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A165721
Integers of the form k*(k+13)/12.
0
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)
FORMULA
From R. J. Mathar, Sep 25 2009: (Start)
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).
G.f.: x*(-4-2*x-x^3+x^5)/((x^2+1)^2*(x-1)^3). (End)
Sum_{n>=1} 1/a(n) = 712/507 - (3 + 4*sqrt(3))*Pi/39. - Amiram Eldar, Jul 26 2024
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
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Definition simplified by R. J. Mathar, Sep 25 2009
STATUS
approved