Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$? Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?
 A: This is A222548 in the online encyclopedia of integer sequences.  They don't provide a closed form, but they do give the following:
$$a(n)\approx\frac{\pi^2}{6}n^2+O(n\log n)$$
A: Let me complement @Marko Riedel's excellent answer by providing a low-tech approach to the leading term.
Let us denote by $\mathbf{1}\{\cdot\}$ the indicator function of the set $\{\cdot\}$. Then using the identity
$$\lfloor n/i \rfloor = \sum_{j=1}^{n} \mathbf{1}\{ij \leq n \}$$
we can rearrange the sum to find:
\begin{align*}
a(n)
&= \sum_{i=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \mathbf{1}\{ij \leq n\}\mathbf{1}\{ik \leq n\}
= \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \sum_{i=1}^{n} \mathbf{1}\Big\{i \leq \frac{n}{\max\{j,k\}}\Big\} \\
&= \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{\max\{j,k\}} \right\rfloor
= \sum_{l=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{l} \right\rfloor \mathbf{1}\{\max\{j,k\} = l\} \\
&= \sum_{l=1}^{n} (2l-1) \left\lfloor \frac{n}{l} \right\rfloor.
\end{align*}
Dividing both sides by $n^2$, the RHS can be identified as Riemann sum and thus
$$ \frac{a(n)}{n^2}
= \sum_{l=1}^{n} \frac{2l-1}{n} \left\lfloor \frac{n}{l} \right\rfloor \frac{1}{n}
\xrightarrow[\ n\to\infty \ ]{}
\int_{0}^{1} 2x \left\lfloor\frac{1}{x}\right\rfloor \, dx
= \zeta(2). $$
Here, the last integral can be easily computed by applying the substitution $x \mapsto 1/x$.
A: To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that
$$a(n+1)-a(n) = 1 + \sum_{k=1}^n 
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\=
1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right)
= \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\=
2\sigma(n+1)-\tau(n+1).$$
It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) =
\sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$
We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series
$$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} =
2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$
We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$
so that by the aforementioned theorem,
$$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$
In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting
$$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +
\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$
which yields
$$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right)
+\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$
This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$
A: A quick proof
We have $\lfloor x\rfloor =x+O(1)$ thus $\left\lfloor\frac nk\right\rfloor = \frac{n}{k} + O(1)$ and $\left\lfloor\frac nk\right\rfloor^2= ( \frac{n}{k} + O(1))^2= (\frac{n}{k})^2 + \frac 1k O(n)$. Finally $\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2=n^2\frac{\pi^2}6+O(n\log(n))}$
