%I #10 Dec 05 2013 19:56:52
%S 1,2,36,4944,61824720,7418966770948320,
%T 86554612327730451932767396638240,
%U 9891955694597765935173416758656692436898621843615462139173105920
%N a(1) = 1; a(n) = n multiplied by the concatenation of all previous terms.
%C a(n) >= 10^(2^(n-2)-1) (can be easily shown by induction).
%e Let n = 4. The previous terms are 1,2 and 36. Their concatenation is 1236. This number is multiplied by 4 and we get a(4) = 4944.
%t a = {1}; For[n=2,n<10,n++,AppendTo[a,n*FromDigits[Flatten[IntegerDigits[a]]]]]; a
%K base,nonn,less
%O 1,2
%A _Amarnath Murthy_, Jun 11 2004
%E Edited, corrected and extended by _Stefan Steinerberger_, Jun 16 2007