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

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

Author

Clark Kimberling, Oct 21 2014

Keywords

• nonn
• easy

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

• Andrew Howroyd, Table of n, a(n) for n = 1..10000

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