

A013936


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


11



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, 36, 38, 40, 41, 42, 43, 46, 47, 48, 49, 53, 54, 55, 56, 58, 59, 60, 61, 63, 65, 66, 67, 70, 72, 74, 75, 77, 78, 80, 81, 83, 84, 85, 86, 88, 89, 90, 92, 96, 97, 98, 99, 101, 102, 103
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OFFSET

1,2


REFERENCES

T. M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, 1976, page 73, problem 24.


LINKS

Franklin T. AdamsWatters, 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)*nzeta(1/2)*n^(1/2)


FORMULA

a(n) = a(n1)+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*na(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. AdamsWatters, 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


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 *)


PROG

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


CROSSREFS

Sequence in context: A191914 A329898 A145353 * A229992 A076437 A028790
Adjacent sequences: A013933 A013934 A013935 * A013937 A013938 A013939


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Henri Lifchitz


STATUS

approved



