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%I #5 Sep 10 2016 16:20:49
%S 7,259,4294967299,340282366920938463463374607431768211459
%N 4^(4^n) + 3.
%C Numbers of the form 4^(4^n) + 3 are divisible by 7. We prove this by induction making use of the expansion (1) a^m - b^m = (a-b)(a^(m-1)+a^(m-2)b+...+b^(m-1). For n = 1, we have 4^4 + 3 = 259 = 7*37. So the statement is true for n=1. Now assume the statement is true for some integer k and show that it is also true for k+1. Thus we have 4^(4^k) + 3 = 7h for some h. Let 4^(4^(k+1))+ 3 = h1. Now consider the difference h1 - 7h. If this is a multiple of 7 then so is h1. So we have 4^(4^(k+1))+ 3 - (4^(4^k) + 3)) = 256^(4^k)-4^(4^k). This is of the form (1) where n = 4^k, a = 256 and b = 4. So the difference, a^n-b^n is divisible by (a-b) = (256-4) = 252 = 7*36. This implies 4^(4^(k+1)+3 is divisible by 7. So we have assumed the statement was true for k and have shown it to be true for k+1. Therefore by the induction hypothesis, the statement is true for all n.
%F This function is derived from the odd case of Fermat numbers of order 3 or F(m,3) = 2^(2^m)+3. Let m = 2n+1, to get 2^(2^(2n+1)) + 3 = 2^(2*2^(2*n))+3 = 4^(2^(2*n))+3 = 4^(4^n)) + 3.
%F a(n) = 3+A137840(n). - _R. J. Mathar_, Sep 10 2016
%e For n = 0, 4^(4^0) + 3 = 4 + 3 = 7, the first entry.
%o (PARI) g(n) = for(x=0,n,y=4^(4^x)+3;print1(y","))
%K nonn
%O 0,1
%A _Cino Hilliard_, Jul 07 2007
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