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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 $b<a<100,000$.

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
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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.

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  • $\begingroup$ I can't see that the the openness of the inequality influence of my conjecture. But thanks for the link! $\endgroup$ – Lehs Oct 7 '19 at 9:40
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    $\begingroup$ @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. $\endgroup$ – Steven Stadnicki Oct 11 '19 at 20:51
  • $\begingroup$ @StevenStadnicki - But my conjecture isn't about large numbers $b$. And it doesn't predict an equality. $\endgroup$ – Lehs Oct 11 '19 at 21:21

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