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%I #38 Aug 19 2021 11:00:07
%S 1,2,3,5,6,7,8,10,12,13,14,16,17,18,19,22,23,25,26,28,29,30,31,33,35,
%T 36,38,40,41,42,43,46,47,48,49,53,54,55,56,58,59,60,61,63,65,66,67,70,
%U 72,74,75,77,78,80,81,83,84,85,86,88,89,90,92,96,97,98,99,101,102,103
%N a(n) = Sum_{k=1..n} floor(n/k^2).
%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 24.
%H Franklin T. Adams-Watters, <a href="/A013936/b013936.txt">Table of n, a(n) for n = 1..10000</a>
%H Benoit Cloitre, <a href="/A013936/a013936.pdf">On the divisor and circle problems</a>
%H Benoit Cloitre, <a href="/A013936/a013936.png">Plot of a(n)-zeta(2)*n-zeta(1/2)*n^(1/2)</a>
%F 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
%F Conjecture : limit n ->infinity (Pi^2/6*n-a(n))/sqrt(n) = c = 1.45... - _Benoit Cloitre_, Jan 10 2003
%F 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
%F 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
%F G.f.: (1/(1 - x))*Sum_{k>=1} x^(k^2)/(1 - x^(k^2)). - _Ilya Gutkovskiy_, Feb 11 2017
%p f := n->sum(floor(n/k^2),k=1..n); [ seq(f(j),j=1..100 ];
%t Table[Sum[Floor[n/k^2],{k,n}],{n,80}] (* _Harvey P. Dale_, Mar 28 2013 *)
%o (PARI) a(n)=sum(k=1,sqrtint(n),floor(n/k^2))
%Y Cf. A046951.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, _Henri Lifchitz_
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