A228483 - OEIS

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A228483

a(n) = 2 - mu(n), where mu(n) is the Moebius function (A008683).

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

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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

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

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