Can you provide a counterexample for a claim given below?
Inspired by Puzzle 937 I have formulated the following claim:
For any $n > 0$ let $B = p_1 \cdot p_2 \cdot .... \cdot p_n$ be the product of the first $n$ primes. Let $X$ be the smallest number, bigger than $B^k/p_{n+1}$ and coprime to $B^k$, where $k$ is a fixed positive integer. Define the number $m_n$ as $X \cdot p_{n+1}-B^k$ , then $m_n$ is either $1$ or prime.
I was searching for counterexample using the following PARI/GP code:
CE(lb,ub,k)={
for(n=lb,ub,
B=prod(i=1,n,prime(i));
X=ceil((B^k)/prime(n+1));
while(gcd(X,B^k)!=1,
X++);
m=X*prime(n+1)-B^k;
if(!(ispseudoprime(m) || m==1),print(m)))
}
After sieving $p_n$, numbers coprime to the primoral $p_n\#$ are prime at least up to $p_{n+1}^2$ ($1$ and square excluded of course)
e.g. $p_n=5$, numbers coprime to $p_n\#=30=B$ (also coprime to $B^k$) are $\{1,7,11,13,17,19,23,29,31,37,41,43,47,49,...\}$ and are prime up to $49$.
These numbers are generally expressed in this form: $p_n\#\cdot i+\{1,p_{n+1},...\}$
e.g. for $p_n=3$ they are expressed as $$6i+\{1,5\}$$ e.g. for $p_n=5$ they are expressed as $$30i+\{1,7,11,13,17,19,23,29\}$$
Obviously, $X$ is coprime to $B^k$ (it was picked in the coprime list) and so is $p_{n+1}$ which means $X\cdot p_{n+1}$ is also coprime to $B^k$(=$30i$ in our example) and since $X\cdot p_{n+1}=B^k+m_n$, we know $m_n$ is in that coprime list.
$m_n=p_{n+1}\cdot x$ where $x=X-\frac{B^k}{p_{n+1}}$ is smaller than the maximum gap between coprimes in the list (since we pick the first smaller coprime greater than $\frac{B^k}{p_{n+1}}$). If this maximum gap is smaller than $p_{n+1}$, then $m_n<p_{n+1}^2$ and is therefore prime or $1$.
I think it was shown that the maximum gap between numbers coprime to $p_n$ is smaller than $p_{n+1}$, but I'll have a look when I have a bit more time.
edit: well....it was not http://oeis.org/A048670