A072985 Coefficient of the highest power of q in the expansion of nu(0)=1, nu(1)=b and for n >= 2, nu(n) = b*nu(n-1) + lambda*(n-1)_q*nu(n-2) with (b,lambda)=(2,3), where (n)_q = (1+q+...+q^(n-1)) and q is a root of unity.

1, 2, 7, 6, 21, 18, 63, 54, 189, 162, 567, 486, 1701, 1458, 5103, 4374, 15309, 13122, 45927, 39366, 137781, 118098, 413343, 354294, 1240029, 1062882, 3720087, 3188646, 11160261, 9565938, 33480783, 28697814, 100442349, 86093442

Offset: 0

Author

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

Keywords

• nonn
• easy

Comments

Instead of listing the coefficients of the highest power of q in each nu(n), if we list the coefficients of the smallest power of q (i.e., constant terms), we get a sequence of weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n >= 2, f(n) = 2*f(n-1) + 3*f(n-2).

Examples

			nu(0) = 1;
nu(1) = 2;
nu(2) = 7;
nu(3) = 20 + 6q;
nu(4) = 61 + 33q + 21q^2;
nu(5) = 182 + 144q + 120q^2 + 78q^3 + 18q^4;
nu(6) = 547 + 570q + 585q^2 + 501q^3 + 381q^4 + 162q^5 + 63q^6; ...
The coefficients of the highest power of q give this sequence.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.
• Index entries for linear recurrences with constant coefficients, signature (0,3).

Crossrefs

Cf. A014983.

Programs

Magma

[1] cat [(1/6)*(13+(-1)^n)*3^Floor(n/2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

Mathematica

CoefficientList[Series[-(1 + 2 x + 4 x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
Join[{1}, LinearRecurrence[{0, 3}, {2, 7}, 33]] (* Jean-François Alcover, Sep 23 2017 *)

PARI

x='x+O('x^30); Vec((1+2*x+4*x^2)/(1-3*x^2)) \\ G. C. Greubel, May 26 2018

Formula

For given b and lambda, the recurrence relation is given by; t(0)=1, t(1)=b, t(2) = b^2+lambda and for n >= 3, t(n) = lambda*t(n-2).

G.f.: (1 + 2*x + 4*x^2)/(1-3*x^2). - R. J. Mathar, Dec 05 2007

a(n) = 3*a(n-2) for n>2. - Ralf Stephan, Jul 19 2013

a(n) = (1/6)*(13 + (-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013

Extensions

More terms from R. J. Mathar, Dec 05 2007