A074085 Coefficient of q^2 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) = (2,1).
0, 0, 0, 0, 5, 24, 91, 308, 978, 2978, 8802, 25440, 72251, 202316, 559941, 1534548, 4170256, 11250630, 30158900, 80389600, 213204513, 562896832, 1480086111, 3877337556, 10123000126, 26347306474, 68378847990, 176994780672
Offset: 0
Keywords
Examples
The first 6 nu polynomials are nu(0) = 1, nu(1) = 2, nu(2) = 5, nu(3) = 12 + 2*q, nu(4) = 29 + 9*q + 5*q^2, nu(5) = 70 + 32q + 24*q^2 + 14*q^3 + 2*q^4, so the coefficients of q^2 are 0,0,0,0,5,24.
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 (6, -9, -4, 9, 6, 1).
Crossrefs
Programs
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Mathematica
b=2; lambda=1; expon=2; 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[{0,0}, LinearRecurrence[{6, -9, -4, 9, 6, 1}, {0, 0, 5, 24, 91, 308}, 30]] (* Jean-François Alcover, Dec 13 2018 *)
Formula
G.f.: (5*x^4 - 6*x^5 - 8*x^6 - 2*x^7)/(1 - 2*x - x^2)^3.
a(n) = 6*a(n-1) - 9*a(n-2) - 4*a(n-3) + 9*a(n-4) + 6*a(n-5) + a(n-6) for n >= 8.
Extensions
Edited by Dean Hickerson, Aug 21 2002
Comments