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

Author

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

Keywords

• nonn
• easy

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).

Examples

			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,...
		

Links

• 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).

Crossrefs

Essentially the same as A162255 and A164073.

Cf. A002605.

Programs

Mathematica

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

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

Extensions

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