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 1-8 of 8 results.

A246129 Decimal expansion of the number whose continued fraction is given by A246127 (limiting block extension of an infinite Fibonacci word).

Original entry on oeis.org

2, 3, 6, 6, 3, 0, 4, 6, 9, 4, 6, 5, 3, 2, 7, 2, 6, 5, 6, 6, 8, 2, 4, 9, 7, 2, 0, 5, 8, 6, 1, 4, 5, 6, 9, 1, 0, 0, 8, 1, 9, 9, 4, 8, 1, 0, 4, 0, 9, 5, 8, 9, 1, 0, 9, 3, 0, 5, 4, 1, 0, 2, 7, 1, 3, 8, 5, 3, 7, 7, 9, 1, 0, 1, 9, 1, 3, 5, 3, 1, 1, 3, 4, 6, 2, 6
Offset: 1

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Comments

The (2,1)-version of the infinite Fibonacci word, A014675, as a sequence, is (2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...). Its limiting block extension, A246128, is the sequence (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2,...), which is the continued fraction for 2.366304...

Examples

			[2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1,...] =  2.3663046946532726566824972058...
		

Crossrefs

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*GoldenRatio], {n, 10000}]]; t = {{2}}; 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]; t1 = Last[t] (*A246127*)
    q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)
    u = N[FromContinuedFraction[t1], 100]
    r = RealDigits[u][[1]] (* A246129 *)

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

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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,...)
		

Crossrefs

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]]

A246128 Index sequence for limit-block extending the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

0, 2, 7, 10, 15, 23, 31, 36, 44, 49, 57, 70, 78, 91, 104, 112, 125, 138, 159, 193, 214, 248, 282, 303, 337, 371, 392, 426, 447, 481, 515, 536, 570, 591, 625, 659, 680, 714, 748, 803, 892, 981, 1036, 1125, 1180, 1269, 1358, 1413, 1502, 1557, 1646, 1735, 1790
Offset: 0

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Author

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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), 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^.
...
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 A246127.

Examples

			S = the infinite Fibonacci word A014675, with 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) = (2,2)
B'(2) = (2,2,1)
B'(3) = (2,2,1,2)
B'(4) = (2,2,1,2,1)
B'(5) = (2,2,1,2,1,2)
S^ = (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...),
with index sequence (0,2,7,10,15,...)
		

Crossrefs

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*GoldenRatio], {n, 10000}]]; t = {{2}}; 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]; t1 = Last[t] (*A246127*)
    q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)

A246144 Limiting block extension of A000002 (Kolakoski sequence) with first term as initial block.

Original entry on oeis.org

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

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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

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*)

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

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Author

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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,...)
		

Crossrefs

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 *)

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

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Author

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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

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

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Author

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,...)
		

Crossrefs

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

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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

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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