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Search: id:A072988
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| A072988 |
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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)=(3,1), 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, 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
(list; graph; listen)
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OFFSET
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0,2
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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)=3f(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.: -(1+3*x+9*x^2)/((x-1)*(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)=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,...
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CROSSREFS
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Cf. A006190.
Sequence in context: A091043 A167790 A010708 this_sequence A131814 A003620 A111229
Adjacent sequences: A072985 A072986 A072987 this_sequence A072989 A072990 A072991
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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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