A158673 a(n) = 60*n^2 + 1.

1, 61, 241, 541, 961, 1501, 2161, 2941, 3841, 4861, 6001, 7261, 8641, 10141, 11761, 13501, 15361, 17341, 19441, 21661, 24001, 26461, 29041, 31741, 34561, 37501, 40561, 43741, 47041, 50461, 54001, 57661, 61441, 65341, 69361, 73501, 77761, 82141, 86641, 91261, 96001

Offset: 0

Author

Vincenzo Librandi, Mar 24 2009

Keywords

• nonn
• easy

Comments

The identity (60*n^2+1)^2 - (900*n^2+30)*(2*n)^2 = 1 can be written as a(n)^2 - A158672(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, A158672.

Programs

Magma

I:=[1, 61, 241]; [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 19 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {1, 61, 241}, 50] (* Vincenzo Librandi, Feb 19 2012 *)
60*Range[0,40]^2+1 (* Harvey P. Dale, Jun 18 2021 *)

PARI

for(n=0, 40, print1(60*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 19 2012

Formula

G.f.: (1+58*x+61*x^2)/(1-x)^3.

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

From Amiram Eldar, Mar 20 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)) + 1)/2.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(15)))*Pi/(2*sqrt(15)) + 1)/2. (End)

Extensions

Comment rewritten, a(0) added and formula replaced by R. J. Mathar, Oct 22 2009