A074360 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)=(2,2).
0, 0, 0, 0, 0, 40, 232, 1072, 4400, 16864, 61728, 218496, 753792, 2547840, 8468608, 27755776, 89886976, 288101888, 915089920, 2883416064, 9021001728, 28042881024, 86672025600, 266472878080, 815347462144, 2483820617728
Offset: 0
Keywords
Examples
The first 6 nu polynomials are nu(0)=1, nu(1)=2, nu(2)=6, nu(3)=16+4q, nu(4)=44+20q+12q^2, nu(5)=120+80q+64q^2+40q^3+8q^4, so the coefficients of q^1 are 0,0,0,0,0,40.
Links
- M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.
Crossrefs
Programs
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Maple
nu := proc(b,lambda,n) global q; local qp,i ; if n = 0 then RETURN(1) ; elif n =1 then RETURN(b) ; fi ; qp:=0 ; for i from 0 to n-2 do qp := qp + q^i ; od ; RETURN( b*nu(b,lambda,n-1)+lambda*qp*nu(b,lambda,n-2)) ; end: A074360 := proc(n) RETURN( coeftayl(nu(2,2,n),q=0,3) ) ; end: for n from 0 to 30 do printf("%d,", A074360(n)) ; od ; # R. J. Mathar, Sep 20 2006
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Mathematica
nu[0] = 1; nu[1] = 2; nu[n_] := nu[n] = 2*nu[n - 1] + 2*Total[q^Range[0, n - 2]]*nu[n - 2] // Expand; a[n_] := Coefficient[nu[n], q, 3]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Nov 18 2017 *)
Formula
Conjecture: O.g.f: 8*x^5*(1+x)*(12*x^4+24*x^3-2*x^2-16*x+5)/(2*x^2+2*x-1)^4. - R. J. Mathar, Jul 22 2009
Extensions
More terms from R. J. Mathar, Sep 20 2006
Comments