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Is it true, that $\forall x \in \mathbb N, \pi(x+200)-\pi(x) \leq 50 $ ?

$$\pi(x)=\text{card}(\{n \in [0,x] \cap \mathbb N, n\text{ is prime}\})$$

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Yes.

Up to $207$ there are $46$ primes. Hence, the inequality is true for $x \le 7$.

Let $$\pi_{210}(x) = \textrm{card}(\{n \in [0,x] \cap \mathbb{N}, \, \gcd(n,210)=1\}).$$ For $x>7$, $\pi(x+200)-\pi(x) \le \pi_{210}(x+200)-\pi_{210}(x)$. Since $\pi_{210}$ is $210$-periodic, it is enough to verify that $\pi_{210}(x+200)-\pi_{210}(x) \le 50$ for $x \le 210$, which can be done by hand or by computer. Here is SageMath code:

L = [int(gcd(i,210)==1) for i in range(420)]

max(sum([numpy.array(L[i:][:210]) for i in range(200)]))

and the last line outputs $47$. So the bound can be improved to this number.

The number $210=2 \times 3 \times 5 \times 7$ was chosen because $\prod_{p \mid 210}(1-1/p) < 50/200$.

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  • $\begingroup$ We have $\pi(x+210)-\pi(x) \leq 48$ $\endgroup$
    – Dattier
    Commented Sep 2, 2020 at 11:30
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    $\begingroup$ Indeed. By considering coprimality with $2310 = 2 \times 3 \times 5 \times 7 \times 11$, the same method produces $\pi(x+210)-\pi(x) \le 46$ for $x>11$; small $x$ can be checked by hand, and in fact there are cases of equality. The second Hardy-Littlewood Conjecture predicts $\pi(x+210)-\pi(x) \le \pi(210)=46$ also (for $x \ge 2$), but the general form of the conjecture is believed to be false. $\endgroup$
    – Ofir Gorodetsky
    Commented Sep 2, 2020 at 11:36
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    $\begingroup$ for x=1, we have $\pi(211)-\pi(1)=47$ $\endgroup$
    – Dattier
    Commented Sep 2, 2020 at 11:41
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    $\begingroup$ Yes, sorry. That's the only counterexample though. $\endgroup$
    – Ofir Gorodetsky
    Commented Sep 2, 2020 at 11:46

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