|
%I #5 Mar 30 2012 18:58:12
%S 4,8,13,18,23,27,33,37,42,47,52,56,62,66,70,76,80,85,90,95,99,105,109,
%T 114,119,124,128,134,138,142,147,152,157,161,167,171,176,181,186,190,
%U 196,200,204,210,214,219,224,229,233,239,243,248,253,258,262
%N Position of 5^n in joint ranking of {2^i}, {3^j}, {5^k}.
%C 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.
%F A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],
%F A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],
%F A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],
%F where []=floor.
%e The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
%e A205812=(1,3,5,7,10,11,14,...)
%e A205813=(2,6,9,12,15,...)
%e A205814=(4,8,13,18,23,...)
%t f[1, n_] := 2^n; f[2, n_] := 3^n;
%t f[3, n_] := 5^n; z = 1000;
%t d[n_, b_, c_] := Floor[n*Log[b, c]];
%t t[k_] := Table[f[k, n], {n, 1, z}];
%t t = Sort[Union[t[1], t[2], t[3]]];
%t p[k_, n_] := Position[t, f[k, n]];
%t Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
%t Table[n + d[n, 3, 2] + d[n, 5, 2],
%t {n, 1, 50}] (* A206812 *)
%t Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
%t Table[n + d[n, 2, 3] + d[n, 5, 3],
%t {n, 1, 50}] (* A206813 *)
%t Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
%t Table[n + d[n, 2, 5] + d[n, 3, 5],
%t {n, 1, 50}] (* A206814 *)
%Y Cf. A206805, A206812, A206813.
%K nonn
%O 1,1
%A _Clark Kimberling_, Feb 17 2012
|
