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%I #26 Sep 08 2022 08:45:30
%S 5,7,19,259,65539,4294967299,18446744073709551619,
%T 340282366920938463463374607431768211459,
%U 115792089237316195423570985008687907853269984665640564039457584007913129639939
%N Fermat numbers of order 3 or F(n,3) = 2^(2^n)+3.
%C This is equivalent to F(n)+2 or 2^(2^n)+ 1 + 2. Conjecture: If n is odd, 7 is a divisor of F(n,3).
%C The conjecture is true: the order of 2 mod 7 is 3, and if n is odd then 2^n == 2 mod 3 so 2^(2^n) + 3 == 2^2 + 3 == 0 mod 7. - _Robert Israel_, Nov 20 2014
%H Vincenzo Librandi, <a href="/A130729/b130729.txt">Table of n, a(n) for n = 0..11</a>
%H Tigran Hakobyan, <a href="http://arxiv.org/abs/1601.04946">On the unboundedness of common divisors of distinct terms of the sequence a(n)=2^2^n+d for d>1</a>, arXiv:1601.04946 [math.NT], 2016.
%F F(n,m): The n-th Fermat number of order m = 2^(2^n)+ m. The traditional Fermat numbers are F(n,1) or Fermat numbers of order 1.
%t Table[(2^(2^n) + 3), {n, 0, 15}] (* _Vincenzo Librandi_, Jan 09 2013 *)
%o (PARI) fplusm(n,m)= { local(x,y); for(x=0,n, y=2^(2^x)+m; print1(y",") ) }
%o (Magma) [2^(2^n) + 3: n in [0..11]]; // _Vincenzo Librandi_, Jan 09 2013
%Y Cf. A063486, A130730.
%K nonn
%O 0,1
%A _Cino Hilliard_, Jul 05 2007
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