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a(n) = floor((1/n) * Sum_{k=1..n} sqrt(k)).
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%I #13 Sep 08 2022 08:46:20

%S 1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%T 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,

%U 5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6

%N a(n) = floor((1/n) * Sum_{k=1..n} sqrt(k)).

%H Mircea Merca, <a href="http://www.jstor.org/stable/10.4169/college.math.j.48.2.129">On the Arithmetic Mean of the Square Roots of the First n Positive Integers</a>, The College Mathematics Journal, Vol. 48, No. 2 (March 2017), pp. 129-133.

%H S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf">On the sum of the square roots of the first n natural numbers</a>, Journal of the Indian Mathematical Society, VIII (1915), pp. 173-175.

%H Thomas P. Wihler, <a href="https://arxiv.org/abs/1803.00362">Rounding the arithmetic mean value of the square roots of the first n integers</a>, arXiv:1803.00362 [math.NT], 2018.

%F a(n) = floor((2/3)*sqrt(n+1)*(1+1/(4*n))). See Theorem 1 of Wihler paper.

%t Table[Floor[1/n Sum[Sqrt[k], {k, n}]], {n, 200}] (* _Vincenzo Librandi_, Mar 02 2018 *)

%o (PARI) a(n) = floor(sum(k=1, n, sqrt(k))/n);

%o (Magma) [Floor(&+[Sqrt(k)/n: k in [1..n]]): n in [1..100]]; // _Bruno Berselli_, Mar 02 2018

%Y Cf. A025224.

%K nonn

%O 1,8

%A _Michel Marcus_, Mar 02 2018