%I #7 Aug 07 2013 10:05:23
%S 1,2,10,10003,100150090027004050247
%N a(n) = a(n-1)^n+n-1; a(1)=1.
%C a(6) has 121 digits. a(n) modulo 2 gives the pattern 1,0,0,1,1,0,0,1,1...
%F a(n) = a(n-1)^n+n-1.
%e a(4) = a(3)^4+3 = 10^4+3 = 10003.
%t RecurrenceTable[{a[1]==1,a[n]==a[n-1]^n+n-1},a[n],{n,6}] (* _Harvey P. Dale_, Oct 17 2011 *)
%K nonn
%O 1,2
%A Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 26 2009