I require a set of primes, all being equal to or greater than $2v+2$.
The product of the primes should be at least $(2^v)+1$.
I have one additional constraint. Each prime minus one must be divisible by some number between $v$ and $2v$. In more technical terms, each prime $p_k$ should be equal to $1+(c_k)(d_k)$, where $d_k$ can be any natural, and $v \le c_k \le 2v$.
My question is: How big can the largest prime be, assuming I want to minimize its size, in terms of $v$?
The whole reasoning behind the additional constraint is that I'm trying to get each prime in the set to have between $v$ and $2v$ roots of unity.
Each prime in the set should be unique; i.e. no repeated primes.
For $v=1$, you want primes greater than 4 with a product at least 3, so the best you can do is $\{5\}$ with greatest prime 5.
For $v=2$ you want odd primes greater than 6 with a product at least 5, so the best you can do is $\{7\}$ with greatest prime 7.
For $v=3$ you want primes which are 1 mod (at least one of) 3, 4, or 5 which are greater than 8 with a product at least 9, so the best you can do is $\{11\}$ with greatest prime 11.
For $v=4$ you want primes which are 1 mod 4, 5, 6, or 7 and which are greater than 10 with a product at least 17, so the best you can do is $\{11,13\}$ with greatest prime 13.
It's not hard to find the set with the minimal size of maximal element. For example, for $v=100$ you get $\{211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277\}$. Generally you just take primes starting from $2v+3$ until the product is sufficiently large; I haven't yet found a case where the prime is in none of the residue classes.
Assuming this continues to hold (or approximately hold) you want to take primes starting from $2v+3$ until you get a product greater than $2^v$. This should take around $v\log2/\log v$ primes with the greatest around $(2+\log2)v$ on average.