A249118 Position of 32n^6 in the ordered union of {h^6, h >=1} and {32*k^6, k >=1}.

2, 5, 8, 11, 13, 16, 19, 22, 25, 27, 30, 33, 36, 38, 41, 44, 47, 50, 52, 55, 58, 61, 63, 66, 69, 72, 75, 77, 80, 83, 86, 89, 91, 94, 97, 100, 102, 105, 108, 111, 114, 116, 119, 122, 125, 127, 130, 133, 136, 139, 141, 144, 147, 150, 152, 155, 158, 161, 164

Offset

1,1

Comments

Let S = {h^6, h >=1} and T = {32*k^6, k >=1}. Then S and T are disjoint. The position of n^6 in the ordered union of S and T is A249117(n), and the position of 32*n^6 is A249118(n). Equivalently, the latter two give the positions of n*2^(2/3) and n*2^(3/2), respectively, when all the numbers h*2^(2/3) and k*2^(3/2) are jointly ranked.

Links

Table of n, a(n) for n=1..59.

Example

{h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
{32*k^6, k >=1} = {32, 2048, 23328, 131072, 500000, ...};
so the ordered union is {1, 32, 64, 729, 2048, 4096, 15625, ...}, and a(2) = 5
because 32*2^6 is in position 5 of the ordered union.

Mathematica

z = 200; s = Table[h^6, {h, 1, z}]; t = Table[32*k^6, {k, 1, z}];
v = Union[s, t]  (* A249116 *)
Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, 100}]]  (* A249117 *)
Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, 100}]]  (* A249118 *)

Crossrefs

Cf. A249116, A249117.

Sequence in context: A330112 A206911 A093609 * A292643 A376954 A140101

Adjacent sequences: A249115 A249116 A249117 * A249119 A249120 A249121

Keyword

nonn,easy

Author

Clark Kimberling, Oct 21 2014

Status

approved