A249117 Position of n^6 in the ordered union of {h^6, h >= 1} and {32*k^6, k >= 1}.

1, 3, 4, 6, 7, 9, 10, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 28, 29, 31, 32, 34, 35, 37, 39, 40, 42, 43, 45, 46, 48, 49, 51, 53, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 71, 73, 74, 76, 78, 79, 81, 82, 84, 85, 87, 88, 90, 92, 93, 95, 96, 98, 99, 101, 103

Offset

1,2

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

Robert Israel, Table of n, a(n) for n = 1..10000

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) = 3
because 2^6 is in position 3 of the ordered union.

Maple

Res:= NULL: count:= 0:
a:= 1: b:= 1:
for pos from 1 while count < 100 do
  if a^6 < 32*b^6 then
    Res:= Res, pos;
    count:= count+1;
    a:= a+1
  else
    b:= b+1
  fi
od:
Res; # Robert Israel, Aug 11 2019

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, A249118.

Sequence in context: A141206 A140100 A292642 * A093610 A206912 A330113

Adjacent sequences: A249114 A249115 A249116 * A249118 A249119 A249120

Keyword

nonn,easy

Author

Clark Kimberling, Oct 21 2014

Status

approved