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%I #4 Dec 05 2013 19:56:32
%S 1,1,449,1,61206166658506517441,359
%N a(1) = 1, a(n) = smallest number such that the n-th partial concatenation has n divisors.
%C Next term is about 6.4*10^150 and is too large to include. - _David Wasserman_, Jun 16 2005
%e 1,11,11449,114491, etc. have 1, 2, 3 and 4 divisors respectively.
%e 11449= 107^2, 114491 = 13*8807.
%K base,easy,nonn
%O 1,3
%A _Amarnath Murthy_, Sep 26 2003
%E One more term from _David Wasserman_, Jun 16 2005
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