A283364 Numbers m such that both numbers 2^m +- 1 have at most 2 distinct prime factors.

1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 17, 19, 23, 31, 61, 101, 127, 167, 199, 347

Offset: 1

Author

Vladimir Shevelev, Mar 06 2017

Keywords

• nonn
• more
• hard

Comments

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.

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]

Any further terms are > 1122. - Lucas A. Brown, Oct 21 2024

Crossrefs

Cf. A001221, A046799, A046800, A067886.

Programs

Mathematica

Select[Range@ 200, Times @@ Boole@ Map[PrimeNu@ # <= 2 &, 2^# + {-1, 1}] == 1 &] (* Michael De Vlieger, Mar 06 2017 *)
Select[Range[350],Max[PrimeNu[2^#+{1,-1}]]<3&] (* Harvey P. Dale, Dec 23 2017 *)

PARI

isok(n) = omega(2^n+1)<=2 && omega(2^n-1)<=2;
for(n=1, 347, if(isok(n)==1, print1(n,", "))); \\ Indranil Ghosh, Mar 06 2017

Extensions

More terms from Peter J. C. Moses, Mar 06 2017