A158660 a(n) = 56*n^2 + 1.

1, 57, 225, 505, 897, 1401, 2017, 2745, 3585, 4537, 5601, 6777, 8065, 9465, 10977, 12601, 14337, 16185, 18145, 20217, 22401, 24697, 27105, 29625, 32257, 35001, 37857, 40825, 43905, 47097, 50401, 53817, 57345, 60985, 64737, 68601, 72577, 76665, 80865, 85177, 89601

Offset: 0

Author

Vincenzo Librandi, Mar 23 2009

Keywords

• nonn
• easy

Comments

The identity (56*n^2 + 1)^2 - (784*n^2 + 28)*(2*n)^2 = 1 can be written as a(n)^2 - A158659(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, A158482, A158657, A158659.

Programs

Magma

I:=[1, 57, 225]; [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 17 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {1, 57, 225}, 50] (* Vincenzo Librandi, Feb 17 2012 *)
56 Range[0,40]^2+1 (* Harvey P. Dale, Jun 14 2022 *)

PARI

for(n=0, 40, print1(56*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 17 2012

Formula

G.f.: -(1 + 54*x + 57*x^2)/(x-1)^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(14)))*Pi/(2*sqrt(14)) + 1)/2.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(14)))*Pi/(2*sqrt(14)) + 1)/2. (End)

From Elmo R. Oliveira, Jan 17 2025: (Start)

E.g.f.: exp(x)*(1 + 56*x + 56*x^2).

a(n) = A158482(2*n) for n > 0. (End)

Extensions

Comment rephrased and redundant formula replaced by R. J. Mathar, Oct 19 2009