A047618 Numbers that are congruent to {0, 2, 5} mod 8.

0, 2, 5, 8, 10, 13, 16, 18, 21, 24, 26, 29, 32, 34, 37, 40, 42, 45, 48, 50, 53, 56, 58, 61, 64, 66, 69, 72, 74, 77, 80, 82, 85, 88, 90, 93, 96, 98, 101, 104, 106, 109, 112, 114, 117, 120, 122, 125, 128, 130, 133, 136, 138, 141, 144, 146, 149, 152, 154, 157, 160, 162

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..3000
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Programs

Magma

[n : n in [0..150] | n mod 8 in [0, 2, 5]]; // Wesley Ivan Hurt, Jun 10 2016

Maple

seq(3*n - floor(n/3) - (n^2 mod 3), n=0..51); # Gary Detlefs, Mar 19 2010

Mathematica

LinearRecurrence[{1, 0, 1, -1}, {0, 2, 5, 8}, 62] (* L. Edson Jeffery, Sep 02 2014 *)
Table[((8*n-9)+2*Sin[(2*n*Pi)/3]/Sqrt[3])/3, {n, 62}] (* L. Edson Jeffery, Sep 02 2014 *)
Table[8 n + {0, 2, 5}, {n, 0, 100}]//Flatten (* Vincenzo Librandi, Jun 11 2016 *)

PARI

a(n) = floor(8*(n-1)/3); \\ Michel Marcus, Sep 03 2014

Formula

a(n) = 3*(n - 1) - floor((n - 1)/3) - ((n - 1)^2 % 3). - Gary Detlefs, Mar 19 2010; corrected by L. Edson Jeffery, Sep 02 2014

a(n) = floor(8*(n-1)/3). - Gary Detlefs, Jan 02 2012

G.f.: x^2*(2+3*x+3*x^2)/((1+x+x^2)*(x-1)^2). - R. J. Mathar, Feb 03 2014

Conjecture: a(n)+a(n+1)+a(n+2) = 8*n-1; or a(n) = 8*(n-2)-a(n-1)-a(n-2)-1, n>3, with a(1)=0, a(2)=2, a(3)=5. - L. Edson Jeffery, Sep 02 2014

a(n) = a(n-1)+a(n-3)-a(n-4), n>4, with a(1)=0, a(2)=2, a(3)=5, a(4)=8. - L. Edson Jeffery, Sep 02 2014

a(n) = ((8*n-9)+2*sin((2*n*Pi)/3)/sqrt(3))/3. - L. Edson Jeffery, Sep 02 2014

a(3k) = 8k-3, a(3k-1) = 8k-6, a(3k-2) = 8k-8. - Wesley Ivan Hurt, Jun 10 2016