OFFSET
0,3
COMMENTS
Also: 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)=(1,3), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.
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)+3f(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+x+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)*(7+(-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013
EXAMPLE
nu(0)=1;
nu(1)=1;
nu(2)=4;
nu(3)=7+3q;
nu(4)=19+15q+12q^2;
nu(5)=40+45q+42q^2+30q^3+9q^4;
nu(6)=97+147q+180q^2+168q^3+147q^4+81q^5+36q^6;
by listing the coefficients of the highest power in each nu(n), we get, 1,1,4,3,12,9,36,....
MATHEMATICA
CoefficientList[Series[-(1 + x + x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{0, 3}, {1, 1, 4}, 40] (* Harvey P. Dale, Mar 13 2016 *)
PROG
(Magma) [1] cat [(1/6)*(7+(-1)^n)*3^Floor(n/2):n in [1..40]]; // Vincenzo Librandi, Jul 20 2013
(PARI) a(n)=3^(n\2)\(3/4)^!bittest(n, 0) \\ M. F. Hasler, Dec 03 2014
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
Simpler definition from M. F. Hasler, Dec 03 2014
STATUS
approved