OFFSET
1,1
COMMENTS
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.
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
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = n + floor( (2*n^6)^(1/6) ). - David A. Corneth, Aug 11 2019
EXAMPLE
{h^6, h >= 1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
{2*k^6, k >= 1} = {2, 128, 1458, 8192, 31250, 93312, ...};
so the ordered union is {1, 2, 64, 128, 729, 1458, 4096, 8192, 15625, ...}, and
a(2) = 4 because 2*2^6 is in position 4.
MATHEMATICA
PROG
(PARI) a(n) = n + sqrtnint(2*n^6, 6) \\ David A. Corneth, Aug 11 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 21 2014
STATUS
approved