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A072985
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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)=(2,3), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.
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1, 2, 7, 6, 21, 18, 63, 54, 189, 162, 567, 486, 1701, 1458, 5103, 4374, 15309, 13122, 45927, 39366, 137781, 118098, 413343, 354294, 1240029, 1062882, 3720087, 3188646, 11160261, 9565938, 33480783, 28697814, 100442349, 86093442
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| 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)+3f(n-2).
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LINKS
| 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
| 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+2*x+4*x^2)/(-1+3*x^2). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
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EXAMPLE
| nu(0)=1, nu(1)=2, nu(2)=7, nu(3)=20+6q, nu(4)=61+33q+21q^2, nu(5)=182+144q+120q^2+78q^3+18q^4, nu(6)=547+570q+585q^2+501q^3+381q^4+162q^5+63q^6. By listing the coefficients of the highest power in each nu(n) we get 1,2,6,4,12,8,24,...
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CROSSREFS
| Cf. A014983.
Sequence in context: A089417 A168205 A082017 * A082187 A021365 A179378
Adjacent sequences: A072982 A072983 A072984 * A072986 A072987 A072988
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KEYWORD
| nonn
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AUTHOR
| Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
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EXTENSIONS
| More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
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