%I #20 Aug 12 2019 10:42:14
%S 1,3,5,7,9,11,13,15,17,18,20,22,24,26,28,30,32,34,35,37,39,41,43,45,
%T 47,49,51,52,54,56,58,60,62,64,66,68,69,71,73,75,77,79,81,83,85,86,88,
%U 90,92,94,96,98,100,102,103,105,107,109,111,113,115,117,119
%N Position of n^6 in the ordered union of {h^6, h >= 1} and {2*k^6, k >= 1}.
%C Let S = {h^6, h >= 1} and T = {2*k^6, k >= 1}. Then S and T are disjoint, and their ordered union is given by A249073. The position of n^6 in is A249123(n), and the position of 2*n^6 is A249124(n). Also, a(n) is the position of n in the joint ranking of the positive integers and the numbers k*2^(1/6), so that A249123 and A249124 are a pair of Beatty sequences.
%H Robert Israel, <a href="/A249123/b249123.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = n + floor(2^(-1/6)*n). - _Robert Israel_, Aug 12 2019
%e {h^6, h >= 1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
%e {2*k^6, k >= 1} = {2, 128, 1458, 8192, 31250, 93312, ...};
%e so the ordered union is {1, 2, 64, 128, 729, 1458, 4096, 8192, 15625, ...}, and
%e a(2) = 3 because 2^6 is in position 3.
%p Res:= NULL: count:= 0:
%p a:= 1: b:= 1:
%p for pos from 1 while count < 100 do
%p if a^6 < 2*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[2*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}]] (* A249123 *)
%t Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]] (* A249124 *)
%o (PARI) a(n) = n + sqrtnint(((n^6) \ 2), 6) \\ _David A. Corneth_, Aug 11 2019
%Y Cf. A249073, A249124.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 21 2014
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