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%I #6 Oct 01 2013 21:35:28
%S 13,123,14643,214358883,45949729863572163,
%T 2111377674535255285545615254209923,
%U 4457915684525902395869512133369841539490161434991526715513934826243
%N Numbers of the form 11^(2^n) + 2.
%C Except for the first term, these numbers are divisible by 3. This follows from the binomial expansion of (9+2)^(2^n)+2 = 9h + 2^(2^n)+2. Now 2^(2^n)+2 can be written as 2*(2^(2^n-1)+1) and 2^(2^n-1)+1 is divisible by 3. This is evident from the identity, a^m+b^m = (a+b)(a^(m-1) - a(m-2)b + ... + b^(m-1)) for odd m and 2^n-1 is odd.
%o (PARI) g(a,n) = if(a%2,b=2,b=1);for(x=0,n,y=a^(2^x)+b;print1(y","))
%K nonn
%O 1,1
%A _Cino Hilliard_, Dec 08 2008