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Position of 2^n when {2^j} and {3^k} are jointly ranked; complement of A206807.
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%I #31 Jan 27 2020 07:05:44

%S 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,

%T 42,44,45,47,48,50,52,53,55,57,58,60,61,63,65,66,68,70,71,73,75,76,78,

%U 79,81,83,84,86,88,89,91,92,94,96,97,99,101,102,104,106,107

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

%C 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.

%H Jinyuan Wang, <a href="/A206805/b206805.txt">Table of n, a(n) for n = 1..1000</a>

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

%e The joint ranking begins with 2,3,4,8,9,16,27,32,64,81,128,243,256, so that

%e this sequence = (1,3,4,6,8,9,11,13,...),

%e A206807 = (2,5,7,10,12,...).

%t f[n_] := 2^n; g[n_] := 3^n; z = 200;

%t c = Table[f[n], {n, 1, z}]; s = Table[g[n], {n, 1, z}];

%t j = Sort[Union[c, s]];

%t p[n_] := Position[j, f[n]]; q[n_] := Position[j, g[n]];

%t Flatten[Table[p[n], {n, 1, z}]] (* A206805 *)

%t Table[n + Floor[n*Log[3, 2]], {n, 1, 50}] (* A206805 *)

%t Flatten[Table[q[n], {n, 1, z}]] (* A206807 *)

%t Table[n + Floor[n*Log[2, 3]], {n, 1, 50}] (* A206807 *)

%o (PARI) a(n) = n + floor(n*log(2)/log(3)); \\ _Jinyuan Wang_, Jan 27 2020

%Y Cf. A182773, A182774, A206807.

%K nonn

%O 1,2

%A _Clark Kimberling_, Feb 16 2012