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a(1)=a(2)=1, a(n) = (a(n-1)+1)*(a(n-2)+1) for n > 2.
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%I #4 Feb 27 2015 16:37:16

%S 1,1,4,10,55,616,34552,21319201,736642386706,15704627843968647814,

%T 11568694537326272321321120595205,

%U 181682042349262169758803442669575561298555791374890

%N a(1)=a(2)=1, a(n) = (a(n-1)+1)*(a(n-2)+1) for n > 2.

%C The terms have an interesting factorization pattern, often sharing factors.

%C The next term (a(13)) has 82 digits. - _Harvey P. Dale_, Feb 27 2015

%t RecurrenceTable[{a[1]==a[2]==1,a[n]==(a[n-1]+1)(a[n-2]+1)},a,{n,15}] (* _Harvey P. Dale_, Feb 27 2015 *)

%K nonn

%O 1,3

%A _Gerald McGarvey_, Aug 08 2004