A013936 a(n) = Sum_{k=1..n} floor(n/k^2).

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Offset: 1

Author

N. J. A. Sloane, Henri Lifchitz

Keywords

• nonn

References

• T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 24.

Links

• Franklin T. Adams-Watters, Table of n, a(n) for n = 1..10000
• Benoit Cloitre, On the divisor and circle problems
• Benoit Cloitre, Plot of a(n)-zeta(2)*n-zeta(1/2)*n^(1/2)

Crossrefs

Cf. A046951.

Programs

Maple

f := n->sum(floor(n/k^2),k=1..n); [ seq(f(j),j=1..100 ];

Mathematica

Table[Sum[Floor[n/k^2],{k,n}],{n,80}] (* Harvey P. Dale, Mar 28 2013 *)

PARI

a(n)=sum(k=1,sqrtint(n),floor(n/k^2))

Formula

a(n) = a(n-1)+A046951(n). Bounded above by n*Pi^2/6: the growth of the differences seems to be roughly proportional to sqrt(n). - Henry Bottomley, Aug 16 2001

Conjecture : limit n ->infinity (Pi^2/6*n-a(n))/sqrt(n) = c = 1.45... - Benoit Cloitre, Jan 10 2003

If lim_{n->infinity} (Pi^2/6*n - a(n)) / sqrt(n) does exist, it converges very slowly. It does appear to be bounded. - Franklin T. Adams-Watters, Nov 17 2006

In fact we have: a(n) = zeta(2)*n+zeta(1/2)*n^(1/2)+O(n^theta) with theta<1/2 and we conjecture that theta=1/4+epsilon is the best possible choice. Also a(n)=sum_{1<=k<=n}floor(sqrt(n/k)). - Benoit Cloitre, Nov 05 2012

G.f.: (1/(1 - x))*Sum_{k>=1} x^(k^2)/(1 - x^(k^2)). - Ilya Gutkovskiy, Feb 11 2017