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%I #13 Aug 22 2019 05:35:14
%S 2,4,6,8,10,12,14,16,19,21,23,25,27,29,31,33,36,38,40,42,44,46,48,50,
%T 53,55,57,59,61,63,65,67,70,72,74,76,78,80,82,84,87,89,91,93,95,97,99,
%U 101,104,106,108,110,112,114,116,118,120,123,125,127,129,131
%N Position of 2*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 is A249123(n), and the position of 2*n^6 is A249124(n). Also, a(n) is the position of n*2^(1/6) 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.
%C Every positive integer m is of the form k + floor( (2*k^6)^(1/6) ) (this sequence) or of the form k + floor( (k^6 / 2)^(1/6) ) (A249123) for some positive integer k but not both. - _David A. Corneth_, Aug 12 2019
%H David A. Corneth, <a href="/A249124/b249124.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = n + floor( (2*n^6)^(1/6) ). - _David A. Corneth_, Aug 11 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) = 4 because 2*2^6 is in position 4.
%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(2*n^6, 6) \\ _David A. Corneth_, Aug 11 2019
%Y Cf. A249073, A249123.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 21 2014
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