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A074083 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)=(1,1).

%I A074083 #22 Mar 03 2024 11:34:42
%S A074083 0,0,0,0,0,4,14,39,97,224,494,1051,2177,4412,8784,17228,33360,63886,
%T A074083 121164,227833,425147,787916,1451198,2657821,4842727,8782230,15857426,
%U A074083 28517864,51095760,91232520,162372682,288115147,509790277,899630376
%N A074083 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)=(1,1).
%C A074083 The coefficient of q^0 in nu(n) is the Fibonacci number F(n+1). The coefficient of q^1 is A023610(n-3).
%H A074083 M. Beattie, S. Dăscălescu and S. Raianu, <a href="https://arxiv.org/abs/math/0204075">Lifting of Nichols Algebras of Type B_2</a>, arXiv:math/0204075 [math.QA], 2002.
%H A074083 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (4, -2, -8, 5, 8, -2, -4, -1).
%F A074083 G.f.: (4x^5-2x^6-9x^7+x^8+6x^9+2x^10)/(1-x-x^2)^4.
%F A074083 a(n) = 4a(n-1)-2a(n-2)-8a(n-3)+5a(n-4)+8a(n-5)-2a(n-6)-4a(n-7)-a(n-8) for n>=11.
%e A074083 The first 6 nu polynomials are nu(0)=1, nu(1)=1, nu(2)=2, nu(3)=3+q, nu(4)=5+3q+2q^2, nu(5)=8+7q+6q^2+4q^3+q^4, so the coefficients of q^3 are 0,0,0,0,0,4.
%t A074083 b=1; 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]
%t A074083 (* Second program: *)
%t A074083 Join[{0, 0, 0}, LinearRecurrence[{4, -2, -8, 5, 8, -2, -4, -1}, {0, 0, 4, 14, 39, 97, 224, 494}, 31]] (* _Jean-François Alcover_, Jan 27 2019 *)
%Y A074083 Coefficients of q^0, q^1 and q^2 are in A000045, A023610 and A074082. Related sequences with different values of b and lambda are in A074084-A074089.
%K A074083 nonn
%O A074083 0,6
%A A074083 Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 19 2002
%E A074083 Edited by _Dean Hickerson_, Aug 21 2002