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A249099 Position of 3*n^6 in the ordered union of {h^6, h >=1} and {3*k^6, k >=1}.

%I A249099 #25 Feb 18 2025 15:37:16
%S A249099 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 A249099 55,57,59,61,63,66,68,70,72,74,77,79,81,83,85,88,90,92,94,96,99,101,
%U A249099 103,105,107,110,112,114,116,118,121,123,125,127,129,132,134
%N A249099 Position of 3*n^6 in the ordered union of {h^6, h >=1} and {3*k^6, k >=1}.
%C A249099 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 a(n).
%C A249099 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 this sequence are a pair of Beatty sequences.
%H A249099 Andrew Howroyd, <a href="/A249099/b249099.txt">Table of n, a(n) for n = 1..10000</a>
%F A249099 a(n) = floor((1+3^(1/6)) * n). - _Kevin Ryde_, Feb 18 2025
%e A249099 {h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
%e A249099 {3*k^6, k >=1} = {3, 192, 2187, 12288, 46875, 139968, ...};
%e A249099 so the ordered union is {1, 3, 64, 192, 729, 2187, 4096, 12288, ...}, and
%e A249099 a(2) = 4 because 3*2^6 is in position 4.
%t A249099 z = 200; s = Table[h^6, {h, 1, z}]; t = Table[3*k^6, {k, 1, z}]; u = Union[s, t];
%t A249099 v = Sort[u]  (* A249073 *)
%t A249099 m = Min[120, Position[v, 2*z^2]]
%t A249099 Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, m}]]  (* A249098 *)
%t A249099 Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]]  (* A249099 *)
%o A249099 (PARI) a(n) = sqrtnint(3*n^6,6) + n; \\ _Kevin Ryde_, Feb 18 2025
%Y A249099 Cf. A246708, A249073, A249097, A249098.
%K A249099 nonn,easy
%O A249099 1,1
%A A249099 _Clark Kimberling_, Oct 21 2014
%E A249099 Incorrect conjectured formulas removed by _Kevin Ryde_, Feb 18 2025