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Search: id:A074324
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| A074324 |
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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)=(1,3), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. |
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+0
4
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| 1, 1, 4, 3, 12, 9, 36, 27, 108, 81, 324, 243, 972, 729, 2916, 2187, 8748, 6561, 26244, 19683, 78732, 59049, 236196, 177147, 708588, 531441, 2125764, 1594323, 6377292, 4782969, 19131876, 14348907, 57395628, 43046721, 172186884, 129140163
(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)=f(n-1)+3f(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+x+x^2)/(-1+3*x^2) . - 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)=1; nu(2)=4; nu(3)=7+3q; nu(4)=19+15q+12q^2; nu(5)=40+45q+42q^2+30q^3+9q^4; nu(6)=97+147q+180q^2+168q^3+147q^4+81q^5+36q^6; by listing the coefficients of the highest power in each nu(n), we get, 1,1,4,3,12,9,36,....
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CROSSREFS
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Cf. A006130.
Sequence in context: A061727 A055527 A055523 this_sequence A162766 A166552 A122804
Adjacent sequences: A074321 A074322 A074323 this_sequence A074325 A074326 A074327
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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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EXTENSIONS
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More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
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