A158463 a(n) = 12*n^2 - 1.

-1, 11, 47, 107, 191, 299, 431, 587, 767, 971, 1199, 1451, 1727, 2027, 2351, 2699, 3071, 3467, 3887, 4331, 4799, 5291, 5807, 6347, 6911, 7499, 8111, 8747, 9407, 10091, 10799, 11531, 12287, 13067, 13871, 14699, 15551, 16427, 17327, 18251, 19199, 20171, 21167, 22187

Offset: 0

Author

Vincenzo Librandi, Mar 19 2009

Keywords

• sign
• easy

Comments

Related to Legendre polynomials - see Mma line. - N. J. A. Sloane, Nov 17 2009

One notices that this sequence produces an inordinate number of semiprimes, perhaps better than mere chance for large values of n. - J. M. Bergot, Jun 30 2011

Sequence found by reading the line from -1, in the direction -1, 11, ..., in the square spiral whose vertices are -1 together with the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

Examples

			G.f. = -1 + 11*x + 47*x^2 + 107*x^3 + 191*x^4 + 299*x^5 + 431*x^6 + 587*x^7 + 767*x^8 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A001082, A080663, A135453.

Programs

Magma

[12*n^2 - 1: n in [0..100]]; // G. C. Greubel, Sep 25 2018

Mathematica

Table[Numerator[LegendreP[2,2n]],{n,0,50}] (* N. J. A. Sloane, Nov 17 2009 *)
a[ n_] := 2 LegendreP[ 2, 2 n] (* Michael Somos, Jul 01 2011 *)
12*Range[0,40]^2-1 (* or *) LinearRecurrence[{3,-3,1},{-1,11,47},50] (* Harvey P. Dale, Jun 22 2019 *)

PARI

{a(n) = 12 * n^2 - 1} /* Michael Somos, Nov 12 2011 */

Formula

a(-n) = a(n).

G.f.: (1-14*x-11*x^2)/(x-1)^3. - R. J. Mathar, Aug 27 2011

a(n) = A135453(n) - 1. - Omar E. Pol, Jul 18 2012

From Amiram Eldar, Feb 04 2021: (Start)

Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(12))*cot(Pi/sqrt(12)))/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(12))*csc(Pi/sqrt(12)) - 1)/2.

Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(12))*csc(Pi/sqrt(12)).

Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(12))*sin(Pi/sqrt(6))/sqrt(2). (End)

From Gerry Martens, Apr 06 2024: (Start)

a(n) = Re((2*n*i-1)^3).

a(n) = -8*(1/4+n^2)^(3/2)*cos(3*arctan(2*n)). (End)

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

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

a(n) = A080663(2*n) for n > 0. (End)

Extensions

Minor edits by N. J. A. Sloane, Nov 17 2009

Comment rewritten by Bruno Berselli, Aug 27 2011