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%I #50 Oct 21 2024 09:00:11
%S 1,2,3,4,5,6,7,9,11,13,17,19,23,31,61,101,127,167,199,347
%N Numbers m such that both numbers 2^m +- 1 have at most 2 distinct prime factors.
%C If a(n) > 9 then a(n) is prime. Proof: If k = 2*m > 9 then 2^(2*m)-1 has at least 3 factors; being 3, (2^m - 1) / 3 and 2^m + 1 which excludes even numbers > 9.
%C If k = 2*m + 1 > 9 is not prime then k = p*q, q, p > 3 so 2^(p*q) + 1 is divisible by 3, 2^p + 1 and 2^q + 1. If p = q then 2^(p^2) + 1 is divisible by 3, 2^p + 1 and (2^p^2 + 1) / (2^p + 1) > 2^p + 1. Which excludes odd composite numbers > 9 and completes the proof. [comments reworded by _David A. Corneth_, Nov 23 2019]
%C Any further terms are > 1122. - _Lucas A. Brown_, Oct 21 2024
%t Select[Range@ 200, Times @@ Boole@ Map[PrimeNu@ # <= 2 &, 2^# + {-1, 1}] == 1 &] (* _Michael De Vlieger_, Mar 06 2017 *)
%t Select[Range[350],Max[PrimeNu[2^#+{1,-1}]]<3&] (* _Harvey P. Dale_, Dec 23 2017 *)
%o (PARI) isok(n) = omega(2^n+1)<=2 && omega(2^n-1)<=2;
%o for(n=1, 347, if(isok(n)==1, print1(n,", "))); \\ _Indranil Ghosh_, Mar 06 2017
%Y Cf. A001221, A046799, A046800, A067886.
%K nonn,more,hard
%O 1,2
%A _Vladimir Shevelev_, Mar 06 2017
%E More terms from _Peter J. C. Moses_, Mar 06 2017