Möbius function of consecutive numbers This question arose from a problem in Niven & Zuckerman's book "Introduction to the Theory of Numbers". In the chapter that the authors introduce the Möbius function, the first exercise is the following: 
Find a positive integer $n$ such that $\mu(n)+\mu(n+1)+\mu(n+2)=3$, i.e, $\mu(n)=\mu(n+1)=\mu(n+2)=1$. A brute-force approach reveals the solution $n=33$.
My question is simply: Are there infinitely many $n$ in the previous conditions?
I really don't know how to approach this problem. I tried various things (factorials, Chinese remainder theorem, etc.) and i didn't come up with nothing. Also, for the first thousand numbers, there are the solutions $n=33,85,93,141,201,213,217,301,393,445,633,697,869,921$. One can also think in the variation of the problem with $\mu(n)=\mu(n+1)=\mu(n+2)=-1$.
Thanks in advance!
 A: This is not an answer, I do some prediction from the probability. Interestingly, the predicted results and the actual results are very similar.
Since $\mu(n)=0$ if $n$ is not squarefree, we should exclude this possibility, 
$$n\not\equiv 0,-1,-2 \pmod {p^2}, \forall \in \mathbb P.$$
$$\prod_{p\in\mathbb P} (1-\frac{3}{p^2})\approx 0.125802 \approx \frac{1}8.$$
Hence we have about $\dfrac{1}8$ possibilities for $n$ such that $\mu(n),\mu(n+1),\mu(n+2)$ are not equal to $0$.
We assume that if $\mu(m)\neq 0$ then the probability of a coin thrown sides are equal, the probability of $\mu(n)=\mu(n+1)=\mu(n+2)=1$ is equal to $$0.125802 \times \frac{1}2\times \frac{1}2 \times \frac{1}2 \approx \frac{1}{64}.$$
Hence there are about $\dfrac{N}{64}$ such integers $n\leq N.$ A063838 says $a_{1000}=67105$, this is very close to our result.
A: Again, this isn't really an answer, but perhaps is an interesting interpretation of the problem that warrants some further discussion (Its utility is not known to me): Since the sum of the primitive $n$th roots of unity is $\mu(n)$, this is equivalent to finding $n$ such that the coefficients of the $x^{\varphi(n)-1}, x^{\varphi(n+1)-1}$, and $x^{\varphi(n+2)-1}$ terms of the cyclotomic polynomials $\Phi_n(x), \Phi_{n+1}(x)$, and $\Phi_{n+2}(x)$ respectively are all equal to $1$.
