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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Sign up to join this communityIs it true, that $\forall x \in \mathbb N, \pi(x+200)-\pi(x) \leq 50 $ ?
$$\pi(x)=\text{card}(\{n \in [0,x] \cap \mathbb N, n\text{ is prime}\})$$
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 $\prod_{p \mid 210}(1-1/p) < 50/200$.