runs of consecutive non squarefree integers This question gained no attention at Math SE.
Call a sequence of $k$ consecutive naturals squary if each one of them is divided by a square > 1. The Chinese Remainder theorem trivially guarantees us squary $k$-sequences for each $k$ (even with pre-defined divisors in a given order).  
Denote by $f(k,n)$ the number of squary sequences $(x+1,x+2,...,x+k)$  contained in $[1,n]$. For $k$ fixed and $n\to\infty$, what is known about the asymptotics of $f(k,n)$?
It is easy to show that $f(1,n)\sim (1-\frac6{\pi^2})n $.
 A: The original question is answered by Lucia. I was thinking on, what if $k$ depends on $n$? I'm interested on question whether $k = \frac{\log n}{\log\log n}$.
In related question (Consecutive non squarefree integers) we have seen that $f\left(\frac{\log n}{\log\log n},n\right) > 0$ infinitely often. 
Now I prove that, for any $\varepsilon$
$$f\left(\frac{\log n}{\log\log n}, n\right) \geq n^{\frac{1}{2}- \varepsilon}$$
infinitely often. 
For some $k$  and for some $\pi$ permutation of $\{k+1, k+2, \ldots ,p_k^2\}$ let the following congruence system:
$$
\begin{array}{ccll}  
 x_{\pi}& \equiv & k& \mod{p_k^2}  \\
  x_{\pi}& \equiv & k+1& \mod{p_{\pi(k+1)}^2}  \\
 & \colon & & \\
 x_{\pi}& \equiv & p_k^2& \mod{p_{\pi(p_k^2)}^2}  
  \end{array} 
$$
The system for any $\pi$ has a solution $x_{\pi}$ smaller than $\Pi_{i=k}^{p_k^2} p_i^2$, and  $\{x_{\pi}-k, \ldots, x_{\pi}-p_k^2\}$ can not contain square-free numbers.
If $\pi_1 \neq \pi_2$, then the difference between $x_{\pi_1}$ and $x_{\pi_2}$ is at least $p_k^2-k$, because in $\{x_{\pi}-k,  \ldots, x_{\pi}-p_k^2\}$ exactly one integer is divisible by $p_i^2$ for $k\leq i \leq p_k^2$, and only $x_{\pi}-k$ is divisible by $p_k^2$, so the intervals $[x_{\pi_1}-p_k^2, x_{\pi_1}-k]$ and $[x_{\pi_2}-p_k^2, x_{\pi_2}-k]$ are not overlapping. From that, there are at least $(p_k^2-k-1)!$ disjoint  intervals of size $p_k^2-k$.
So let $m = p_k^2 \sim k^2\log^2k$,  there are $(m-o(m^{1/2}))!$ disjoint intervals of size $m-o(m^{1/2})$ which doesn't contain square-free integers  until $\Pi_{i=1}^{m} p_i^2 \leq e^{2(1+\varepsilon)m\log m}$. By letting $m = \frac{\log n}{\log\log n}$, the Stirling formula gives the solution.
Remarks: With more carefully written the congruence system, the same holds for $\frac{\pi^2}{6}\frac{\log n}{\log\log n}$. I think this also motivate that, $\frac{\pi^2}{6}\frac{\log n}{\log\log n}$ is not the best lower bound, but the fact $f\left((1+\varepsilon)\frac{\pi^2}{6}\frac{\log n}{\log\log n},n\right) > 0$ happens infinitely often, is not known.
A: At the request of Wolfgang, I'll provide a few more details of my comment.  Let $K=\{k_1,\ldots, k_{\ell}\}$ be any set of $\ell$ distinct natural numbers, and put $\nu(K,p)$ to be the number of distinct residue classes among the elements of $K$ taken $\pmod{p^2}$.  Then Mirsky's result says that 
$$ 
\sum_{n\le x} \prod_{j=1}^{\ell}\mu(n+k_j)^2 \sim C_0(K) x
$$ 
where 
$$ 
C_0(K) = \prod_{p} \Big(1 -\frac{\nu(K,p)}{p^2}\Big). 
$$ 
This is analogous to the Hardy-Littlewood conjecture for prime $k$-tuples, but much easier to prove.  Mirsky worked out good error terms in the asymptotic formula above; many others have worked on precise asymptotic formulae in such problems, for example Heath-Brown looked at the special case when $K=\{1,2\}$ carefully.  
The problem in question asks for 
$$ 
\sum_{n\le x} \prod_{j=1}^{k} (1-\mu(n+j)^2) 
$$ 
and expanding out the product we get that this is 
$$ 
\sum_{n\le x} \sum_{K \subset \{1,\ldots, k\}} (-1)^{|K|} \prod_{j\in K} \mu(n+j)^2, 
$$ 
which by Mirsky's result is 
$$ 
\sim x \sum_{K\subset \{1,\ldots, k\}} (-1)^{|K|} C_0(K) \sim C(k) x, 
$$ 
for a suitable constant $C(k)$.  
As noted in my comment it is now easy to see that 
$$ 
C(2) = 1- 2 \frac{6}{\pi^2} + C_0(\{1,2\}) = 1-\frac{12}{\pi^2}+ \prod_p \Big(1-\frac{2}{p^2}\Big).
$$ 
Further 
$$ 
C(3) =1-\frac{18}{\pi^2} + 3 \prod_{p} \Big(1-\frac{2}{p^2} \Big) - \prod_p \Big(1-\frac{3}{p^2}\Big);
$$ 
$$
C(4)=1-\frac{24}{\pi^2}+6\prod_p \Big(1-\frac{2}{p^2}\Big) -4\prod_p\Big(1-\frac{3}{p^2}\Big).
$$
From $C(5)$ on, things will be more messy.
Added: In my comment to the question I noted that $C(k)$ seems to decay like $k^{-(c+o(1))k}$ for large $k$ and this is in keeping with the $\log n/\log \log n$ behavior in the related question on long gaps between square-free numbers.  In this context, I recently came across a paper by Geoffrey Grimmett which proves that $C(k)=k^{-(6/\pi^2+o(1))k}$ as $k \to \infty$.  The paper appeared in Math. Proc. Camb. Phil. Soc. in 1997, and the link is to a preprint from the author's homepage.
