A025224 -id:A025224 - cp's OEIS frontend

A300287 a(n) = floor((1/n) * Sum_{k=1..n} sqrt(k)).

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, 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, 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

Offset: 1

Michel Marcus, Mar 02 2018

nonn

Mircea Merca, On the Arithmetic Mean of the Square Roots of the First n Positive Integers, The College Mathematics Journal, Vol. 48, No. 2 (March 2017), pp. 129-133.
S. Ramanujan, On the sum of the square roots of the first n natural numbers, Journal of the Indian Mathematical Society, VIII (1915), pp. 173-175.
Thomas P. Wihler, Rounding the arithmetic mean value of the square roots of the first n integers, arXiv:1803.00362 [math.NT], 2018.

Cf. A025224.

[Floor(&+[Sqrt(k)/n: k in [1..n]]): n in [1..100]]; // Bruno Berselli, Mar 02 2018

Table[Floor[1/n Sum[Sqrt[k], {k, n}]], {n, 200}] (* Vincenzo Librandi, Mar 02 2018 *)

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

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

A336112 a(n) is the least number k such that the Sum_{i=0..k} sqrt(k) equals or exceeds n.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23

Offset: 0

Robert G. Wilson v, Jul 08 2020

nonn

Inspired by A045880.
Let c = (9/4)^(1/3) = (3/2)^(2/3) ~ 1.310370697..., then a(n) ~ c*n^(2/3).
a(10^k) for k>= 0: 1, 6, 28, 131, 608, 2823, 13104, 60822, 282311, 1310371, 6082202, 28231081, 131037070, 608220200, ..., .

			a(0) = 0 since the sqrt(0) = 0;
a(1) = 1 since the sqrt(0) + sqrt(1) = 1;
a(2) = 2 since the sqrt(0) + sqrt(1) + sqrt(2) ~ 2.41421... which exceeds 2;
a(3) = 3 since the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which easily exceeds 3;
a(4) = 3 because the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which barely exceeds 4; etc.

Cf. A025224, A045880, A073333.

f[n_] := Block[{k = s = 0}, While[s < n, k++; s = s + Sqrt@k]; k]; Array[f, 75, 0]

a(n) = my(s=0, k=0); while ((s+=sqrt(k)) < n, k++); k; \\ Michel Marcus, Jul 09 2020

a(k*n) ~ k^(2/3)*a(n).

cp's OEIS Frontend

A300287 a(n) = floor((1/n) * Sum_{k=1..n} sqrt(k)).

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