

A249123


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


4



1, 3, 5, 7, 9, 11, 13, 15, 17, 18, 20, 22, 24, 26, 28, 30, 32, 34, 35, 37, 39, 41, 43, 45, 47, 49, 51, 52, 54, 56, 58, 60, 62, 64, 66, 68, 69, 71, 73, 75, 77, 79, 81, 83, 85, 86, 88, 90, 92, 94, 96, 98, 100, 102, 103, 105, 107, 109, 111, 113, 115, 117, 119
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OFFSET

1,2


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 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.


LINKS



FORMULA



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) = 3 because 2^6 is in position 3.


MAPLE

Res:= NULL: count:= 0:
a:= 1: b:= 1:
for pos from 1 while count < 100 do
if a^6 < 2*b^6 then
Res:= Res, pos;
count:= count+1;
a:= a+1
else
b:= b+1
fi
od:


MATHEMATICA

z = 200; s = Table[h^6, {h, 1, z}]; t = Table[2*k^6, {k, 1, z}]; u = Union[s, t];
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



CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



