A074086 Coefficient of q^3 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, 0, 14, 71, 282, 997, 3298, 10439, 32012, 95834, 281494, 814131, 2324422, 6564135, 18362810, 50947395, 140329400, 384031508, 1044880222, 2828084399, 7618214354, 20432838121, 54585196818, 145287466799, 385397215108
Offset: 0
Keywords
Examples
The first 6 nu polynomials are nu(0)=1, nu(1)=2, nu(2)=5, nu(3)=12+2q, nu(4)=29+9q+5q^2, nu(5)=70+32q+24q^2+14q^3+2q^4, so the coefficients of q^3 are 0,0,0,0,0,14.
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 (8, -20, 8, 26, -8, -20, -8, -1).
Crossrefs
Programs
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Mathematica
b=2; lambda=1; expon=3; 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, 0}, LinearRecurrence[{8, -20, 8, 26, -8, -20, -8, -1}, {0, 0, 14, 71, 282, 997, 3298, 10439}, 25]] (* Jean-François Alcover, Jan 27 2019 *)
Formula
G.f.: (14x^5-41x^6-6x^7+49x^8+30x^9+5x^10)/(1-2x-x^2)^4.
a(n) = 8a(n-1)-20a(n-2)+8a(n-3)+26a(n-4)-8a(n-5)-20a(n-6)-8a(n-7)-a(n-8) for n>=11.
Extensions
Edited by Dean Hickerson, Aug 21 2002
Comments