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%I #38 Jun 17 2022 16:02:14
%S 2,3,5,17,65537
%N a(n) = (2^2^...^2) (with n 2's) + 1.
%C a(0) could equally well be taken to be 1 rather than 2, which gives A007516. - _N. J. A. Sloane_, Sep 14 2009
%C A subsequence of the Fermat numbers 2^2^n + 1 = A000215.
%C a(0) through a(4) are primes; a(5) = 2^65536 + 1 is divisible by 825753601.
%C a(5) = 20035299...19156737 has 19729 decimal digits. - _Alois P. Heinz_, Jun 15 2022
%C It is unknown if a(6) = A000215(65536) is composite. - _Jeppe Stig Nielsen_, Jun 15 2022
%D P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 73.
%H Y. Bugeaud and M. Queffélec, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Bugeaud/bugeaud3.html">On Rational Approximation of the Binary Thue-Morse-Mahler Number</a>, Journal of Integer Sequences, 16 (2013), #13.2.3.
%H Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Prime factors k.2^n + 1 of Fermat numbers F_m</a>
%F a(0) = 2, a(n) = 2^a(n-1)/2 + 1 for n >= 1.
%F a(n) = A014221(n) + 1. - _Leroy Quet_, Jun 10 2009, updated by _Jeppe Stig Nielsen_, Jun 15 2022
%Y Cf. Fermat numbers 2^2^n + 1 = A000215. A007516 is another version.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, _Robert G. Wilson v_, _David W. Wilson_
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