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a(1) = 1; a(n+1) = a(n)*(a(n) + n).
0

%I #17 Jan 19 2015 04:07:23

%S 1,2,8,88,8096,65585696,4301483913318592,

%T 18502763856538658405109886092608,

%U 342352270330833327273885765082712290623512199361069533040982528

%N a(1) = 1; a(n+1) = a(n)*(a(n) + n).

%F a(n) ~ c^(2^n), where c = 1.32476133150649295684153231492231865907402550177962959967618... . - _Vaclav Kotesovec_, Jan 19 2015

%p a[1]:=1: for n from 1 to 10 do a[n+1]:=a[n]*(a[n]+n) od: seq(a[n],n=1..10); # _Emeric Deutsch_, Jul 31 2005

%t s=1;lst={};Do[s*=n+s;AppendTo[lst, s], {n, 0, 3!, 1}];lst (* _Vladimir Joseph Stephan Orlovsky_, Nov 10 2008 *)

%t RecurrenceTable[{a[1]==1,a[n]==a[n-1](a[n-1]+n-1)},a[n],{n,10}] (* _Harvey P. Dale_, May 04 2011 *)

%K easy,nonn

%O 1,2

%A _Amarnath Murthy_, Jul 25 2005

%E Corrected and extended by _Emeric Deutsch_, Jul 31 2005