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a(1) = 1; a(n) = n multiplied by the concatenation of all previous terms.
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%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