A070548 a(n) = Cardinality{ k in range 1 <= k <= n such that Moebius(k) = 1 }.

1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 11, 11, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23

Offset: 1

Author

Benoit Cloitre, May 02 2002

Keywords

• easy
• nonn

Comments

Moebius(k)=1 iff k is the product of an even number of distinct primes (cf. A008683). See A057627 for Moebius(k)=0.

There was an old comment here that said a(n) was equal to A072613(n) + 1, but this is false (e.g., at n=210). - N. J. A. Sloane, Sep 10 2008

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..10000
• Ed Pegg Jr., The Möbius Function (and squarefree numbers), Math Games, November 3, 2003.
• Eric Weisstein's World of Mathematics, Mertens Conjecture.

Crossrefs

Cf. A008683, A002321, A013928, A059956, A070549.

Programs

Maple

with(numtheory); M:=10000; c:=0; for n from 1 to M do if mobius(n) = 1 then c:=c+1; fi; lprint(n,c); od; # N. J. A. Sloane, Sep 14 2008

Mathematica

a[n_] := If[MoebiusMu[n] == 1, 1, 0]; Accumulate@ Array[a, 100] (* Amiram Eldar, Oct 01 2023 *)

PARI

for(n=1,150,print1(sum(i=1,n,if(moebius(i)-1,0,1)),","))

Formula

Asymptotics: Let N(i) = number of k in the range [1,n] with mu(k) = i, for i = 0, 1, -1. Then we know N(1) + N(-1) ~ 6n/Pi^2 (see A059956). Also, assuming the Riemann hypothesis, | N(1) - N(-1) | < n^(1/2 + epsilon) (see the Mathworld Mertens Conjecture link). Hence a(n) = N(1) ~ 3n/Pi^2 + smaller order terms. - Stefan Steinerberger, Sep 10 2008

a(n) = (1/2)*Sum_{i=1..n} (mu(i)^2 + mu(i)) = (1/2)*(A013928(n+1) + A002321(n)). - Ridouane Oudra, Oct 19 2019

From Amiram Eldar, Oct 01 2023: (Start)

a(n) = A013928(n+1) - A070549(n).

a(n) = A070549(n) + A002321(n). (End)