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Search: id:A072946
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| A072946 |
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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)=(2,2), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. |
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+0
3
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| 1, 2, 6, 4, 12, 8, 24, 16, 48, 32, 96, 64, 192, 128, 384, 256, 768, 512, 1536, 1024, 3072, 2048, 6144, 4096, 12288, 8192, 24576, 16384, 49152, 32768, 98304, 65536, 196608, 131072, 393216, 262144, 786432, 524288, 1572864, 1048576, 3145728, 2097152
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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)=2f(n-1)+2f(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+2*x+4*x^2)/(-1+2*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)=2, nu(2)=6, nu(3)=16+4q, nu(4)=44+20q+12q^2, nu(5)=120+80q+64q^2+40q^3+8q^4, nu(6)=328+288q+280q^2+232q^3+168q^4+64q^5+24q^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
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Cf. A002605.
Sequence in context: A059909 A145177 A007517 this_sequence A134000 A127730 A118416
Adjacent sequences: A072943 A072944 A072945 this_sequence A072947 A072948 A072949
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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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