1)
S^ = (0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,...),
with index sequence (0,3,6,12,20,30,36,68,...)

Crossrefs

Cf. A246146, A246127, A246140, A246142, A246144, A010060.

Programs

  • Mathematica
    seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 14]; (* A010060 *)
Take[s, 60]
t = {{0}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (* A246146 *)
-1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (* A246147 *)

A246141 Index sequence for limit-block extending A006337 (difference sequence of the Beatty sequence for sqrt(2)) with first term as initial block.

Original entry on oeis.org

1, 3, 6, 8, 15, 20, 27, 32, 37, 49, 66, 78, 90, 107, 119, 136, 148, 160, 177, 189, 206, 235, 247, 276, 305, 317, 346, 375, 404, 416, 445, 474, 486, 515, 556, 585, 614, 655, 684, 725, 754, 783, 824, 853, 894, 923, 964, 993, 1022, 1063, 1092, 1133, 1162, 1191
Offset: 1

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 17 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 A006337 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128. If the sequence S is given with offset 1, then the role played by s(0) in the above definitions is played by s(1) instead, as in the case of A246140 and A246141.
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. 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 = A006337, with B = (s(1)); that is, (m,k) = (1,0)
S = (1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2,...)
B'(0) = (1)
B'(1) = (1,2)
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, 2, 1, 1, 2, 1,...),
with index sequence (1,3,6,8,15,...)

Links

Crossrefs

Cf. A246140, A246127, A246142, A246144, A246147, A006337.

Programs

  • Mathematica
    seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s =  Differences[Table[Floor[n*Sqrt[2]], {n, 10000}]]; Take[s, 60]
t = {{1}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # +Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (* A246140 *)
Accumulate[Table[p[k], {k, 0, n - 1}]] (* A246141 *)

A246143 Index sequence for limit-block extending A004539 (base-2 representation of sqrt(2)) with first term as initial block.

Original entry on oeis.org

1, 3, 17, 18, 35, 45, 239, 341, 470, 1180, 1230, 2205, 5318, 45652, 68042, 73350, 119458, 388804, 475300, 773496, 836779, 845397, 1133816, 2670010
Offset: 1

Views

Author

Clark Kimberling and Peter J. C. Moses, Aug 17 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 A004539 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).
The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128. If the sequence S is given with offset 1, then the role played by s(0) in the above definitions is played by s(1) instead, as in the case of A246142 and A246143.
Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. 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 = A004539, with B = (s(1)); that is, (m,k) = (1,0)
S = (1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, ...)
B'(0) = (1)
B'(1) = (1,1)
B'(2) = (1,1,1)
B'(3) = (1,1,1,0)
B'(4) = (1,1,1,0,0)
B'(5) = (1,1,1,0,0,1)
S^ = (1,1,1,0,0,1,1,0,1,0,0,1,1,0,...),
with index sequence (1,3,17,18,35,45,239,341,...)

Crossrefs

Cf. A246142, A246127, A246144, A246146, A004539.

Programs

  • Mathematica
    seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 1]]]] &[seqtofind]; s = RealDigits[Sqrt[2], 2, 10000000][[1]]; Take[s, 60]
t = {{1}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (*A246142*)
Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246143*)
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