# $\pi(x+200)-\pi(x)\leq 50$?

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

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

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 $$\prod_{p \mid 210}(1-1/p) < 50/200$$.

• We have $\pi(x+210)-\pi(x) \leq 48$ – Dattier Sep 2 at 11:30
• 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. – Ofir Gorodetsky Sep 2 at 11:36
• for x=1, we have $\pi(211)-\pi(1)=47$ – Dattier Sep 2 at 11:41
• Yes, sorry. That's the only counterexample though. – Ofir Gorodetsky Sep 2 at 11:46