A074087 Coefficient of q^1 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=b*nu(n-1)+lambda*(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(2,3).
0, 0, 0, 6, 33, 144, 570, 2118, 7587, 26448, 90420, 304470, 1013061, 3338112, 10911150, 35423862, 114342855, 367242336, 1174368360, 3741029094, 11876859369, 37591894320, 118659631650, 373630740966, 1173847761003
Offset: 0
Examples
The first 6 nu polynomials are 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, so the coefficients of q^1 are 0,0,0,6,33,144.
Links
- G. C. Greubel, 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 (4,2,-12,-9).
Crossrefs
Programs
-
Magma
I:=[0,0,6,33]; [0] cat [n le 4 select I[n] else 4*Self(n-1) + 2*Self(n-2) -12*Self(n-3) -9*Self(n-4): n in [1..30]]; // G. C. Greubel, May 26 2018
-
Mathematica
b=2; lambda=3; expon=1; nu[0]=1; nu[1]=b; nu[n_] := nu[n]=Together[b*nu[n-1]+lambda(1-q^(n-1))/(1-q)nu[n-2]]; a[n_] := Coefficient[nu[n], q, expon] (* Second program: *) Join[{0}, LinearRecurrence[{4,2,-12,-9}, {0,0,6,33}, 50]] (* G. C. Greubel, May 26 2018 *)
-
PARI
x='x+O('x^30); concat([0,0,0], Vec((6*x^3 +9*x^4)/(1-2*x-3*x^2)^2)) \\ G. C. Greubel, May 26 2018
Formula
G.f.: (6*x^3 +9*x^4)/(1-2*x-3*x^2)^2.
a(n) = 4*a(n-1) +2*a(n-2) -12*a(n-3) -9*a(n-4) for n>=5.
Extensions
Edited by Dean Hickerson, Aug 21 2002
Comments