A144133 Gegenbauer polynomial C_n^2(3).
1, 12, 106, 828, 6051, 42408, 288788, 1925736, 12637733, 81897876, 525360702, 3341936196, 21109664455, 132544828560, 827948567080, 5148653356944, 31891223012553, 196848686563164, 1211273655997202, 7432579805359884
Offset: 0
Examples
1 + 12*x + 106*x^2 + 828*x^3 + 6051*x^4 + 42408*x^5 + ...
Links
- Eric Weisstein's World of Mathematics, Gegenbauer Polynomial.
- Wikipedia, Gegenbauer polynomials
- Index entries for linear recurrences with constant coefficients, signature (12,-38,12,-1)
Programs
-
Mathematica
lst={};Do[AppendTo[lst,GegenbauerC[n,2,3]],{n,0,8^3}];lst LinearRecurrence[{12, -38, 12, -1}, {1, 12, 106, 828}, 100] (* Emanuele Munarini, Mar 07 2018 *) -
Maxima
makelist(ultraspherical(n,2,3),n,0,24); /* Emanuele Munarini, Mar 07 2018 */
-
PARI
{a(n) = local(s=1); if( n<0, n = -4 - n; s=-1); s * polcoeff( 1 / (1 - 6*x + x^2)^2 + x * O(x^n), n)} /* Michael Somos, May 11 2012 */ -
PARI
a(n)=([0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1; -1, 12, -38, 12]^n*[1; 12; 106; 828])[1, 1] \\ Charles R Greathouse IV, Feb 07 2022
Formula
From Michael Somos, May 11 2012: (Start)
G.f.: 1 / (1 - 6*x + x^2)^2.
a(-4 - n) = -a(n).
Convolution square of A001109. (End)
From Emanuele Munarini, Mar 07 2018: (Start)
a(n) = (1/4)*Sum_{k=0..n} p(2*k+1)*p(2*n-2*k+1) = (1/32)*(14*n+13)*p(2*n+1) + (3/16)*(n+1)*p(2*n), where the p(n) = A000129(n+1) are Pell numbers.
a(n+4) - 12*a(n+3) + 38*a(n+2) - 12*a(n+1) + a(n) = 0. (End)