A206805 Position of 2^n when {2^j} and {3^k} are jointly ranked; complement of A206807.

1, 3, 4, 6, 8, 9, 11, 13, 14, 16, 17, 19, 21, 22, 24, 26, 27, 29, 30, 32, 34, 35, 37, 39, 40, 42, 44, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 61, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 79, 81, 83, 84, 86, 88, 89, 91, 92, 94, 96, 97, 99, 101, 102, 104, 106, 107

Offset: 1

Author

Clark Kimberling, Feb 16 2012

Keywords

• nonn

Comments

The joint ranking is for j >= 1 and k >= 1, so that the sets {2^j} and {3^k} are disjoint. Not identical to A182774; e.g., A206805 contains 318 but A182774 does not.

Examples

			The joint ranking begins with 2,3,4,8,9,16,27,32,64,81,128,243,256, so that
this sequence = (1,3,4,6,8,9,11,13,...),
A206807       = (2,5,7,10,12,...).
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A182773, A182774, A206807.

Programs

Mathematica

f[n_] := 2^n; g[n_] := 3^n; z = 200;
c = Table[f[n], {n, 1, z}]; s = Table[g[n], {n, 1, z}];
j = Sort[Union[c, s]];
p[n_] := Position[j, f[n]]; q[n_] := Position[j, g[n]];
Flatten[Table[p[n], {n, 1, z}]] (* A206805 *)
Table[n + Floor[n*Log[3, 2]], {n, 1, 50}] (* A206805 *)
Flatten[Table[q[n], {n, 1, z}]]  (* A206807 *)
Table[n + Floor[n*Log[2, 3]], {n, 1, 50}] (* A206807 *)

PARI

a(n) = n + floor(n*log(2)/log(3)); \\ Jinyuan Wang, Jan 27 2020

Formula

a(n) = n + floor(n*log_2(3)) (while A206807(n) = n + floor(n*log_3(2))).