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%I #15 Dec 23 2015 22:40:10
%S 0,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,1,1,1,1,2,2,1,1,1,1,0,0,1,1,1,1,1,
%T 2,1,2,2,2,2,2,2,3,2,2,2,2,2,1,1,1,0,1,0,1,0,0,1,1,1,1,1,1,0,0,1,1,1,
%U 1,1,2,1,2,2,1,1,1,2,2,1,1,1,2,1,1,2,2,2,2,1,2,1,1,2,2,2,2,1,1,0,1,1,1,1,1
%N (Number of squarefree numbers <= n) minus round(n/zeta(2)).
%C Race between the number of squarefree numbers and round(n/zeta(2)).
%C First term < 0: a(172) = -1.
%D G. H. Hardy and S. Ramanujan, The normal number of prime factors of a number n, Q. J. Math., 48 (1917), pp. 76-92.
%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth edition (1979), Clarendon Press, pp. 269-270.
%H Daniel Forgues, <a href="/A158819/b158819.txt">Table of n, a(n) for n=1..100000</a>
%H A. Granville, <a href="http://www.dms.umontreal.ca/~andrew/Postscript/polysq3.ps">ABC means we can count squarefree</a>
%F Since zeta(2) = Sum_{i>=1} 1/(i^2) = (Pi^2)/6, we get:
%F a(n) = A013928(n+1) - n/Sum_{i>=1} 1/(i^2) = O(sqrt(n));
%F a(n) = A013928(n+1) - 6*n/(Pi^2) = O(sqrt(n)).
%Y Cf. A008966 1 if n is squarefree, else 0.
%Y Cf. A013928 Number of squarefree numbers < n.
%Y Cf. A100112 If n is the k-th squarefree number then k else 0.
%Y Cf. A057627 Number of nonsquarefree numbers not exceeding n.
%Y Cf. A005117 Squarefree numbers.
%Y Cf. A013929 Nonsquarefree numbers.
%K sign
%O 1,7
%A _Daniel Forgues_, Mar 27 2009
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