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A158819 a(n) = (number of squarefree numbers <= n) minus round(n/zeta(2)).

%I A158819 #22 Jan 20 2025 03:51:57
%S A158819 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 A158819 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 A158819 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 A158819 a(n) = (number of squarefree numbers <= n) minus round(n/zeta(2)).
%C A158819 Race between the number of squarefree numbers and round(n/zeta(2)).
%C A158819 First term < 0: a(172) = -1.
%D A158819 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 A158819 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth edition, Clarendon Press, 1979, pp. 269-270.
%H A158819 Daniel Forgues, <a href="/A158819/b158819.txt">Table of n, a(n) for n=1..100000</a>
%H A158819 Andrew Granville, <a href="https://doi.org/10.1155/S1073792898000592">ABC allows us to count squarefrees</a>, International Mathematics Research Notices, Vol. 1998, No. 19 (1998), pp. 991-1109; <a href="https://dms.umontreal.ca/~andrew/PDF/polysq3.pdf">alternative link</a>.
%F A158819 Since zeta(2) = Sum_{i>=1} 1/(i^2) = (Pi^2)/6, we get:
%F A158819 a(n) = A013928(n+1) - n/Sum_{i>=1} 1/(i^2) = O(sqrt(n));
%F A158819 a(n) = A013928(n+1) - 6*n/(Pi^2) = O(sqrt(n)).
%t A158819 seq[lim_] := Accumulate[Boole[SquareFreeQ /@ Range[lim]]] - Round[Range[lim]/Zeta[2]]; seq[105] (* _Amiram Eldar_, Jan 20 2025 *)
%Y A158819 Cf. A008966 (1 if n is squarefree, else 0).
%Y A158819 Cf. A013928 (number of squarefree numbers < n).
%Y A158819 Cf. A100112 (if n is the k-th squarefree number then k else 0).
%Y A158819 Cf. A057627 (number of nonsquarefree numbers not exceeding n).
%Y A158819 Cf. A005117 (squarefree numbers).
%Y A158819 Cf. A013929 (nonsquarefree numbers).
%Y A158819 Cf. A013661 (zeta(2)).
%K A158819 sign
%O A158819 1,7
%A A158819 _Daniel Forgues_, Mar 27 2009