A128880 Triangular numbers congruent to 1 or 5 mod 6.

1, 55, 91, 253, 325, 595, 703, 1081, 1225, 1711, 1891, 2485, 2701, 3403, 3655, 4465, 4753, 5671, 5995, 7021, 7381, 8515, 8911, 10153, 10585, 11935, 12403, 13861, 14365, 15931, 16471, 18145, 18721, 20503, 21115, 23005, 23653, 25651, 26335, 28441

Offset: 1

Author

Zak Seidov, Apr 18 2007, Apr 25 2007

Keywords

• nonn
• easy

Comments

Or, except for the first term, triangular numbers the least prime factor of which is >=5.

There are no triangular numbers that are congruent to 5 mod 6. - Amiram Eldar, Aug 18 2022

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Intersection of A000217 and A007310.

Programs

Mathematica

c=0;Do[tr=n(n+1)/2;If[Abs[Mod[tr,6]]==1,c++;a[c]=tr],{n,300}];Table[a[i],{i,c}]
Select[Accumulate[Range[500]],MemberQ[{1,5},Mod[#,6]]&] (* Harvey P. Dale, Sep 28 2013 *)

PARI

Vec(-x*(1+54*x+34*x^2+54*x^3+x^4)/((1+x)^2*(x-1)^3) + O(x^100)) \\ Colin Barker, Jan 26 2016

Formula

a(1)=Tr(1), a(2)=Tr(10), where Tr(k)=k(k+1)/2 is triangular number; for n>=3 a(n)=Tr(k(n)), where k(n)=k(n-2)+12 with k(1)=1, k(2)=10.

G.f.: -x*(1+54*x+34*x^2+54*x^3+x^4) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Jul 07 2015

From Colin Barker, Jan 26 2016: (Start)

a(n) = (36*n^2+18*(-1)^n*n-36*n-9*(-1)^n+11)/2.

a(n) = 18*n^2-9*n+1 for n even.

a(n) = 18*n^2-27*n+10 for n odd.

(End)

Sum_{n>=1} 1/a(n) = Pi/3. - Amiram Eldar, Aug 18 2022