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%I #13 Dec 29 2016 05:24:51
%S 2,4,6,8,11,13,15,17,19,22,24,26,28,30,33,35,37,39,41,44,46,48,50,52,
%T 55,57,59,61,63,66,68,70,72,74,77,79,81,83,85,88,90,92,94,96,99,101,
%U 103,105,107,110,112,114,116,118,121,123,125,127,129,132,134
%N Position of 3*n^6 in the ordered union of {h^6, h >=1} and {3*k^6, k >=1}.
%C Let S = {h^6, h >=1} and T = {3*k^6, k >=1}. Then S and T are disjoint, with ordered union given by A249097. The position of n^6 is A249098(n), and the position of 3*n^6 is A249099(n). Also, a(n) is the position of n in the joint ranking of the positive integers and the numbers k*3^(1/6), so that A249098 and A249099 are a pair of Beatty sequences.
%F Empirical g.f.: x*(3*x^4+2*x^3+2*x^2+2*x+2) / ((x-1)^2*(x^4+x^3+x^2+x+1)). - _Colin Barker_, Oct 22 2014
%F Conjecture: a(n) = 2n + floor(n/5). - _Wesley Ivan Hurt_, Dec 28 2016
%e {h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
%e {3*k^6, k >=1} = {3, 192, 2187, 12288, 46875, 139968, ...};
%e so the ordered union is {1, 3, 64, 192, 729, 2187, 4096, 12288, ...}, and
%e a(2) = 4 because 3*2^6 is in position 4.
%t z = 200; s = Table[h^6, {h, 1, z}]; t = Table[3*k^6, {k, 1, z}]; u = Union[s, t];
%t v = Sort[u] (* A249073 *)
%t m = Min[120, Position[v, 2*z^2]]
%t Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, m}]] (* A249098 *)
%t Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]] (* A249099 *)
%Y Cf. A249097, A249099.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 21 2014
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