%I #17 Jan 01 2024 11:44:51

%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(n-1)-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!*(3-e)) for n>0.

%F a(n) = n! - floor(e*n!) + 1, n>0. - _Gary Detlefs_, Jun 06 2010

%e a(5) = 5*(a(4)-1) = 5*(8-1) = 35.

%t a=3;Table[a=a*n-n,{n,1,2*4!}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 22 2010 *)

%t RecurrenceTable[{a[0]==3,a[n]==n(a[n-1]-1)},a,{n,20}] (* _Harvey P. Dale_, Jul 17 2018 *)

%K easy,nonn

%O 0,1

%A _Henry Bottomley_, Apr 10 2006