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 A110874 Number of prime factors of 2 + n^(n+1) counted with multiplicity. 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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). LINKS Sean A. Irvine, Table of n, a(n) for n = 1..100 FORMULA a(n) = A001222(1 + A110567(n)) = A001222(2 + A007778(n)) = A001222(2 + n^(n+1)). EXAMPLE 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? MATHEMATICA Table[PrimeOmega[2+n^(n+1)], {n, 41}] (* Harvey P. Dale, Nov 08 2020 *) CROSSREFS Cf. A001222, A007778, A110567, A110676. Sequence in context: A308569 A350073 A174978 * A010253 A065274 A260325 Adjacent sequences: A110871 A110872 A110873 * A110875 A110876 A110877 KEYWORD nonn AUTHOR Jonathan Vos Post, Sep 18 2005 EXTENSIONS More terms from Sean A. Irvine, Sep 17 2023 STATUS approved

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Last modified December 5 05:37 EST 2023. Contains 367575 sequences. (Running on oeis4.)