# Yet a conjecture about the prime counting function.

This is a conjecture emanating from random tests with my set routines written in Forth.

$$a>b \;\wedge \pi(a+b)=\pi(a)+\pi(b)\implies b<11$$

Furthermore, if $$b\in\mathbb N$$, then $$b\in\{0,1,2,3,4,9,10 \}$$

I would like (partly) proofs or counterexamples.

Tested for $$b.

Testing the numbers of hits for the non trivial values of $$b$$ and for some upper limits:

               1       2       3       4       9      10
100       74      48       8      14       2       2
1000      832     332      34      66       4       6
10000     8771    2454     204     406      11      20
100000    90408   19180    1223    2444      37      72
1000000   921502  156992    8168   16334     165     328


Your claim is a consequence of the Secondly Hardy Littlewood conjecture which has been open since 1923. We don't even know if $$\pi(a+b) \le \pi(a) + \pi(b)$$ holds in general so proving $$\pi(a+b) = \pi(a) + \pi(b)$$ implies $$b < 11$$ is beyond the current state of art.
• @Lehs If there are large $a, b$ with $\pi(a+b)\gt \pi(a)+\pi(b)$ — that is, if the conjecture doesn't hold — then by increasing $b$ monotonically we can get a $b_0\gt b$ with $\pi(a+b_0)=\pi(a)+\pi(b_0)$. (This is because of the asymptotics of the counting function along with the fact that if $f(a,b)=\pi(a+b)-\pi(a)-\pi(b)$, then $|f(a,b+1)-f(a,b)|\leq 1$). So your conjecture would, in turn, imply the Hardy-Littlewood conjecture. – Steven Stadnicki Oct 11 '19 at 20:51
• @StevenStadnicki - But my conjecture isn't about large numbers $b$. And it doesn't predict an equality. – Lehs Oct 11 '19 at 21:21