TITLE: Is there an unlimited number of sequences of consecutive numbers with $\mu(n)=0$ of any length?
QUESTION [5 upvotes]: EDIT
I have received interesting comments to my post. Especially the comment of @Martin Hopf showed that this problem is a "classic" and by no means new.
Here is Eric Weissten's article on "squareful (nice term!) numbers" https://mathworld.wolfram.com/Squareful.html from which you can find the topic exposed.
This article https://arxiv.org/pdf/1210.3829.pdf is an exhaustive study of the problem.
Original post
Consider the subset of the natural numbers $n\ge 1$ which have at least one square prime factor. These can be formally defined by $\mu (n) = 0$, where $\mu$ is the Möbius function.
The first 40 of these are
$$s_{0}=\{4,8,9,12,16,18,20,24,25,27,28,32,36,40,44,45,48,49,50,52,54,56,60,63,64,68,72,75,76,80,81,84,88,90,92,96,98,99,100,104\}$$
Closer inspection of $s_{0}$ shows that there are sequences of consecutive such numbers. We call them compact sequences. If sorted by the length $m$ of the compact sequence we find in $s_{0}$ the following
$$m=2: \{8,9\}, \{24,25\}, \{27,28\},\{44,45\},\{48,49\},\{49,50\},\{63,64\},\{75,76\},\{80,81\},\{98,99\},\{99,100\}$$
$$m=3:  \{48,49,50\}, \{98,99,100\}$$
We see that for a given length $m$ there is more than one sequence, and that for the given length 40 of $s_{0}$ there are no sequences with length $m\gt3$.
Prolonging the list $s_0$ we find also longer compact sequences, and, again, for a given length there is more than one sequence.
Quoting only the first member of the first appearance of the corresponding sequence
I found numerically in the format $\{m,\text{first term of first appearance}\}$
$$c=\left(
\begin{array}{cc}
 2 & 8 \\
 3 & 48 \\
 4 & 242 \\
 5 & 844 \\
 6 & 22'020 \\
 7 & 217'070 \\
 8 & 1'092'747 \\
 9 & 8'870'024 \\
10 & 221'167'422\\
\end{array}
\right)$$
Remark: this sequence is contained in OEIS (https://oeis.org/A045882). Thanks to @Martin Hopf for pointing this out in a comment.
Conjecture
(1)  there is a compact sequences for any given length and
(2)  there are infinitely many compact sequences for any given length
Unfortunately, I was not able to prove or disprove the conjecture. Can you do better?
Additional question
(3) can you devise a formula for the first appearance of the compact series for given $m$?

REPLY [2 votes]: There is an elementary way to show that for all $m$ there are infinitely many integers such that $$|\mu(n)|=|\mu(n+m+1)|=1,\qquad  \mu(n+1)=\ldots=\mu(n+m)=0$$

Let $N_m=\prod_{j\le m} p_{j+m}^2$. With the CRT take $$a_m\in [1,N_m], \qquad a_m\equiv -j \bmod p_{j+m}^2,\qquad j \in 1\ldots m$$ so that $\mu(a_m+j+lN_m)=0$ for $j\in 1\ldots m$.
Look at the Dirichlet series $$\begin{eqnarray}F_m(s)&\\ =&\sum_{l\ge 0} |\mu(a_m+lN_m)| (a_m+lN_m)^{-s}+|\mu(a_m+m+1+lN_m)| (a_m+m+1+lN_m)^{-s}\end{eqnarray}$$
Note that $a_m,a_m+m+1$ are coprime with $N_m$. By the orthogonality of Dirichlet characters we have $$F_m(s)= \sum_{\chi \bmod N_m}\frac{\overline{\chi(a_m)}+\overline{\chi(a_m+m+1)}}{\varphi(N_m)}\sum_{n\ge 1}\chi(n)|\mu(n)| n^{-s} $$
For $\chi$ a non trivial Dirichlet character, $\sum_{n\ge 1}\chi(n)|\mu(n)| n^{-s}$ is holomorphic at $s=1$.
Whence the asymptotic as $s\to 1$ is determined by the trivial character $1_{\gcd(n,N_m)=1}$:
$$\begin{eqnarray}F_m(s)&\sim& \frac{2}{\varphi(N_m)}\sum_{n\ge 1,\gcd(n,N_m)=1} |\mu(n)| n^{-s}\\&=&\frac{2}{\varphi(N_m)} \frac1{\prod_{p\ |\ N_m} (1+p^{-s})} \frac{\zeta(s)}{\zeta(2s)}\\&\sim& \frac{2}{N_m} \frac1{\prod_{p\ | \ N_m} (1-p^2)} \frac{1}{\zeta(2)}\zeta(s)\end{eqnarray}$$
Since $\frac1{\prod_{p\ | \ N_m} (1-p^2)}\frac{1}{\zeta(2)}>1/2$ this implies that for infinitely many $l$ $$|\mu(a_m+lN_m)|=|\mu(a_m+m+1+lN_m)|=1$$
so that $a_m+lN_m,\ldots, a_m+m+1+lN_m$ is our chain of length $m$.