|
%I #11 Oct 06 2016 14:46:13
%S 1,1,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,
%T 5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,
%U 5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5,2,5
%N 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)=(2,1), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.
%C 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)=2f(n-1)+f(n-2).
%H M. Beattie, S. Dăscălescu and S. Raianu, <a href="https://arxiv.org/abs/math/0204075">Lifting of Nichols Algebras of Type B_2</a>, arXiv:math/0204075 [math.QA], 2002.
%F 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).
%F O.g.f.: -(x+1+4*x^2)/((x-1)*(x+1)) = -4-3/(x-1)+2/(x+1). - _R. J. Mathar_, Dec 05 2007
%e nu(0)=1, nu(1)=2, nu(2)=5, nu(3)=12+2q, nu(4)=29+9q+5q^2, nu(5)=70+32q+24q^2+14q^3+2q^4, nu(6)=169+102q+91q^2+42q^3+38q^4+9q^5+5q^6. By listing the coefficients of the highest power in each nu(n) we get 1,2,5,2,5,2,5,...
%Y Cf. A000129.
%K nonn
%O 0,3
%A Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
|
