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%I #22 Sep 21 2022 18:33:09
%S 2,2,2,2,2,7072833120,2072005925466,240164550712338756
%N a(n) is the smallest number m greater than 1 such that m^(2^k)+1 for k=0,1,...,n are primes.
%C The first five terms of this sequence correspond to Fermat primes.
%C Note that 7072833120 is not the smallest base to give at least six possibly nonconsecutive k values. For example, 292582836^(2^k) + 1 is prime for k = 0,1,2,3,4,7. - _Jeppe Stig Nielsen_, Sep 18 2022
%H PrimeGrid, <a href="http://www.primegrid.com/forum_thread.php?id=9261">Generalized Fermat Progression Search</a>
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_137.htm">Puzzle 137. Product of primes + 1, a square</a>, The Prime Puzzles and Problems Connection.
%H Carlos Rivera, <a href="http://primepuzzles.net/puzzles/puzz_399.htm">Prime Puzzle 399. Some more terms</a>, The Prime Puzzles and Problems Connection.
%e a(5)=7072833120 because 7072833120^2^k+1 for k=0,1,2,3,4,5 are primes.
%Y Cf. A019434, A000215.
%Y Cf. A090873, A090874, A090875.
%Y All solutions for fixed n: A006093 (n=0), A070689 (n=1), A070325 (n=2), A070655 (n=3), A070694 (n=4), A235390 (n=5), A335805 (n=6), A337364 (n=7).
%K nonn,more
%O 0,1
%A _Farideh Firoozbakht_, Jan 31 2004
%E a(6) from _Jens Kruse Andersen_, May 06 2007
%E a(7) from _Kellen Shenton_, Aug 13 2020
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