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.

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

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