# Numbers $n$ with $n,n+2$ coprime to $p_k\#$ on $$In the context of sieving for twin primes (p\# is the primorial function) the following seems true. The number of n such that n, (n+2) are coprime to p_k\# for n=1,2,…,~p_k\# is \prod_{i=2}^k (p_i-2). Calculation of the first 7 products and a bit of luck in OEIS suggest this is true but it does not seem so easy to prove. I notice that n+2 is a simple permutation of GCD(n,p_k\#) and so maybe using characters? Partly I am posting this because I haven’t seen it before so someone else might find it interesting. It does not lead (directly) to a useful sieve because p_k\# grows too fast but probably it has been considered somewhere. #### Solutions Collecting From Web of "Numbers n with n,n+2 coprime to p_k\# on$$"

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.