A010015 a(0) = 1, a(n) = 25*n^2 + 2 for n > 0.

1, 27, 102, 227, 402, 627, 902, 1227, 1602, 2027, 2502, 3027, 3602, 4227, 4902, 5627, 6402, 7227, 8102, 9027, 10002, 11027, 12102, 13227, 14402, 15627, 16902, 18227, 19602, 21027, 22502, 24027, 25602, 27227, 28902, 30627, 32402, 34227, 36102, 38027, 40002

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Subsequence of A160842. - Bruno Berselli, Feb 06 2012

The identity (25*n^2 + 1)^2 - (25*n^2 + 2)*(5*n)^2 = 1 can be written as (A016850(n+1) + 1)^2 - a(n+1)*A008587(n+1)^2 = 1. - Vincenzo Librandi, Feb 08 2012

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A206399.

Programs

Mathematica

Join[{1}, 25 Range[40]^2 + 2] (* Bruno Berselli, Feb 06 2012 *)
Join[{1}, LinearRecurrence[{3, -3, 1}, {27, 102, 227}, 50]] (* Vincenzo Librandi, Feb 08 2012 *)

PARI

A010015(n)=25*n^2+2-!n \\ M. F. Hasler, Feb 14 2012

Formula

G.f.: (1+x)*(1 + 23*x + x^2)/(1-x)^3. - Bruno Berselli, Feb 06 2012

E.g.f.: (x*(x+1)*25 + 2)*e^x - 1. - Gopinath A. R., Feb 14 2012

Sum_{n>=0} 1/a(n) =3/4+sqrt(2)/20*Pi*coth(Pi*sqrt(2)/5) = 1.062575323280590.. - R. J. Mathar, May 07 2024

a(n) = A262221(n)+A262221(n+1). - R. J. Mathar, May 07 2024