T/F: The interval of width $n$ containing the most amount of primes is $[2,n+2]$? Given $n\in\mathbb{N},$ the interval of width $n$ containing the most amount of primes is $[2,n+2]$ ( rather than $[2+x, n+2+x]$ ).
This sounds like it should be true since the primes spread out more in general as $x$ increases. But general trends don't prove specific statements.
This is obviously true for $n=1$ as there is no even prime greater than $2,$ and so for any $x\geq 1,\ [x+2,x+3]$ contains at most one prime whereas $[2,3]$ contains two primes.
I think simple arguments like this can be made for small $n$.
For fixed even $n\geq 2, n$ even, and for all $x\geq 1:$ (number of primes in $[2, n+2]$ ) $\geq$ (number of primes in $[x+2, n+2+x]$ ) implies that (number of primes in $[2, (n+1)+2]$ ) $\geq$ (number of primes in $[x+2, (n+1)+2+x]$ ) because $\underset{x\geq 1}{\max}$(number of primes in $[x+2, n+3+x]$ ) $=\underset{x \text{ odd}}{\max}$ (number of primes in $[x+2, n+2+x]$ ) . Therefore, we only need to check even $n'$s going forward (because if it's true for even $n$ then it's true for $n+1$.)
$n=2:\ $ for any $x\geq 1,\ [x+2,x+4]$ contains at most two primes, which can only happen if $x$ is odd, which is not more than the number of primes in $[2,3].$
$n=4$ and $n=6$ are easy to check.
$n=8:\ [2,10]$ contains four primes. In order for $[x,x+8]$ to contain five primes, $x$ must be odd and $x, x+2, x+4, x+6, x+8$ must all be prime. But one of $x, x+2, x+4$ is divisible by three and so this is not possible.
 A: I am surprised that nobody has mentioned this yet, but this is essentially the Second Hardy-Littlewood conjecture and it is most likely false, since it contradicts the theoretically much more sound First Hardy-Littlewood conjecture. There might not be a counter-example below $10^{100}$ or even $10^{1000}$, however.
A: For $n\ge 1$ a positive integer, let $m_n$ be the smallest positive integer for which $\pi(n+m_n)-\pi(m_n-1)$ is optimal.
Lemma For $n\ge 59$, we have $m_n\le n+1$.
Proof: For $n\ge 59$, we have the inequalities
$$
\frac{n}{\log n - 1/2}\le \pi(n)\le \frac{n}{\log n-9/8}.
$$
Therefore, if $m_n\ge n+2$, we have
$$
\log(m_n+n)-9/8\ge \log(2)+\log(n+1)-9/8> \log(n+1)-1/2\quad\text{and}\quad \log(m_n-1)-1/2\ge \log(n+1)-1/2,
$$
whence
$$
\begin{align*}
\pi(n+m_n)-\pi(m_n-1) &\le \frac{n+m_n}{\log(n+m_n)-9/8} - \frac{m_n-1}{\log(m_n-1)-1/2}\\
&\le \frac{n+m_n}{\log(n+1)-1/2} - \frac{m_n-1}{\log(n+1)-1/2}\\
&=\frac{n+1}{\log(n+1)-1/2}\le \pi(n+1).
\end{align*}
$$
So the interval $[2,n+2]$ contains more primes than $[m_n,m_n+n]$. This contradicts the definition of $m_n$. We conclude that $m_n\le n+1$. $\square$
Theorem If $m_n=2$ for all $n\le 58$, then $m_n=2$ for all $n\ge 1$.
Proof: By induction on $n$. The base case $n=1$ is clear. Let $n\ge 2$ and assume that $m_k=2$ for all $k=1,\ldots,n-1$. If $n\le 58$, we are done. Otherwise, $n\ge 59$ and by the lemma, $m_n\le n+1$. Now,
$$
\begin{align*}
\left(\pi(n+2)-\pi(1)\right) - \left(\pi(n+m_n)-\pi(m_n-1)\right)
=& \left(\pi(m_n-1)-\pi(1)\right) - \left(\pi(n+m_n)-\pi(n+2)\right)\ge 0,
\end{align*}
$$
because $m_{m_n-2}=2$, by the induction hypothesis. $\square$
A: This is far from a proof, I just wanted to share some code to find potential counterexamples. Though so far, the statement in the question seems to be true.

I wrote a Python program to check all primes up to $n=1,000,000$, and all interval sizes up to $x=500$, and in every case, the interval $[2,n+2]$ contained the most primes.
The way it works is it lists all the consecutive gaps between primes (dPrimes), then for every possible interval length I = 2...500 it adds up the size of all I consecutive  gaps. In each case, this sum is clearly the difference between the first and last prime in the interval, and if this difference is smaller than the sum of the first I gaps, then we have found an interval more dense than $[2,n+2]$.
def main(N = 1000000, X = 500):
    
    import math

    primes = [i for i in range(2,N+1) if all(i % j != 0 for j in range(2,int(math.sqrt(i))+1))]
    foundDenserInterval = False
    
    def diff(L):
        return [L[i+1]-L[i] for i in range(len(L)-1)]

    dPrimes = diff(primes)

    for I in range(2,X+1):
        print(I) # not necessary, just shows progress
        orig_gapsums = sum(dPrimes[:I])
        all_gapsums = [sum(dPrimes[j:j+I]) for j in range(1,len(dPrimes)-I)]
        for g in range(len(all_gapsums)):
            if all_gapsums[g] <= orig_gapsums:
                print("FOUND DENSER INTERNVAL:", [p for p in primes[g+1:] if p <= primes[g+1]+I])
                foundDenserInterval = True

    if not foundDenserInterval:
        print("DID NOT FIND A DENSER INTERVAL THAN [2,N+2]")

if __name__ == "__main__":
    main()

It took around $4$ minutes to run this on my PC (with a dual-core $2 \times 2.8$ GHz processor), so if anyone has a beefier computer, feel free to run it for even larger $N$ and $X$! Or feel free to improve on my code. :)
A: This is a strategy on how (I think) this can be proved not a complete answer.
The wikipedia page on the prime counting function has an explicit bound on the difference between the prime counting function and the logarithmic integral function.
This can be used to get an explicit interval bound for the number of primes in any interval $[a,b]$. As the error term is decaying faster than the number of primes, this should suffice to prove that there can be no counter example for $a,b$ sufficiently big with explicit bounds on how big $a,b$ need to be. One can then in principle brute force the smaller cases numerically.
