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

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

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 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

• Kevin Ryde, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to Beatty sequences

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