A158546 a(n) = 144*n^2 + 12.

12, 156, 588, 1308, 2316, 3612, 5196, 7068, 9228, 11676, 14412, 17436, 20748, 24348, 28236, 32412, 36876, 41628, 46668, 51996, 57612, 63516, 69708, 76188, 82956, 90012, 97356, 104988, 112908, 121116, 129612, 138396, 147468, 156828, 166476, 176412, 186636, 197148

Offset: 0

Author

Vincenzo Librandi, Mar 21 2009

Keywords

• nonn
• easy

Comments

The identity (24*n^2 + 1)^2 - (144*n^2 + 12)*(2*n)^2 = 1 can be written as A158547(n)^2 - a(n)*A005843(n)^2 = 1.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005843, A158480, A158547.

Programs

Magma

I:=[12, 156, 588]; [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 14 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {12, 156, 588}, 50] (* Vincenzo Librandi, Feb 14 2012 *)
144*Range[0,40]^2+12 (* Harvey P. Dale, Jul 14 2025 *)

PARI

for(n=0, 40, print1(144*n^2 + 12", ")); \\ Vincenzo Librandi, Feb 14 2012

Formula

G.f.: 12*(1 + 10*x + 13*x^2)/(1-x)^3.

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

From Amiram Eldar, Mar 06 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)) + 1)/24.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)) + 1)/24. (End)

From Elmo R. Oliveira, Jan 15 2025: (Start)

E.g.f.: 12*exp(x)*(1 + 12*x + 12*x^2).

a(n) = 12*A158480(n). (End)

Extensions

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009