A072988 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)=(3,1), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.

1, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3, 10, 3

Offset

0,2

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 weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n>=2, f(n)=3f(n-1)+f(n-2).

Links

Table of n, a(n) for n=0..83.

M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.

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).

O.g.f.: -(1+3*x+9*x^2)/((x-1)*(x+1)). - R. J. Mathar, Dec 05 2007

Example

nu(0)=1, nu(1)=3, nu(2)=10, nu(3)=33+3q, nu(4)=109+19q+10q^2, nu(5)=360+93q+66q^2+36q^3+3q^4, nu(6)=1189+407q+336q^2+246q^3+147q^4+29q^5+10q^6. By listing the coefficients of the highest power in each nu(n) we get 1,3,10,3,10,3,10,...

Crossrefs

Cf. A006190.

Sequence in context: A321118 A167790 A010708 * A170855 A338683 A361943

Adjacent sequences: A072985 A072986 A072987 * A072989 A072990 A072991

Keyword

nonn

Author

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

Status

approved