%I
%S 1,3,4,6,7,9,10,12,14,15,17,18,20,21,23,24,26,28,29,31,32,34,35,37,39,
%T 40,42,43,45,46,48,49,51,53,54,56,57,59,60,62,64,65,67,68,70,71,73,74,
%U 76,78,79,81,82,84,85,87,88,90,92,93,95,96,98,99,101,103
%N Position of n^6 in the ordered union of {h^6, h >= 1} and {32*k^6, k >= 1}.
%C Let S = {h^6, h >= 1} and T = {32*k^6, k >= 1}. Then S and T are disjoint. The position of n^6 in the ordered union of S and T is A249117(n), and the position of 32*n^6 is A249118(n). Equivalently, the latter two give the positions of n*2^(2/3) and n*2^(3/2), respectively, when all the numbers h*2^(2/3) and k*2^(3/2) are jointly ranked.
%H Robert Israel, <a href="/A249117/b249117.txt">Table of n, a(n) for n = 1..10000</a>
%e {h^6, h >= 1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
%e {32*k^6, k >= 1} = {32, 2048, 23328, 131072, 500000, ...};
%e so the ordered union is {1, 32, 64, 729, 2048, 4096, 15625, ...}, and a(2) = 3
%e because 2^6 is in position 3 of the ordered union.
%p Res:= NULL: count:= 0:
%p a:= 1: b:= 1:
%p for pos from 1 while count < 100 do
%p if a^6 < 32*b^6 then
%p Res:= Res, pos;
%p count:= count+1;
%p a:= a+1
%p else
%p b:= b+1
%p fi
%p od:
%p Res; # _Robert Israel_, Aug 11 2019
%t z = 200; s = Table[h^6, {h, 1, z}]; t = Table[32*k^6, {k, 1, z}];
%t v = Union[s, t] (* A249116 *)
%t Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, 100}]] (* A249117 *)
%t Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, 100}]] (* A249118 *)
%Y Cf. A249116, A249118.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 21 2014
