MHB Set of 2015 Consecutive Positive Ints with 15 Primes

AI Thread Summary
The discussion centers on whether there exists a set of 2015 consecutive positive integers containing exactly 15 prime numbers. It is established that as the starting integer n increases from 1 to 2016! + 2, the number of primes in the set, denoted as f(S_n), changes by at most 1 at each step. Starting from f(S_1) = 305 and ending at f(S_n) = 0, the intermediate value theorem suggests that f(S_n) must equal 15 at some point. The calculations demonstrate that the transition from a high to a low number of primes occurs smoothly, confirming the existence of such a set. The conclusion affirms the mathematical validity of the claim.
lfdahl
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Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
 
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lfdahl said:
Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
[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]
 
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]

Thankyou so much, Opalg, for your excellent solution and participation!
 
lfdahl said:
Thankyou so much, Opalg, for your excellent solution and participation!

Yes, we can always count on Chris (Opalg) to post a robust, lucid solution. (Yes)
 
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