A206814 Position of 5^n in joint ranking of {2^i}, {3^j}, {5^k}.

4, 8, 13, 18, 23, 27, 33, 37, 42, 47, 52, 56, 62, 66, 70, 76, 80, 85, 90, 95, 99, 105, 109, 114, 119, 124, 128, 134, 138, 142, 147, 152, 157, 161, 167, 171, 176, 181, 186, 190, 196, 200, 204, 210, 214, 219, 224, 229, 233, 239, 243, 248, 253, 258, 262

Offset

1,1

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.

Links

Table of n, a(n) for n=1..55.

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.

Example

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,...)

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 *)

Crossrefs

Cf. A206805, A206812, A206813.

Sequence in context: A312123 A312124 A312125 * A312126 A312127 A312128

Adjacent sequences: A206811 A206812 A206813 * A206815 A206816 A206817

Keyword

nonn

Author

Clark Kimberling, Feb 17 2012

Status

approved