%I #27 Jan 28 2025 15:28:31
%S 0,0,0,2,8,22,60,146,352,814,1860,4170,9256,20326,44300,95874,206320,
%T 441758,941780,2000058,4233144,8932310,18796700,39457522,82643328,
%U 172743182,360399460,750625066,1560902472,3241109574,6720828460,13918875490,28792188176
%N 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)=(1,2).
%C Coefficient of q^0 is A001045(n+1).
%H Colin Barker, <a href="/A074352/b074352.txt">Table of n, a(n) for n = 0..1000</a>
%H 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 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,3,-4,-4).
%F a(0) = 0; for n>0, a(n) = (1/27)*(2^n*(6*n-11) + (-1)^n*(6*n-16)).
%F From _Colin Barker_, Nov 18 2017: (Start)
%F G.f.: 2*x^3*(1 + 2*x) / ((1 + x)^2*(1 - 2*x)^2).
%F a(n) = 2*a(n-1) + 3*a(n-2) - 4*a(n-3) - 4*a(n-4) for n>4. (End)
%e The first 6 nu polynomials are nu(0)=1, nu(1)=1, nu(2)=3, nu(3)=5+2q, nu(4)=11+8q+6q^2, nu(5)=21+22q+20q^2+14q^3+4q^4, so the coefficients of q^1 are 0,0,0,2,8,22.
%t LinearRecurrence[{2, 3, -4, -4}, {0, 0, 0, 2, 8}, 50] (* _Paolo Xausa_, Jan 28 2025 *)
%o (PARI) a(n)=if(n<1,0,(1/27)*(2^n*(6*n-11)+(-1)^n*(6*n-16)))
%o (PARI) a(n)=if(n<1,0,(1/81)*(2^(n-1)*(6*n^2-43)+ (-1)^n*(6*n^2-24*n+62)))
%o (PARI) concat(vector(3), Vec(2*x^3*(1 + 2*x) / ((1 + x)^2*(1 - 2*x)^2) + O(x^40))) \\ _Colin Barker_, Nov 18 2017
%Y Coefficient of q^0, q^2 and q^3 are in A001045, A074353 and A074354. Related sequences with other values of b and lambda are in A074082-A074089, A074353-A074363.
%K nonn,easy
%O 0,4
%A Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
%E More terms and formula from _Benoit Cloitre_, Jan 12 2003
%E Corrected by _Franklin T. Adams-Watters_, Oct 25 2006
%E Corrected by _T. D. Noe_, Oct 25 2006