The discussion confirms that there exists a set of 2015 consecutive positive integers containing exactly 15 prime numbers. By defining the set \( S_n \) as the 2015 integers starting from \( n \) and the function \( f(S_n) \) as the count of primes in that set, it is established that as \( n \) progresses from 1 to \( 2016! + 2 \), \( f(S_n) \) transitions from 305 to 0. This guarantees that \( f(S_n) \) must equal 15 at some point due to the nature of prime distribution and the intermediate value theorem.
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[sp]For each natural number $n$ let $S_n$ be the set of $2015$ consecutive positive integers starting at $n$, and let $f(S_n)$ be the number of primes in that set.lfdahl said:Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
Opalg said:[sp]For each natural number $n$ let $S_n$ be the set of $2015$ consecutive positive integers starting at $n$, and let $f(S_n)$ be the number of primes in that set.
To get from $S_n$ to $S_{n+1}$, you have to remove $n$ from $S_n$, and add $n+2015$ to the set. If either both or neither of those two numbers are prime then $f(S_{n+1}) = f(S_n)$. If one of them is prime and the other is not then $f(S_{n+1})$ and $f(S_n)$ will differ by $1$.
When $n=1$, $S_1$ is greater than $15$ (in fact, I think that $S_1 = 305$). When $n = 2016! + 2$, $S_n = 0$, because $2016! + k$ is divisible by $k$ whenever $2\leqslant k \leqslant 2016$.
So as $n$ increases from $1$ to $2016! + 2$, $f(S_n)$ changes by at most $1$ at each step, and has to go from $305$ to $0$. By a sort of integer-valued "intermediate value theorem", it must take the value $15$ at some point.[/sp]
lfdahl said:Thankyou so much, Opalg, for your excellent solution and participation!