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%I #4 Oct 28 2014 00:08:38
%S 2,5,8,11,13,16,19,22,25,27,30,33,36,38,41,44,47,50,52,55,58,61,63,66,
%T 69,72,75,77,80,83,86,89,91,94,97,100,102,105,108,111,114,116,119,122,
%U 125,127,130,133,136,139,141,144,147,150,152,155,158,161,164
%N Position of 32n^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.
%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) = 5
%e because 32*2^6 is in position 5 of the ordered union.
%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, A249117.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 21 2014
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