This question gained no attention at [Math SE](https://math.stackexchange.com/questions/562229/sequences-with-square-divisors-everywhere).


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](http://en.wikipedia.org/wiki/Square-free_integer#Distribution) that $f(1,n)\sim (1-\frac6{\pi^2})n $.



runs of consecutive non squarefree integers
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](http://www.statslab.cam.ac.uk/~grg/papers/USsieve.pdf) 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.


