%I
%S 3,2,2,3,8,35,204,1421,11360,102231,1022300,11245289,134943456,
%T 1754264915,24559708796,368395631925,5894330110784,100203611883311,
%U 1803665013899580,34269635264092001,685392705281840000
%N a(n) = n*(a(n1)1) starting with a(0)=3.
%C Starting with a(0)=0 would give A007526(n); starting with a(0)=1 would give A038156(n). In general, for this recurrence, a(n) = ceiling(1 + n!*(a(0)e)) for n>0; this is the first case with positive terms.
%F a(n) = ceiling(1 + n!*(3e)) for n>0.
%F a(n) = 3*n!  n! * sum_{k=0..n} 1/(k1)!, with n>=0.  _Paolo P. Lava_, Oct 07 2008
%F a(n) = n!  floor(e*n!) + 1, n>0.  _Gary Detlefs_, Jun 06 2010
%e a(5) = 5*(a(4)1) = 5*(81) = 35.
%t a=3;Table[a=a*nn,{n,1,2*4!}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 22 2010 *)
%t RecurrenceTable[{a[0]==3,a[n]==n(a[n1]1)},a,{n,20}] (* _Harvey P. Dale_, Jul 17 2018 *)
%K easy,nonn
%O 0,1
%A _Henry Bottomley_, Apr 10 2006
