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

1,1

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 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 A249098 and A249099 are a pair of Beatty sequences.

Links

Table of n, a(n) for n=1..61.

Formula

Empirical g.f.: x*(3*x^4+2*x^3+2*x^2+2*x+2) / ((x-1)^2*(x^4+x^3+x^2+x+1)). - Colin Barker, Oct 22 2014

Conjecture: a(n) = 2n + floor(n/5). - Wesley Ivan Hurt, Dec 28 2016

Example

{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.

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 *)

Crossrefs

Cf. A249097, A249099.

Sequence in context: A054683 A327255 A287777 * A329838 A190327 A241176

Adjacent sequences: A249096 A249097 A249098 * A249100 A249101 A249102

Keyword

nonn,easy

Author

Clark Kimberling, Oct 21 2014

Status

approved