OFFSET
0,2
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 sequence of weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n >= 2, f(n) = 2*f(n-1) + 3*f(n-2).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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 (0,3).
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 + 2*x + 4*x^2)/(1-3*x^2). - R. J. Mathar, Dec 05 2007
a(n) = 3*a(n-2) for n>2. - Ralf Stephan, Jul 19 2013
a(n) = (1/6)*(13 + (-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013
EXAMPLE
nu(0) = 1;
nu(1) = 2;
nu(2) = 7;
nu(3) = 20 + 6q;
nu(4) = 61 + 33q + 21q^2;
nu(5) = 182 + 144q + 120q^2 + 78q^3 + 18q^4;
nu(6) = 547 + 570q + 585q^2 + 501q^3 + 381q^4 + 162q^5 + 63q^6; ...
The coefficients of the highest power of q give this sequence.
MATHEMATICA
CoefficientList[Series[-(1 + 2 x + 4 x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
Join[{1}, LinearRecurrence[{0, 3}, {2, 7}, 33]] (* Jean-François Alcover, Sep 23 2017 *)
PROG
(Magma) [1] cat [(1/6)*(13+(-1)^n)*3^Floor(n/2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013
(PARI) x='x+O('x^30); Vec((1+2*x+4*x^2)/(1-3*x^2)) \\ G. C. Greubel, May 26 2018
CROSSREFS
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