A074353 Coefficient of q^2 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).
0, 0, 0, 0, 6, 20, 70, 196, 542, 1396, 3526, 8628, 20766, 49092, 114598, 264356, 603998, 1368148, 3076166, 6870740, 15256158, 33696804, 74073510, 162127940, 353460766, 767816500, 1662394310, 3588252916, 7723318942, 16580031876, 35506388646, 75864499428
Offset: 0
Examples
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^2 are 0,0,0,0,6,20.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- 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 (3,3,-11,-6,12,8).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{3, 3, -11, -6, 12, 8}, {0, 0, 0, 0, 6, 20, 70, 196}, 50] (* Paolo Xausa, Jan 28 2025 *)
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PARI
concat(vector(4), Vec(2*x^4*(3 + x - 4*x^2 - 4*x^3) / ((1 + x)^3*(1 - 2*x)^3) + O(x^40))) \\ Colin Barker, Nov 18 2017
Formula
a(0)=0 for n>0, a(n) = (1/81)*(2^(n-1)*(6*n^2-43) + (-1)^n*(6*n^2-24*n+62)). - Benoit Cloitre, Jan 16 2003
From Colin Barker, Nov 18 2017: (Start)
G.f.: 2*x^4*(3 + x - 4*x^2 - 4*x^3) / ((1 + x)^3*(1 - 2*x)^3).
a(n) = 3*a(n-1) + 3*a(n-2) - 11*a(n-3) - 6*a(n-4) + 12*a(n-5) + 8*a(n-6) for n>7.
(End)
Extensions
More terms from Benoit Cloitre, Jan 16 2003
Corrected by T. D. Noe, Oct 25 2006
Comments