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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)=(1,2).
5

%I #19 Feb 20 2022 07:50:41

%S 0,0,0,0,0,14,64,218,692,1982,5496,14562,37692,95142,236032,576074,

%T 1387780,3304078,7787656,18190386,42151116,96972534,221651472,

%U 503650970,1138286740,2559944414,5731095704,12776843138,28374100572

%N 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)=(1,2).

%C Coefficient of q^0 is A001045(n+1).

%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.

%F Conjectures from _Colin Barker_, Nov 18 2017: (Start)

%F G.f.: 2*x^5*(1 + 2*x)*(7 - 10*x - 13*x^2 + 12*x^3 + 12*x^4) / ((1 + x)^4*(1 - 2*x)^4).

%F a(n) = 4*a(n-1) + 2*a(n-2) - 20*a(n-3) - a(n-4) + 40*a(n-5) + 8*a(n-6) - 32*a(n-7) - 16*a(n-8) for n>10.

%F (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 0,0,0,0,0,14.

%Y Coefficients of q^0, q^1 and q^2 are in A001045, A074352 and A074353. Related sequences with other values of b and lambda are in A074082-A074089, A074355-A074363.

%K nonn

%O 0,6

%A Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

%E More terms from _Benoit Cloitre_, Jan 16 2003

%E Corrected by _T. D. Noe_, Oct 25 2006