cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A184809 a(n) = n + floor(sqrt(3/2)*n).

Original entry on oeis.org

2, 4, 6, 8, 11, 13, 15, 17, 20, 22, 24, 26, 28, 31, 33, 35, 37, 40, 42, 44, 46, 48, 51, 53, 55, 57, 60, 62, 64, 66, 68, 71, 73, 75, 77, 80, 82, 84, 86, 88, 91, 93, 95, 97, 100, 102, 104, 106, 109, 111, 113, 115, 117, 120, 122, 124, 126, 129, 131, 133, 135
Offset: 1

Views

Author

Clark Kimberling, Jan 22 2011

Keywords

Comments

This is the Beatty sequence for 1 + sqrt(3/2).
Also, a(n) is the position of 3*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)) = 3. - Clark Kimberling, Jul 15 2015

Links

Crossrefs

Complement of A184808.
Cf. A184811.

Programs

Formula

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

A249096 {2*h^2, h >=1} union {3*k^2, k >=1}, in increasing order.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Oct 20 2014

Keywords

Comments

Let S = {2*h^2, h >=1} and T = {3*k^2, k >=1}. Then S and T are disjoint. The position of 2*n^2 in (S union T) is A184808(n), and the position of 3*n^2 is A184809(n).

Examples

			{2*h^2, h >=1} = {2, 8, 18, 32, 50, 72, 98, 128, 162, 200, ...};
{3*k^2, k >=1} = {3, 12, 27, 48, 75, 108, 147, 192, 243, ...};
so the union is {2, 3, 8, 12, 18, 27, 32, 48, 50, 72, 75, ...}

Links

Crossrefs

Cf. A184808, A184809.
A249367 is essentially the same sequence.

Programs

  • Mathematica
    z = 120; s = Table[2 h^2, {h, 1, z}]; t = Table[3 k^2, {k, 1, z}]; v = Sort[Union[s, t]]

A184810 Numbers m such that prime(m) has the form floor(k*r), where r=sqrt(2/3); complement of A184811.

Original entry on oeis.org

2, 3, 4, 8, 9, 10, 13, 14, 15, 17, 18, 19, 22, 23, 24, 26, 27, 28, 31, 34, 38, 39, 41, 42, 45, 46, 48, 49, 52, 53, 55, 56, 59, 60, 61, 66, 68, 72, 75, 76, 78, 79, 81, 82, 85, 86, 88, 89, 90, 92, 95, 96, 98, 99, 100, 102, 103, 106, 108, 109, 110, 112, 113, 114, 116, 117, 119, 120, 121, 122, 123, 124, 126, 128, 130, 131, 134, 135, 137, 139, 141, 142, 146, 147, 148, 149, 151, 152, 156, 157, 159, 162, 164, 165, 167, 168, 169, 170, 171, 173, 174, 175, 176, 177, 180
Offset: 1

Views

Author

Clark Kimberling, Jan 22 2011

Keywords

Crossrefs

Cf. A184808, A184809, A184811.

Programs

  • Mathematica
    r=(2/3)^(1/2);s=(3/2)^(1/2); (* complementary because of joint ranking of i*sqrt(2) and j*sqrt(3) *)
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 *)
t1={};Do[If[PrimeQ[a[n]], AppendTo[t1,a[n]]],{n,1,600}]
t2={};Do[If[PrimeQ[a[n]], AppendTo[t2,n]],{n,1,600}]
t3={};Do[If[MemberQ[t1,Prime[n]],AppendTo[t3,n]],{n,1,300}];t3
t4={};Do[If[PrimeQ[b[n]], AppendTo[t4,b[n]]],{n,1,600}]
t5={};Do[If[PrimeQ[b[n]], AppendTo[t5,n]],{n,1,600}]
t6={};Do[If[MemberQ[t4,Prime[n]],AppendTo[t6,n]],{n,1,300}];t6
(* t3 and t6 match A184810 and A184811 *)
Showing 1-3 of 3 results.

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