A158604 a(n) = 42*n^2 + 1.

1, 43, 169, 379, 673, 1051, 1513, 2059, 2689, 3403, 4201, 5083, 6049, 7099, 8233, 9451, 10753, 12139, 13609, 15163, 16801, 18523, 20329, 22219, 24193, 26251, 28393, 30619, 32929, 35323, 37801, 40363, 43009, 45739, 48553, 51451, 54433, 57499, 60649, 63883, 67201

Offset: 0

Author

Vincenzo Librandi, Mar 22 2009

Keywords

• nonn
• easy

Comments

The identity (42*n^2 + 1)^2 - (441*n^2 + 21)*(2*n)^2 = 1 can be written as a(n)^2 - A158603(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, A158603.

Programs

Magma

I:=[1, 43, 169]; [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

Mathematica

LinearRecurrence[{3, -3, 1}, {1, 43, 169}, 50] (* Vincenzo Librandi, Feb 16 2012 *)

PARI

for(n=0, 40, print1(42*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 16 2012

Formula

G.f.: -(1 + 40*x + 43*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/sqrt(42))*Pi/sqrt(42) + 1)/2.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(42))*Pi/sqrt(42) + 1)/2. (End)

Extensions

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