The discussion revolves around the question of whether there exists a set of 2015 consecutive positive integers that contains exactly 15 prime numbers. Participants explore the properties of such sets and the behavior of the prime count as the starting integer varies.
Participants do not reach a consensus on whether a set of 2015 consecutive integers with exactly 15 primes exists, as the discussion includes various claims and reasoning without definitive resolution.
The discussion relies on assumptions about the distribution of prime numbers and the behavior of the function $f(S_n)$, which may not be universally applicable. The transition of prime counts is described without resolving the underlying mathematical complexities.
[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!