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a(n) = a(n-1)^n+n-1; a(1)=1.
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%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