I want to know thenumber of primes that can be there in $k$ consecutive natural numbers. For example we can have 0,1 or 2 primes in 3 consecutive natural numbers.
Let $N_{i,k}$ be the number of primes in $k$ consecutive natural numbers starting from $i$.
We can easily prove that $N_{i,k}=N_{i+1,k}$ or $N_{i,k}=N_{i+1,k}+1$ or $N_{i,k}=N_{i+1,k}-1$. So we can conclude that when $i$ increased by $1$ the number of primes in $k$ consecutive natural numbers changes by atmost $1$.
It is easy to see that $N_{2, k} \ge N_{1,k}$ and $N_{(k+1)!+1,k}=0$. So we can say that when the value of $i$ in $N_{i,k}$ changes from $2$ to $(k+1)!+1$ then $N_{j,k}$ takes all natural values from $0$ to $N_{2, k}$ for some $2\le j \le (k+1)! +1~~($ because when $i$ increased by $1$ the number of primes in $k$ consecutive natural numbers changes by atmost $1)$
Even though I was able to prove the lower bound, I was not able to prove that the maximum value of $N_{i, k}$ is
$N_{2, k}$. Any help to prove the upper bound would be appreciated.
Best Answer
You have rediscoveres a famous conjecture from analytic number theory! Let's translate to the standard notation $\pi(x)$ which denotes the number of primes less than or equal to $x$, so that $N_{i,k} = \pi(i-1+k)-\pi(i-1)$. Hardy and Littlewood conjectured that $\pi(x+k)-\pi(x) \le \pi(k)$ for all $x,y\ge2$.
Interestingly, it is now believed that this conjecture is false, as it is known to contradict the prime $k$-tuples conjecture. The refined conjecture, which is more likely to be true, would be that $\pi(x+k)-\pi(x) \le 2\pi(\frac k2)$. (In other words, the most primes in an interval of length $k$ should occur on the interval $[-\frac k2,\frac k2]$ give or take $1$, where we count negatives of primes as primes.)