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%I #10 Oct 04 2024 23:08:50
%S 7,27,627,390627,152587890627,23283064365386962890627,
%T 542101086242752217003726400434970855712890627
%N a(n) = 5^(2^(n-1)) + 2.
%C Except for the first term, these numbers are divisible by 3. This follows from the identity I: a^n-b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)) for odd values of n. In this example, by expanding the binomial (3+2)^(2^n)+2, we get 3h + 2^(2^n)+2 for some h. Now 2^(2^n)+2 = 2*(2^(2^n-1)+1). Since 2^n-1 is odd, by identity I, 3 divides 2^(2^n)+2 + 3h. Therefore 3 divides 5^(2^n)+2 for n > 0.
%o (PARI) a(n) = 5^(2^(n-1)) + 2
%K nonn,easy
%O 1,1
%A _Cino Hilliard_, Dec 08 2008