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A072946 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. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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)+2f(n-2).

LINKS

Table of n, a(n) for n=0..41.

M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.

Index entries for linear recurrences with constant coefficients, signature (0, 2).

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)/(1-2*x^2). - R. J. Mathar, Dec 05 2007

EXAMPLE

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,...

MATHEMATICA

LinearRecurrence[{0, 2}, {1, 2, 6}, 50] (* Harvey P. Dale, Dec 31 2015 *)

CROSSREFS

Essentially the same as A162255 and A164073.

Cf. A002605.

Sequence in context: A059909 A145177 A007517 * A134000 A127730 A118416

Adjacent sequences:  A072943 A072944 A072945 * A072947 A072948 A072949

KEYWORD

nonn,easy

AUTHOR

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

EXTENSIONS

More terms from R. J. Mathar, Dec 05 2007

STATUS

approved

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Last modified July 22 23:04 EDT 2017. Contains 289676 sequences.