

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(n1)+lambda*(n1)_q*nu(n2) with (b,lambda)=(3,1), where (n)_q=(1+q+...+q^(n1)) and q is a root of unity.


0



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
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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(n1)+f(n2).


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(n2).
O.g.f.: (1+3*x+9*x^2)/((x1)*(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 A320583 A131814
Adjacent sequences: A072985 A072986 A072987 * A072989 A072990 A072991


KEYWORD

nonn


AUTHOR

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


STATUS

approved



