|
| |
|
|
A070548
|
|
a(n) = Card{ k in range 1<=k<=n such that Moebius(k)=1 }.
|
|
3
|
|
|
|
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
|
|
|
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.
Sequence in context: A029269 A194621 A088004 * A209628 A132011 A054893
Adjacent sequences: A070545 A070546 A070547 * A070549 A070550 A070551
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
Benoit Cloitre, May 02 2002
|
|
|
STATUS
|
approved
|
| |
|
|
