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A070548
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a(n) = Cardinality{ k in range 1 <= k <= n such that Moebius(k) = 1 }.
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5
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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
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OFFSET
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1,6
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COMMENTS
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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
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LINKS
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FORMULA
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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
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MAPLE
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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
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MATHEMATICA
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a[n_] := If[MoebiusMu[n] == 1, 1, 0]; Accumulate@ Array[a, 100] (* Amiram Eldar, Oct 01 2023 *)
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PROG
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(PARI) for(n=1, 150, print1(sum(i=1, n, if(moebius(i)-1, 0, 1)), ", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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