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In the context of sieving for twin primes ($p\#$ is the primorial function) the following seems true.

The number of $n$ such that $n, (n+2)$ are coprime to $p_k\#$ for $n=1,2,...,~p_k\#$ is $\prod_{i=2}^k (p_i-2).$

Calculation of the first 7 products and a bit of luck in OEIS suggest this is true but it does not seem so easy to prove. I notice that $n+2$ is a simple permutation of $GCD(n,p_k\#)$ and so maybe using characters? Partly I am posting this because I haven't seen it before so someone else might find it interesting. It does not lead (directly) to a useful sieve because $p_k\#$ grows too fast but probably it has been considered somewhere.

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Let me consider $p_k$ primes.

According to the Chinese Remainder Theorem every number $n$ from $1$ to $p_1p_2\cdots p_k$ is uniquely determined by the set of its $k$ remainders $r_i$ when $n$ is divided by $p_i$.

In practice to every choice of remainders $r_i$ corresponds one and only one number $n$ from $1$ to $p_1p_2\cdots p_k$.

So, we can ask, in how many ways can we choose remainder $r_i$ ? We want it different from 0, otherwise $n$ is divisible by $p_i$ and we want it different from $p_i-2$, otherwise $n+2$ is divisible by $p_i$.

So, apart the case of $p=2$ in which only the remainder 1 is admissible, in all the other cases there are $p_k-2$ possible remainders.

Hence, the number of sets of "good" remainders (and thus the number of "good" pairs) is just $$(p_2-2)(p_3-2)\cdots (p_k-2) $$

as you imagined.

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