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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 $m$?

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    $\begingroup$ With the help of the chinese remainder theorem, you can construct an arbitary long chain of consecutive postive integers, none of them squarefree. Just use residue $0$ modulo $4$ , $-1$ modulo $9$ , $-2$ modulo $25$ , $-3$ modulo $49$ and so on. $\endgroup$
    – Peter
    Mar 11 '21 at 16:00
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    $\begingroup$ More interesting is to find the smallest solution for a given length $n$. I think there is nothing better than brute force. $\endgroup$
    – Peter
    Mar 11 '21 at 16:02
  • $\begingroup$ @ Peter "More interesting ..." This is my additional question (3). Brute force indeed, it took my weak PC more than two hours to find $c(10)$. $\endgroup$ Mar 11 '21 at 16:17
  • $\begingroup$ I noticed that. I just pointed out that it is the more interesting part of the question. Are you interested in the smallest solutions for larger chains ? $\endgroup$
    – Peter
    Mar 11 '21 at 19:09
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    $\begingroup$ Dear downvoter, I'd appreciate to know the reason. $\endgroup$ Mar 11 '21 at 20:16
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Letting $p_1,\dots,p_m$ be distinct primes. Then, using Chinese remainder there, find $x$ such that:

$$x\equiv -i \pmod{p_i^2}$$

Then $p_i^2\mid x+i$ for $i=1,\dots,m.$ and there are infinitely many such $x.$

(This assumes you don’t require $\mu(x)\neq 0$ and $\mu(x+m+1)\neq 0.)$

But it does mean we can get arbitrarily long consecutive sequences.

The smallest for a particular $m$ is probably tricky. This is similar to the case where $x+1,x+2,\dots,x+m$ are all non-primes. It is easy to show that such $x$ exists, but it is hard in general to find the smallest $x$ for a given $m.$

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  • $\begingroup$ Circle method with $\sum_{n\ge 1} a_n(k) e^{2i\pi n z}$, $a_k=|\mu(n)| |\mu(n+k+1)|\prod_{m=1}^k (1-|\mu(n+m)|) $ ? Intuitively it would be easier than weak Goldbach and it should give the asymptotic of $\sum_{n\le x} a_n(k)$ $\endgroup$
    – reuns
    Mar 11 '21 at 17:22
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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 $m$.

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  • $\begingroup$ Sounds about right. I suspect there is a way to rephrase this argument in terms of relative density of sequences, so that it feels more elementary. $\endgroup$
    – Erick Wong
    Mar 13 '21 at 19:44
  • $\begingroup$ Amazingly the same approach fails completely to find arbitrary long runs of squarefree numbers. $\endgroup$
    – reuns
    Mar 13 '21 at 19:45
  • $\begingroup$ Lol, but you can at least find arbitrarily long strings of admissible “constellations”, analogous to how a prime triplet is defined as $(p,p+\cdot, p+6)$ rather than $(p,p+2,p+4)$. $\endgroup$
    – Erick Wong
    Mar 13 '21 at 19:53
  • $\begingroup$ I think the circle method solves all the admissible motives, though it is unclear to me where admissible comes into play. $\endgroup$
    – reuns
    Mar 13 '21 at 19:59
  • $\begingroup$ @ reuns +1 Looks like a great proof for chains of exactly lengths $m$, i.e. $\mu\ne 0$ beyond ends of the chain. But, honestly, it is not elementary enough for my current level of knowledge. $\endgroup$ Mar 15 '21 at 13:04

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