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Numbers k such that 2^k - 1 and 2^k + 1 have the same number of prime factors, counted with multiplicity.
2

%I #17 Sep 08 2022 08:46:22

%S 2,10,11,14,21,23,29,39,47,50,53,55,63,71,73,74,75,82,86,95,101,105,

%T 113,115,121,142,147,150,167,169,179,181,182,190,199,203,209,233,235,

%U 253,277,285,303,307,311,317,335,337,339,342,343,347,349,353,355,358

%N Numbers k such that 2^k - 1 and 2^k + 1 have the same number of prime factors, counted with multiplicity.

%e a(1) = 2: 2^2 - 1 = 3 and 2^2 + 1 are both prime,

%e a(2) = 10: 2^10 - 1 = 1023 = 3 * 11 * 31 and 2^10 + 1 = 1025 = 5^2 * 41 both have 3 prime factors.

%t Select[Range[200], PrimeOmega[2^# - 1 ] == PrimeOmega[2^# + 1 ] &] (* _Amiram Eldar_, Aug 24 2019 *)

%o (PARI) for(k=1, 209, my(f=bigomega(2^k-1),g=bigomega(2^k+1));if(f==g,print1(k,", ")))

%o (Magma) [m:m in [2..400]| &+[p[2]: p in Factorization(2^m-1)] eq &+[p[2]: p in Factorization(2^m+1)]]; // _Marius A. Burtea_, Aug 24 2019

%Y Cf. A000051, A000225, A046051, A054992, A067886.

%K nonn

%O 1,1

%A _Hugo Pfoertner_, Aug 24 2019

%E More terms from _Amiram Eldar_, Aug 24 2019