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Search: id:A072996
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| A072996 |
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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. |
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+0
1
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| 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; listen)
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OFFSET
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0,3
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COMMENT
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Instead of listing the coefficients of the highest power of q in each nu(n), if we listed 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).
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LINKS
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M. Beattie, S. D\u{a}sc\u{a}lescu and S. Raianu, Lifting of Nichols Algebras of Type $B_2$
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FORMULA
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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 (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
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EXAMPLE
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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,...
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CROSSREFS
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Cf. A000129.
Sequence in context: A111129 A059688 A073054 this_sequence A153107 A020855 A116558
Adjacent sequences: A072993 A072994 A072995 this_sequence A072997 A072998 A072999
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KEYWORD
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nonn
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AUTHOR
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Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
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