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%I #11 Oct 06 2016 15:29:25
%S 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,
%T 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,
%U 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
%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)=(3,1), 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)=3f(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.: -(1+3*x+9*x^2)/((x-1)*(x+1)). - _R. J. Mathar_, Dec 05 2007
%e 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,...
%Y Cf. A006190.
%K nonn
%O 0,2
%A Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
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