The squarefree numbers form a set which is well-known to have density $6/\pi^2$ in the integers, in other words their characteristic function $\mu^2$ satisfy $\sum_{n\leq x}\mu^2(n)=6/\pi^2x+o(x)$. In this set all your dream theorems about the count of linear patterns are true and easy.

Hardy-Littlewood squarefree tuples

The Hardy-Littlewood conjecture consists, in its strongest form, in an asymptotic, for every $\mathcal{H}=\{h_1,\ldots,h_k\}$, of the form

$\displaystyle{\lvert\{n\leq x\mid n+h_1,\ldots,n+h_k\text{ are all prime}\}\rvert=(c(\mathcal{H})+o(1))\frac{x}{\log^kx}}$

where $c(\mathcal{H})$ is a constant possibly equal to 0. It is quite far from being proven, although interesting steps towards it have been made in the last ten years, by Goldston-Pintz-Yildirim and Maynard.

In the squarefree numbers, the analogous problem is rather easy and was settled by Mirsky in 1947. He proved that

$\displaystyle{\lvert\{n\leq x\mid n+h_1,\ldots,n+h_k\text{ are all squarefree}\}\rvert=c(\mathcal{H})x+O(x^{2/3+\epsilon})}$

where

$\displaystyle{c(\mathcal{H})=\prod_p(1-\frac{\nu_{\mathcal{H}}(p^2)}{p^2})}$

and $\nu_{\mathcal{H}}(p^2)$ is the cardinality of the size of the projection of the set $\mathcal{H}$ in $\mathbb{Z}/p^2\mathbb{Z}$, that is the number of forbidden classes modulo $p^2$ for $n$.

I show a quick way of deriving it with a much worse error term $O(x/\log x)$. Fix $w$, a large number; ultimately, $w=C\log x$ should be ok. Fix $W=\prod_{p\leq w}p^2$. Then the number $X_w$ of $n\leq x$ such that none of $n+h_1,\ldots,n+h_k$ have a square factor $p^2$ with $p\leq w$ is easy to compute: in $[W]$ or any interval of length $W$ there are by chinese remainder theorem this many of them
$\displaystyle{ \prod_{p\leq w}(p^2-\nu_{\mathcal{H}}). }$
Thus in $[x]$, there are this many of them

$\displaystyle{x/W^2\prod_{p\leq w}(p^2-\nu_{\mathcal{H}})+O(W^2)}.$

There are also people who are not squarefree, but escape our sieving by primes less than $w$. These are the guys such that at least one of the $n+h_i$ have a divisor $p^2$ with $p>w$ prime. Now there are are at most $x/p^2$ multiples of $p^2$ in $[x]$. So at most
$\displaystyle{ \sum_{w
should be removed. Let’s try to balance the error terms $O(x/w)$ and $O(W^2)=\exp(2w(1+o(1))$; take $w=\frac{1}{2}(\log x-\log\log x)$.
This way both seem to be $O(x/\log x)$.

Green-Tao type theorem

We can also easily get asymptotics of the form

$\displaystyle{\sum_{n\in\mathbb{Z}^d\cap K}\prod_{i\in[t]}\mu^2(\psi_i(t))=\text{Vol}(K)(\prod_p\beta_p+o(1))}$

where $K\subset [0,N]^d$ is a convex body and $\Psi$ a system of affine-linear forms of which no two are affinely related. Moreover $\beta_p$ is again the proportion of non-forbidden vectors $a=(a_1,\ldots,a_d)$ of residues modulo $p^2$. One way to do it would be to simply use the Hardy-Littlewood type asymptotics proved above coordinate by coordinate. We can also prove it quite easily, by first observing that the set of n where at least one of the $\psi_i$ vanishes has size $O(N^{d-1})$ because the 0-set of an affine-linear form is an affine hyperplane. Then as $\psi_i(n)=O(N)$, the divisors $a_i$ will have to satisfy $a_i\ll \sqrt{N}$. So now we restrict the set of $n$ to the ones where the forms don’t vanish. Then we partition the sum into one where $a_1$ is restricted to be at most $N^{\delta}$ (we fix some small $\delta >0$), and the remaining one. Using the well-known fact that $\mu^2(n)=\sum_{d^2\mid n}\mu(d)$, we remark that the remaining one equals

$\displaystyle{ \sum_{N^{\delta}\leq a_1\ll \sqrt{N}}\mu(a_1)\sum_{n:a_1^2\mid \psi_1(n)}\mu^2(\psi_2(n))\cdots\mu^2(\psi_t(n))}$

Now the number of $n$ in the sum is $O(N^d/a_1^2+N^{d-1})$. Given that $\sum_{N^{\delta}\leq a_1} a_1^{-2}\ll N^{-\delta}$, we can bound this sum by $O(N^{d-\delta})$. We proceed similarly for $i=2,\ldots,t$, so that the sum to estimate is

$\displaystyle{ \sum_{a_1,\ldots,a_t\leq x^\delta}\prod_i\mu(a_i)\lvert\{n\in\mathbb{Z}^d\cap K\mid \forall i \quad \psi_i(n)\equiv 0\mod a_i^2\}\rvert}$

Quite gross volume packing arguments indicate that the number of integral points (lattice points) to estimate above is $\text{Vol}(K)\alpha_{\Psi}(a_1^2,\ldots,a_t^2)+O(N^{d-1+2t\delta})$, where

$\displaystyle{ \alpha_{\Psi}(k_1,\ldots,k_t)=\mathbb{E}_{n\in [\text{lcm}(k_i)]^d}\prod_i1_{k_i\mid\psi_i(n)}}$

is the density of zeroes modulo the $a_i^2$. Moreover we can extend the sum beyond $N^\delta$ at the cost of a mere $N^{O(\delta)}$. Hence finally for any $\epsilon >0$ we can write that

$\displaystyle{\sum_{n\in\mathbb{Z}^d\cap K}\prod_{i\in[t]}\mu^2(\psi_i(t))=\text{Vol}(K)\prod_p\beta_p+o_\epsilon(N^{d-1+\epsilon}).}$

Indeed by multiplicativity it is easy to transform $\sum_{a_1,\ldots,a_t}\alpha_{\Psi}(a_1^2,\ldots,a_t^2)\prod_i\mu(a_i)$ into a product over primes.