A249096 -id:A249096 - cp's OEIS frontend

A249073 Ordered union of the sets {h^6, h >=1} and {2*k^6, k >=1}.

1, 2, 64, 128, 729, 1458, 4096, 8192, 15625, 31250, 46656, 93312, 117649, 235298, 262144, 524288, 531441, 1000000, 1062882, 1771561, 2000000, 2985984, 3543122, 4826809, 5971968, 7529536, 9653618, 11390625, 15059072, 16777216, 22781250, 24137569, 33554432

Offset: 1

Clark Kimberling, Oct 21 2014

Let S = {h^6, h >=1} and T = {2*k^6, k >=1}. Then S and T are disjoint. The position of n^6 in the ordered union of S and T is A249123(n), and the position of 2*n^6 is A249124(n).

			{h^6, h >=1} = {1, 64, 729, 4096, 15625, 46656, 117649, ...};
{2*k^6, k >=1} = {2, 128, 1458, 8192, 31250, 93312, ...};
so the union is {1, 2, 64, 128, 729, 1458, 4096, 8192, 15625, ...}.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A249096, A249123, A249124.

z = 120; s = Table[h^6, {h, 1, z}]; t = Table[2 k^6, {k, 1, z}]; v = Union[s, t]
Flatten[Table[{n^6,2n^6},{n,20}]]//Union (* Harvey P. Dale, Dec 19 2015 *)

A184808 n + floor(r*n), where r = sqrt(2/3); complement of A184809.

Original entry on oeis.org

1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 19, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 59, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 79, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 99, 101, 103, 105, 107, 108, 110, 112, 114, 116

Offset: 1

Clark Kimberling, Jan 22 2011

nonn

This is the Beatty sequence for 1 + sqrt(2/3).
Also, a(n) is the position of 2*n^2 in the sequence obtained by arranging all the numbers in the sets {2*h^2, h >= 1} and {3*k^2, k >= 1} in increasing order. - Clark Kimberling, Oct 20 2014
Also, numbers n such that floor((n+1)*sqrt(6)) - floor(n*sqrt(6)) = 2. - Clark Kimberling, Jul 15 2015

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A184809, A182760 (comment about joint ranking),

A184810, A184811, A249096.

[n+Floor(n*Sqrt(2/3)): n in [1..70]]; // Vincenzo Librandi, Oct 23 2014

r=(2/3)^(1/2); s=(3/2)^(1/2);
a[n_]:=n+Floor [n*r];
b[n_]:=n+Floor [n*s];
Table[a[n],{n,1,120}]  (* A184808 *)
Table[b[n],{n,1,120}]  (* A184809 *)

main(size)={return(vector(size, n, n+floor(sqrt(2/3)*n)))} /* Anders Hellström, Jul 15 2015 */

a(n) = n + floor(r*n), where r = sqrt(2/3).

A249367 Numbers of the form 2n^2 or 3n^2.

Original entry on oeis.org

0, 2, 3, 8, 12, 18, 27, 32, 48, 50, 72, 75, 98, 108, 128, 147, 162, 192, 200, 242, 243, 288, 300, 338, 363, 392, 432, 450, 507, 512, 578, 588, 648, 675, 722, 768, 800, 867, 882, 968, 972, 1058, 1083, 1152, 1200, 1250, 1323, 1352, 1452, 1458, 1568, 1587, 1682

Offset: 1

M. F. Hasler, Oct 26 2014

Union of 2*A000290 = A001105 and 3*A000290 = A033428.

Essentially a duplicate of A249096.

Cf. A000290, A001105, A033428.

A249096 is essentially the same sequence.

N:= 10000: # to get all terms <= N
{seq(2*i^2,i=0..floor(sqrt(N/2)))} union {seq(3*i^2,i=0..floor(sqrt(N/3)))};
# if using Maple 11 or earlier, uncomment the following line:
# sort(convert(%,list));
# Robert Israel, Oct 27 2014

for(n=0,5000,if(issquare(n/3)||issquare(n/2),print1(n,", "))) \\ Derek Orr, Oct 26 2014

{2, 3} * A000290 = A001105 U A033428.

cp's OEIS Frontend

A249073 Ordered union of the sets {h^6, h >=1} and {2*k^6, k >=1}.

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A184808 n + floor(r*n), where r = sqrt(2/3); complement of A184809.

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A249367 Numbers of the form 2n^2 or 3n^2.

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