Is there an unlimited number of sequences of consecutive numbers with $\mu(n)=0$ of any length?

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.

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?

(3) can you devise a formula for the first appearance of the compact series for given $$m$$?

• 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. Mar 11 '21 at 16:00
• More interesting is to find the smallest solution for a given length $n$. I think there is nothing better than brute force. Mar 11 '21 at 16:02
• @ Peter "More interesting ..." This is my additional question (3). Brute force indeed, it took my weak PC more than two hours to find $c(10)$. Mar 11 '21 at 16:17
• 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 ? Mar 11 '21 at 19:09
• Dear downvoter, I'd appreciate to know the reason. Mar 11 '21 at 20:16

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 $$m.$$

• 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 $\sum_{n\le x} a_n(k)$ Mar 11 '21 at 17:22

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 $$m$$.

• 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. Mar 13 '21 at 19:44
• Amazingly the same approach fails completely to find arbitrary long runs of squarefree numbers. Mar 13 '21 at 19:45
• 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 $(p,p+2,p+4)$. Mar 13 '21 at 19:53
• I think the circle method solves all the admissible motives, though it is unclear to me where admissible comes into play. Mar 13 '21 at 19:59
• @ 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. Mar 15 '21 at 13:04