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a(n) = Sum_{k=1..n} floor(n^2/k^2).
9

%I #19 May 26 2021 12:01:15

%S 1,5,12,22,35,53,72,96,123,153,184,222,260,304,351,402,453,510,568,

%T 633,697,765,839,916,994,1077,1164,1252,1342,1443,1535,1641,1747,1856,

%U 1969,2083,2200,2321,2447,2579,2705,2844,2979,3123,3269,3417,3570,3726,3881

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

%C 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]

%H Seiichi Manyama, <a href="/A153818/b153818.txt">Table of n, a(n) for n = 1..10000</a>

%H Benoit Cloitre, <a href="/A153818/a153818.png">Plot of (a(n)-zeta(2)*n^2-zeta(1/2)*n)/(n^0.5/log(n))</a>

%F From _Benoit Cloitre_, Jan 22 2013: (Start)

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

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

%e 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]

%p 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

%o (PARI) a(n)=sum(k=1,n,n^2\k^2) \\ _Benoit Cloitre_, Jan 22 2013

%Y Cf. A006218, A344675.

%K easy,nonn

%O 1,2

%A _Ctibor O. Zizka_, Jan 02 2009

%E Definition edited by _Emeric Deutsch_, Jan 13 2009

%E Extended by _Emeric Deutsch_, Jan 13 2009