A206813 Position of 3^n in joint ranking of {2^i}, {3^j}, {5^k}.

2, 6, 9, 12, 15, 19, 22, 25, 29, 31, 35, 39, 41, 45, 48, 51, 54, 58, 61, 64, 68, 71, 74, 78, 81, 84, 87, 91, 93, 97, 101, 103, 107, 110, 113, 117, 120, 123, 126, 130, 132, 136, 140, 143, 146, 149, 153, 156, 159, 163, 165, 169, 173, 175, 179, 182, 185, 188

Offset: 1

Author

Clark Kimberling, Feb 17 2012

Keywords

• nonn

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (A206812 for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
A205812=(1,3,5,7,10,11,14,...)
A205813=(2,6,9,12,15,...)
A205814=(4,8,13,18,23,...)
		

Crossrefs

Cf. A206805, A206812, A206814.

Programs

Mathematica

f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)

Formula

A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],

A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],

A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],

where []=floor.