10
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The sequence in question is:

$$S=\left\{\int_0^1\pi(x)\pi(1-x)dx,\int_0^2\pi(x)\pi(2-x)dx,...\right\},$$

where $\pi(x)$ is the prime counting function.

I don't know how to check this for an infinite sequence but I've tried computing many values.

Here's the first prime number in the sequence:

$$\int_0^{13}\pi(x)\pi(13-x)dx=73.$$

and the second in the sequence: $$\int_0^{57}\pi(x)\pi(57-x)dx=3803.$$

This is what I know: The primes thin out as higher numbers are reached. My conclusion is that this sequence will continue to find fewer primes compared to all values computed.

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  • $\begingroup$ Please show at least a few of the values you've computed, including how many primes (if any) you've found, plus any thoughts about how the later values might behave. Thanks. $\endgroup$ – John Omielan Feb 27 at 1:48
  • $\begingroup$ Thanks for adding the extra information. I assume you noticed that, due to symmetry with the $\pi$ step function values, that the integrals for the even upper limits are even and, thus, not prime if $\gt 2$. Unless there's some special significance to the values, I also would expect the fraction of values which are primes found would also decrease for larger values. Out of curiosity, where does this problem come from, i.e., is there any particular significance to the sequence $S$ as you've described it? $\endgroup$ – John Omielan Feb 27 at 2:05
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    $\begingroup$ I haven't checked very many of your values but from my, admittedly very limited, knowledge about prime numbers, I suspect there won't be very many patterns or other info you can glean from your sequence $S$. Although there are likely an infinite # of primes which occur in $S$, I doubt that you, or anybody else, will be able to prove this. $\endgroup$ – John Omielan Feb 27 at 2:18
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    $\begingroup$ For $k<541$ , $I(k)$ is a prime for $k=13,57,119,167,171,173,175,341,395,397,427,431,473,515,519$ $\endgroup$ – Claude Leibovici Feb 27 at 9:54
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    $\begingroup$ Why the integral? Wouldn't a sum be a more natural way of representing it? $\endgroup$ – Peter Taylor Feb 27 at 11:23

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