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Number of distinct prime factors in 2^n + 3.
1

%I #14 Sep 08 2022 08:45:05

%S 1,1,1,1,2,1,1,2,2,2,2,1,3,2,1,1,3,1,3,3,3,3,3,2,2,2,2,1,4,1,3,2,4,3,

%T 2,3,3,2,2,2,3,2,4,4,2,2,3,2,3,3,4,2,5,2,1,3,3,3,4,3,2,2,4,2,5,2,1,5,

%U 3,3,3,3,3,4,4,2,4,3,4,5,4,4,5,1,4,4,2,3,5,4,4,4,4,5,4,4,6,3,2,2

%N Number of distinct prime factors in 2^n + 3.

%H Amiram Eldar, <a href="/A067390/b067390.txt">Table of n, a(n) for n = 1..699</a>

%F a(n) = A001221(A062709(n)). - _Amiram Eldar_, Feb 06 2020

%e a(5) = 2 since 2^5 + 3 = 35 = 5 * 7 has 2 distinct prime factors.

%t PrimeNu @ Table[2^n + 3, {n, 1, 50}] (* _Amiram Eldar_, Feb 06 2020 *)

%o (Magma) [#PrimeDivisors(2^n+3):n in [1..100]]; // _Marius A. Burtea_, Feb 06 2020

%Y Cf. A001221, A062709.

%K nonn

%O 1,5

%A _Benoit Cloitre_, Feb 23 2002