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A228483
a(n) = 2 - mu(n), where mu(n) is the Moebius function (A008683).
3
1, 3, 3, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 1, 2, 3, 2, 3, 2, 1, 1, 3, 2, 2, 1, 2, 2, 3, 3, 3, 2, 1, 1, 1, 2, 3, 1, 1, 2, 3, 3, 3, 2, 2, 1, 3, 2, 2, 2, 1, 2, 3, 2, 1, 2, 1, 1, 3, 2, 3, 1, 2, 2, 1, 3, 3, 2, 1, 3, 3, 2, 3, 1, 2, 2, 1, 3, 3, 2, 2, 1, 3, 2, 1, 1, 1
OFFSET
1,2
COMMENTS
1 <= a(n) <= 3: a(n) = 1 when n is both squarefree and has an even number of distinct prime factors (or if n = 1). So a(n) = 1 when mu(n) = 1. a(n) = 2 when n is square-full. a(n) = 3 when n is both squarefree and has an odd number of distinct prime factors.
When n is semiprime, a(n) is equal to the ratio of the number of prime factors of n (with multiplicity) to the number of its distinct prime factors. Analogously, when n is semiprime, a(n) is equal to the ratio of the sum of the prime factors of n (with repetition) to the sum of its distinct prime factors.
LINKS
FORMULA
a(n) = 2 - mu(n) = 2 - A008683(n).
a(A001358(n)) = 5 - tau(A001358(n)) = 3 - omega(A001358(n)) = 3 + 2*A001358(n) - sigma(A001358(n)) - phi(A001358(n)) = Omega(A001358(n))/omega(A001358(n))= sopfr(A001358(n))/sopf(A001358(n)).
EXAMPLE
a(19) = 3 because mu(19) = -1 and 2 - (-1) = 3.
a(20) = 2 because mu(20) = 0 and 2 - 0 = 2.
a(21) = 1 because mu(21) = 1 and 2 - 1 = 1.
MAPLE
with(numtheory); seq(2-mobius(k), k=1..70);
MATHEMATICA
2 - MoebiusMu[Range[100]] (* Alonso del Arte, Aug 22 2013 *)
PROG
(Magma) [2-MoebiusMu(n): n in [1..100]]; // Vincenzo Librandi, Aug 23 2013
(PARI) a(n) = 2 - moebius(n); \\ Michel Marcus, Apr 26 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Aug 22 2013
STATUS
approved