A165721 Integers of the form k*(k+13)/12.

4, 14, 22, 25, 35, 55, 69, 74, 90, 120, 140, 147, 169, 209, 235, 244, 272, 322, 354, 365, 399, 459, 497, 510, 550, 620, 664, 679, 725, 805, 855, 872, 924, 1014, 1070, 1089, 1147, 1247, 1309, 1330, 1394, 1504, 1572, 1595, 1665, 1785, 1859, 1884, 1960, 2090

Offset

1,1

Comments

Integers of the form k+k*(k+1)/12 = k+A000217(k)/6 (see A069497). - R. J. Mathar, Sep 25 2009

Are all terms composite numbers?

Contribution from Zak Seidov, Sep 25 2009: (Start)

Integers of form n(13+n)/12, n=0,1,2,...

Each four terms of the sequence are composite numbers of forms:

{(4+3 m) (1+4 m), (2+3 m) (7+4 m), (2+m) (11+12 m), m (13+12 m)}, m=0,1,2,...

m=0: {4,14,22,25}; m=1: {35,55,69,74}; m=2: {90,120,140,147}, etc. (End)

Links

Table of n, a(n) for n=1..50.

Index entries for linear recurrences with constant coefficients, signature (3, -5, 7, -7, 5, -3, 1).

Formula

a(n)= 3*a(n-1) -5*a(n-2) +7*a(n-3) -7*a(n-4) +5*a(n-5) -3*a(n-6) +a(n-7). - R. J. Mathar, Sep 25 2009

G.f.: x*(-4-2*x-x^3+x^5)/((x^2+1)^2*(x-1)^3). - R. J. Mathar, Sep 25 2009

Mathematica

q=6; s=0; lst={}; Do[s+=((n+q)/q); If[IntegerQ[s], AppendTo[lst, s]], {n, 6!}]; lst
Select[Table[k (k+13)/12, {k, 200}], IntegerQ] (* or *) LinearRecurrence[ {3, -5, 7, -7, 5, -3, 1}, {4, 14, 22, 25, 35, 55, 69}, 50] (* Harvey P. Dale, Jan 30 2013 *)

Crossrefs

Cf. A165717, A165718, A165719, A165720.

Sequence in context: A297358 A267768 A185008 * A197638 A051806 A034051

Adjacent sequences: A165718 A165719 A165720 * A165722 A165723 A165724

Keyword

nonn

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009

Status

approved