A153818 a(n) = Sum_{k=1..n} floor(n^2/k^2).

1, 5, 12, 22, 35, 53, 72, 96, 123, 153, 184, 222, 260, 304, 351, 402, 453, 510, 568, 633, 697, 765, 839, 916, 994, 1077, 1164, 1252, 1342, 1443, 1535, 1641, 1747, 1856, 1969, 2083, 2200, 2321, 2447, 2579, 2705, 2844, 2979, 3123, 3269, 3417, 3570, 3726, 3881

Offset: 1

Author

Ctibor O. Zizka, Jan 02 2009

Keywords

• easy
• nonn

Comments

How can Sum_{k=1..n} floor(n^2/k^2) be expressed as a function of Sum_{k=1..n} floor(n/k)? [Ctibor O. Zizka, Feb 14 2009]

Examples

			a(4)=22 because floor(16/1) + floor(16/4) + floor(16/9) + floor(16,16) = 16 + 4 + 1 + 1 = 22. [_Emeric Deutsch_, Jan 13 2009]
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Benoit Cloitre, Plot of (a(n)-zeta(2)*n^2-zeta(1/2)*n)/(n^0.5/log(n))

Crossrefs

Cf. A006218, A344675.

Programs

Maple

a := proc (n) options operator, arrow: sum(floor(n^2/k^2), k = 1 .. n) end proc: seq(a(n), n = 1 .. 50); # Emeric Deutsch, Jan 13 2009

PARI

a(n)=sum(k=1,n,n^2\k^2) \\ Benoit Cloitre, Jan 22 2013

Formula

From Benoit Cloitre, Jan 22 2013: (Start)

Asymptotic formula: a(n) = zeta(2)*n^2 + zeta(1/2)*n + O(n^(1/2)).

Conjecture: a(n) = zeta(2)*n^2 + zeta(1/2)*n + O(n^0.5/log(n)) (see link). (End)

Extensions

Definition edited by Emeric Deutsch, Jan 13 2009

Extended by Emeric Deutsch, Jan 13 2009