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 1-4 of 4 results.

A026371 a(n) = least k such that s(k) = n, where s = A026370.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 20, 21, 22, 24, 25, 27, 28, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 43, 44, 46, 47, 49, 50, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 65, 66, 68, 69, 71, 72, 73, 74, 76, 77, 79, 80, 82, 83, 85, 86
Offset: 1

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Author

Keywords

Comments

Complement of A026372; also the rank transform (as at A187224) of (A004526 after removal of its first term, leaving 0,1,1,2,2,3,3,4,4,5,5,6,6,...). - Clark Kimberling, Mar 10 2011

Crossrefs

Programs

  • Mathematica
    seqA = Table[Floor[n/2], {n, 1, 180}]  (* A004526 *)
    seqB = Table[n, {n, 1, 80}];           (* A000027 *)
    jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
    Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
    {#1, 2} & /@ seqB}, 1]];
    limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                   (* A026371 *)
    Complement[Range[Length[seqA]], limseqU]  (* A026372 *)
    (* by Peter J. C. Moses, Mar 10 2011 *)

A187478 Rank transform of the sequence floor(3(n-2)/2); complement of A187479.

Original entry on oeis.org

1, 2, 3, 6, 8, 9, 11, 13, 15, 17, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35, 36, 39, 40, 42, 44, 46, 48, 49, 51, 53, 55, 57, 58, 60, 62, 64, 66, 68, 69, 71, 73, 75, 77, 79, 80, 82, 84, 86, 88, 90, 91, 93, 95, 97, 98, 101, 102, 104, 106, 108, 109, 111, 113, 115, 117, 119, 120
Offset: 1

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Author

Clark Kimberling, Mar 10 2011

Keywords

Comments

See A187224. Although the first term of floor(3(n-2)/2) is negative, we can replace it by 0 without affecting the same joint rankings; thus, the procedure described at A187224 applies.

Crossrefs

Programs

  • Mathematica
    seqA = Table[Floor[3(n-2)/2], {n, 1, 180}]
      seqB = Table[n, {n, 1, 80}];        (* A000027 *)
    jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
    Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
    {#1, 2} & /@ seqB}, 1]];
    limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                     (* A187478 *)
    Complement[Range[Length[seqA]], limseqU]  (* A187479 *)
    (* by Peter J. C. Moses, Mar 10 2011 *)

A187484 Rank transform of the sequence A004526=(0,0,1,1,2,2,3,3,4,4,...); complement of A187475.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 17, 18, 20, 21, 23, 24, 25, 26, 28, 29, 31, 32, 33, 34, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 55, 56, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 77, 78, 80, 81, 83, 84, 85, 86, 88, 89, 91, 92, 94, 95, 97, 98, 99, 100, 102, 103, 105, 106, 107, 108, 110, 111, 113, 114, 116, 117, 119, 120, 121, 122, 124, 125, 127, 128, 129, 130
Offset: 1

Views

Author

Clark Kimberling, Mar 10 2011

Keywords

Comments

See A187224 and A004526.

Crossrefs

Programs

  • Mathematica
    seqA = Table[Floor[(n-1)/2], {n, 1, 180}] (* A004526 *)
    seqB = Table[n, {n, 1, 80}];            (* A000027 *)
    jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
    Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
    {#1, 2} & /@ seqB}, 1]];
    limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                     (* A187484 *)
    Complement[Range[Length[seqA]], limseqU]  (* A187475 *)
    (* by Peter J. C. Moses, Mar 10 2011 *)

A026372 a(n) = greatest k such that s(k) = n, where s = A026370.

Original entry on oeis.org

4, 7, 10, 15, 18, 23, 26, 31, 34, 37, 40, 45, 48, 53, 56, 59, 62, 67, 70, 75, 78, 81, 84, 89, 92, 97, 100, 105, 108, 113, 116, 119, 122, 127, 130, 135, 138, 141, 144, 149, 152, 157, 160, 165, 168, 173, 176, 179, 182, 187, 190, 195, 198
Offset: 1

Views

Author

Keywords

Comments

Complement of A026371. See A187422 and A026371. [Clark Kimberling, Mar 10 2011]

Crossrefs

Programs

Showing 1-4 of 4 results.