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 A072996 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. 0
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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)=2f(n-1)+f(n-2). LINKS 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.: -(x+1+4*x^2)/((x-1)*(x+1)) = -4-3/(x-1)+2/(x+1). - R. J. Mathar, Dec 05 2007 EXAMPLE 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,... CROSSREFS Cf. A000129. Sequence in context: A111129 A168464 A059688 * A244892 A278066 A153107 Adjacent sequences:  A072993 A072994 A072995 * A072997 A072998 A072999 KEYWORD nonn AUTHOR Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002 STATUS approved

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Last modified May 13 10:49 EDT 2021. Contains 343839 sequences. (Running on oeis4.)