A158602 a(n) = 40*n^2 + 1.

1, 41, 161, 361, 641, 1001, 1441, 1961, 2561, 3241, 4001, 4841, 5761, 6761, 7841, 9001, 10241, 11561, 12961, 14441, 16001, 17641, 19361, 21161, 23041, 25001, 27041, 29161, 31361, 33641, 36001, 38441, 40961, 43561, 46241, 49001, 51841, 54761, 57761, 60841, 64001

Offset: 0

Author

Vincenzo Librandi, Mar 22 2009

Keywords

• nonn
• easy

Comments

The identity (40*n^2 + 1)^2 - (400*n^2 + 20)*(2*n)^2 = 1 can be written as a(n)^2 - A158601(n)*A005843(n)^2 = 1.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005843, A158187, A158601.

Programs

Magma

I:=[1,41,161]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 16 2012

Maple

A158602:=n->40*n^2; seq(A158602(k), k=0..100); # Wesley Ivan Hurt, Sep 27 2013

Mathematica

40*Range[0,40]^2+1 (* or *) LinearRecurrence[{3,-3,1},{1,41,161},40] (* Harvey P. Dale, Jul 25 2011 *)
Table[40n^2+1, {n,0,100}] (* Wesley Ivan Hurt, Sep 27 2013 *)

PARI

for(n=0, 40, print1(40*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 16 2012

Formula

G.f.: -(1 + 38*x + 41*x^2)/(x-1)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

From Amiram Eldar, Mar 16 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) + 1)/2.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) + 1)/2. (End)

From Elmo R. Oliveira, Jan 16 2025: (Start)

E.g.f.: exp(x)*(1 + 40*x + 40*x^2).

a(n) = A158187(2*n). (End)

Extensions

Comment rewritten, formula replaced by R. J. Mathar, Oct 28 2009