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Ordered union of the sets {h^6, h >=1} and {32*k^6, k >=1}.
3

%I #5 Oct 28 2014 00:08:15

%S 1,32,64,729,2048,4096,15625,23328,46656,117649,131072,262144,500000,

%T 531441,1000000,1492992,1771561,2985984,3764768,4826809,7529536,

%U 8388608,11390625,16777216,17006112,24137569,32000000,34012224,47045881,56689952,64000000

%N Ordered union of the sets {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 Clark Kimberling, <a href="/A249116/b249116.txt">Table of n, a(n) for n = 1..1000</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 union is {1, 32, 64, 729, 2048, 4096, 15625, ...}

%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. A249117, A249118.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 21 2014