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A070548 a(n) = Cardinality{ k in range 1 <= k <= n such that Moebius(k) = 1 }. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,6

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 Mobius function (and squarefree numbers)

Eric Weisstein's World of Math., Merten's Conjecture

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 Merten's 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

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

PROG

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

CROSSREFS

Cf. A008683, A013928, A002321.

Sequence in context: A272187 A194621 A088004 * A209628 A132011 A054893

Adjacent sequences:  A070545 A070546 A070547 * A070549 A070550 A070551

KEYWORD

easy,nonn

AUTHOR

Benoit Cloitre, May 02 2002

STATUS

approved

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Last modified September 18 22:56 EDT 2020. Contains 337174 sequences. (Running on oeis4.)