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 1-2 of 2 results.

A249098 Position of n^6 in the ordered union of {h^6, h >=1} and {3*k^6, k >=1}.

Original entry on oeis.org

1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 43, 45, 47, 49, 51, 53, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 82, 84, 86, 87, 89, 91, 93, 95, 97, 98, 100, 102, 104, 106, 108, 109, 111, 113, 115
Offset: 1

Views

Author

Clark Kimberling, Oct 21 2014

Keywords

Comments

Let S = {h^6, h >=1} and T = {3*k^6, k >=1}. Then S and T are disjoint, with ordered union given by A249097. The position of n^6 is a(n), and the position of 3*n^6 is A249099(n).
Also, a(n) is the position of n in the joint ranking of the positive integers and the numbers k*3^(1/6), so that this sequence and A249099 are a pair of Beatty sequences.

Examples

			{h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
{3*k^6, k >=1} = {3, 192, 2187, 12288, 46875, 139968, ...};
so the ordered union is {1, 3, 64, 192, 729, 2187, 4096, 12288, ...}, and
a(2) = 3 because 2^6 is in position 3.

Links

Crossrefs

Cf. A249097, A249099.

Programs

  • Mathematica
    z = 200; s = Table[h^6, {h, 1, z}]; t = Table[3*k^6, {k, 1, z}]; u = Union[s, t];
v = Sort[u]  (* A249097 *)
m = Min[120, Position[v, 2*z^2]]
Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, m}]]  (* A249098 *)
Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]]  (* A249099 *)
  • PARI
    a(n) = sqrtnint(n^6\3,6) + n; \\ Kevin Ryde, Feb 19 2025

Formula

a(n) = floor((1+1/3^(1/6)) * n). - Kevin Ryde, Feb 19 2025

A249099 Position of 3*n^6 in the ordered union of {h^6, h >=1} and {3*k^6, k >=1}.

Original entry on oeis.org

2, 4, 6, 8, 11, 13, 15, 17, 19, 22, 24, 26, 28, 30, 33, 35, 37, 39, 41, 44, 46, 48, 50, 52, 55, 57, 59, 61, 63, 66, 68, 70, 72, 74, 77, 79, 81, 83, 85, 88, 90, 92, 94, 96, 99, 101, 103, 105, 107, 110, 112, 114, 116, 118, 121, 123, 125, 127, 129, 132, 134
Offset: 1

Views

Author

Clark Kimberling, Oct 21 2014

Keywords

Comments

Let S = {h^6, h >=1} and T = {3*k^6, k >=1}. Then S and T are disjoint, with ordered union given by A249097. The position of n^6 is A249098(n), and the position of 3*n^6 is a(n).
Also, a(n) is the position of n in the joint ranking of the positive integers and the numbers k*3^(1/6), so that A249098 and this sequence are a pair of Beatty sequences.

Examples

			{h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
{3*k^6, k >=1} = {3, 192, 2187, 12288, 46875, 139968, ...};
so the ordered union is {1, 3, 64, 192, 729, 2187, 4096, 12288, ...}, and
a(2) = 4 because 3*2^6 is in position 4.

Links

Crossrefs

Cf. A246708, A249073, A249097, A249098.

Programs

  • Mathematica
    z = 200; s = Table[h^6, {h, 1, z}]; t = Table[3*k^6, {k, 1, z}]; u = Union[s, t];
v = Sort[u]  (* A249073 *)
m = Min[120, Position[v, 2*z^2]]
Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, m}]]  (* A249098 *)
Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]]  (* A249099 *)
  • PARI
    a(n) = sqrtnint(3*n^6,6) + n; \\ Kevin Ryde, Feb 18 2025

Formula

a(n) = floor((1+3^(1/6)) * n). - Kevin Ryde, Feb 18 2025

Extensions

Incorrect conjectured formulas removed by Kevin Ryde, Feb 18 2025
Showing 1-2 of 2 results.

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