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a(0)=a(1)=a(2)=0; thereafter a(n) = n*a(n-1) + n*a(n-2)/(n-2) + (-1)^(n-1)*4/(n-2).
2

%I #22 Mar 17 2023 09:55:38

%S 0,0,0,4,14,78,488,3526,28858,264256,2678632,29787932,360669542,

%T 4723907966,66555492656,1003783052878,16136592266226,275459689319104,

%U 4976428074043376,94860000118416084,1902729366895036542,40062161968084543054,883460565601444487384,20363470614798268185558,489687069917632739530538,12264310955130816605856448

%N a(0)=a(1)=a(2)=0; thereafter a(n) = n*a(n-1) + n*a(n-2)/(n-2) + (-1)^(n-1)*4/(n-2).

%C Muir gives this recurrence without specifying the initial values.

%C In general, for the same recurrence a(n) = n*a(n-1) + n*a(n-2)/(n-2) + (-1)^(n-1)*4/(n-2), with a(1)=0, a(2)=0, a(3)=m, is a(n) ~ c * n!, where c = exp(-2) + (BesselI(0,2)-BesselI(1,2))*(m-1)/3. Set m=4 for this sequence and m=1 for A000179. - _Vaclav Kotesovec_, May 05 2015

%D T. Muir, A Treatise on the Theory of Determinants. Dover, NY, 1960, Sect. 132, p. 112.

%H Vincenzo Librandi, <a href="/A186638/b186638.txt">Table of n, a(n) for n = 0..69</a>

%F Recurrence (for n>2): (n-2)*a(n) = (n^2 - 3*n + 3)*a(n-1) + (n^2 - 3*n + 3)*a(n-2) + (n-1)*a(n-3). - _Vaclav Kotesovec_, May 05 2015

%F a(n) ~ c * n!, where c = exp(-2) + BesselI(0,2) - BesselI(1,2) = 0.8242837309353508959489495107843515087389944891994982884067... . - _Vaclav Kotesovec_, May 05 2015

%p W:=proc(n) option remember; if n <= 2 then 0 else

%p n*W(n-1)+n*W(n-2)/(n-2)+(-1)^(n-1)*4/(n-2); fi; end;

%t Flatten[{0,0,RecurrenceTable[{a[2]==0,a[3]==4,a[n]==n*a[n-1]+n*a[n-2]/(n-2)+(-1)^(n-1)*4/(n-2)},a,{n,2,20}]}] (* _Vaclav Kotesovec_, May 05 2015 *)

%o (Maxima) a[0]:0$ a[1]:0$ a[2]:0$ a[n]:=n*a[n-1]+n*a[n-2]/(n-2)+4*(-1)^(n-1)/(n-2)$ makelist(a[n], n, 0, 25); /* _Bruno Berselli_, May 23 2011 */

%Y A000179 satisfies essentially the same recurrence.

%K nonn

%O 0,4

%A _N. J. A. Sloane_, Feb 24 2011