[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!
A set of 2015 consecutive positive integers with 15 primes is a sequence of 2015 numbers in a row, where 15 of those numbers are prime numbers. This means that they can only be divided by 1 and themselves, with no remainder.
To find a set of 2015 consecutive positive integers with 15 primes, you can start by choosing any number and counting 2015 numbers after it. Then, check if each number in the set is a prime number. If there are 15 prime numbers in the set, then you have found a set of 2015 consecutive positive integers with 15 primes.
The smallest possible set of 2015 consecutive positive integers with 15 primes would start with the number 2 and end with the number 2016, since 2 is the smallest prime number and 2016 is the 2015th consecutive number after 2. However, there may be smaller sets with 15 primes that do not start with 2.
There is no specific pattern to finding a set of 2015 consecutive positive integers with 15 primes, as it depends on the starting number chosen. However, there are certain strategies and techniques that can be used to increase the chances of finding such a set.
Yes, there can be multiple sets of 2015 consecutive positive integers with 15 primes. This is because there are infinite numbers, and as long as the set of 2015 numbers contains 15 primes, it can be considered a set of 2015 consecutive positive integers with 15 primes.