

A249124


Position of 2*n^6 in the ordered union of {h^6, h >= 1} and {2*k^6, k >= 1}.


3



2, 4, 6, 8, 10, 12, 14, 16, 19, 21, 23, 25, 27, 29, 31, 33, 36, 38, 40, 42, 44, 46, 48, 50, 53, 55, 57, 59, 61, 63, 65, 67, 70, 72, 74, 76, 78, 80, 82, 84, 87, 89, 91, 93, 95, 97, 99, 101, 104, 106, 108, 110, 112, 114, 116, 118, 120, 123, 125, 127, 129, 131
(list;
graph;
refs;
listen;
history;
text;
internal format)



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

z = 200; s = Table[h^6, {h, 1, z}]; t = Table[2*k^6, {k, 1, z}]; u = Union[s, t];
v = Sort[u] (* A249073 *)
m = Min[120, Position[v, 2*z^2]]
Flatten[Table[Flatten[Position[v, s[[n]]]], {n, 1, m}]] (* A249123 *)
Flatten[Table[Flatten[Position[v, t[[n]]]], {n, 1, m}]] (* A249124 *)


PROG

(PARI) a(n) = n + sqrtnint(2*n^6, 6) \\ David A. Corneth, Aug 11 2019


CROSSREFS

Cf. A249073, A249123.
Sequence in context: A248195 A094041 A058066 * A322405 A118081 A152483
Adjacent sequences: A249121 A249122 A249123 * A249125 A249126 A249127


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Oct 21 2014


STATUS

approved



