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

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, 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, 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

Offset

1,7

Comments

Race between the number of squarefree numbers and round(n/zeta(2)).

First term < 0: a(172) = -1.

References

G. H. Hardy and S. Ramanujan, The normal number of prime factors of a number n, Q. J. Math., 48 (1917), pp. 76-92.

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth edition, Clarendon Press, 1979, pp. 269-270.

Links

Daniel Forgues, Table of n, a(n) for n=1..100000

Andrew Granville, ABC allows us to count squarefrees, International Mathematics Research Notices, Vol. 1998, No. 19 (1998), pp. 991-1109; alternative link.

Formula

Since zeta(2) = Sum_{i>=1} 1/(i^2) = (Pi^2)/6, we get:

a(n) = A013928(n+1) - n/Sum_{i>=1} 1/(i^2) = O(sqrt(n));

a(n) = A013928(n+1) - 6*n/(Pi^2) = O(sqrt(n)).

Mathematica

seq[lim_] := Accumulate[Boole[SquareFreeQ /@ Range[lim]]] - Round[Range[lim]/Zeta[2]]; seq[105] (* Amiram Eldar, Jan 20 2025 *)

Crossrefs

Cf. A008966 (1 if n is squarefree, else 0).

Cf. A013928 (number of squarefree numbers < n).

Cf. A100112 (if n is the k-th squarefree number then k else 0).

Cf. A057627 (number of nonsquarefree numbers not exceeding n).

Cf. A005117 (squarefree numbers).

Cf. A013929 (nonsquarefree numbers).

Cf. A013661 (zeta(2)).

Sequence in context: A357900 A357732 A356428 * A031279 A124778 A037831

Adjacent sequences: A158816 A158817 A158818 * A158820 A158821 A158822

Keyword

sign

Author

Daniel Forgues, Mar 27 2009

Status

approved