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%I #13 Jun 19 2018 05:43:59
%S 1,2,3,4,5,9,17,25,81,241
%N Numbers n such that n, n+1 and n+2 all have primitive roots.
%C Start of run of 3 consecutive numbers in A033948.
%C The next term is 3^541 - 2, which is too large to be included here. No more terms below 3^100000, or approximately 1.33*10^47712.
%C There is a multiple of 4 in every four consecutive positive integers and it clearly has no primitive roots if it is larger than 4. Again, there is a multiple of 3 in every three consecutive positive integers, so it must be a power of 3 or two times a power of 3, and the other two numbers must be odd prime powers or two times odd prime powers.
%C According to Pillai's conjecture, there're only finitely many solutions to |3^a - p^b| = 2, |3^a - 2*p^b| = 1, |p^a - 2*3^b| = 1 with a,b >= 2, p odd primes (no solution other than 3^3 - 5^2 = 2, 3^5 - 2*11^2 = 1 below 3^100000). So beyond (25, 26, 27) and (241, 242, 243), it's very likely that all three consecutive numbers with primitive roots are of the form (3^i, 3^i + 1, 3^i + 2), (3^j - 2, 3^j - 1, 3^j), (2*3^k - 1, 2*3^k, 2*3^k + 1) such that (3^i + 1)/2, 3^i + 2, 3^j - 2, (3^j - 1)/2, 2*3^k - 1, 2*3^k + 1 are primes, which only produces one more solution (3^541 - 2, 3^541 - 1, 3^541) below 3^1000000.
%e 81, 82, 83 all have primitive roots (in fact, their least common primitive root is 47), so 81 is a term.
%e Note that A014224 and A028491 have a term 541 in common, so 3^541 - 2, 3^541 - 1 and 3^541 all have primitive roots, so 3^541 - 2 is a term.
%Y Cf. A003306, A003307, A014224, A028491, A051783, A171381.
%K nonn,hard,more
%O 1,2
%A _Jianing Song_, Jun 04 2018