A110874 Number of prime factors of 2 + n^(n+1) counted with multiplicity.

1, 2, 1, 5, 2, 2, 4, 5, 2, 5, 4, 4, 5, 3, 1, 4, 5, 3, 4, 6, 3, 8, 4, 5, 4, 4, 2, 6, 3, 6, 5, 5, 5, 6, 6, 8, 6, 6, 4, 5, 4, 6, 4, 5, 3, 8, 4, 3, 5, 5, 5, 7, 7, 11, 4, 5, 4, 13, 4, 6, 2, 5, 2, 6, 6, 5, 8, 9, 5, 9, 4, 7, 4, 4, 5, 7, 6, 7, 6, 9, 4, 9, 5, 8, 5, 8

Offset: 1

Author

Jonathan Vos Post, Sep 18 2005

Keywords

• nonn

Comments

Compared with A110676, number of prime factors with multiplicity of 2 + n^(n+1), this seems to have an unlimited number of primes (n = 1, 3, 15, ...) and semiprimes (n = 2, 5, 6, 9, 27, ...). Of course, n even gives n | a(n).

Examples

			a(1) = 1 because 2 + 1^2 = 3 is prime (one prime factor).
a(2) = 2 because 2 + 2^3 = 10 = 2 * 5 is semiprime (two prime factors).
a(3) = 1 because 2 + 3^4 = 83 is prime.
a(4) = 5 because 2 + 4^5 = 1026 = 2 * 3^3 * 19 has five prime factors (3 has multiplicity of 3).
a(5) = 2 because 2 + 5^6 = 15627 = 3 * 5209 is semiprime (two prime factors).
a(6) = 2 because 2 + 6^7 = 279938 = 2 * 139969 is semiprime (two prime factors).
a(15) = 1 because 2 + 15^16 = 6568408355712890627 is prime. What is the next prime?
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A001222, A007778, A110567, A110676.

Programs

Mathematica

Table[PrimeOmega[2+n^(n+1)],{n,41}] (* Harvey P. Dale, Nov 08 2020 *)

Formula

a(n) = A001222(1 + A110567(n)) = A001222(2 + A007778(n)) = A001222(2 + n^(n+1)).

Extensions

More terms from Sean A. Irvine, Sep 17 2023