A074323 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)=(1,2), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.

1, 1, 3, 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

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 list 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)+2f(n-2).

The highest powers are given by the quarter-squares A002620(n-1). - Paul Barry, Mar 11 2007

Examples

			nu(0)=1;
nu(1)=1;
nu(2)=3;
nu(3)=5+2q;
nu(4)=11+8q+6q^2;
nu(5)=21+22q+20q^2+14q^3+4q^4;
nu(6)=43+60q+70q^2+64q^3+54q^4+28q^5+12q^6;
by listing the coefficients of the highest power in each nu(n), we get 1,1,3,2,6,4,12,...
		

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

Cf. A001045.

Programs

Mathematica

Join[{1}, LinearRecurrence[{0, 2}, {1, 3}, 41]] (* Jean-François Alcover, Sep 22 2017 *)

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).

G.f.: (1+x+x^2)/(1-2*x^2); a(n)=2^floor(n/2)+2^((n-2)/2)*(1+(-1)^n)/2-0^n/2. - Paul Barry, Mar 11 2007

a(0)=0, a(2n+1) = A000079, a(2n+2) = 3a(2n+1). a(2n)-a(2n+1) = A131577. - Paul Curtz, Mar 05 2008

a(2n+1) = 2^n = A000079(n), a(2n+2) = 3*A000079(n). Also a(2n)-a(2n+1) = A131577. a(2n+1)-a(2n)=2^n for n>0. - Paul Curtz, Apr 09 2008

a(n+1) = A010684(n)*A016116(n). - R. J. Mathar, Jul 08 2009

Extensions

More terms from Paul Barry, Mar 11 2007