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Lets take all odd numbers up to some N, now we want to find some even number E where following is true:

$E-1$ is not a multiple of any of the first N odd numbers

$E+1$ is not a multiple of any of the first N odd numbers

on average, there should be 1 of such numbers in N, but obviously there is not such a number amongst first N numbers, therefore it is not regularly dispersed. Can i be sure there will be such a number up to $2\times N$, $3\times N$, or some other $c\times N$?

In other words, how long until I can find such a number for N? I can prove it for N-factorial but that is huge and nowhere near to $c\times N$.

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    $2+1$ is a multiple of $3$. – 2017-01-01
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    Why do you think there should be "on average, $1$ of such numbers in $N$"? How do you establish the claim for $N!$? If there were no twin primes above $N$ then at least one of $E\pm 1$ is composite hence divisible by some odd number $≤\sqrt E$. Thus, $E$ would have to exceed $N^2$. – 2017-01-01
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    On average one in N is from this formula: 1/3 * 3/5 * 5/7 * 7/9 * ... (N-2)/N which gives us 1/N, as basically E mod X can't be 1 and can't be X-1 for each odd X up to N. That should be the worst estimate, as all but two of the modulo values are okay. N-factorial is also based on that, as there has to be at least one in 3*5*7*9*...*N, which is certainly less then N!. – 2017-01-01
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    Not following. The divisibilities involved are not independent. Regardless, do you follow my argument? If you could find a constant $c$ that did the job that would prove the Twin Prime Conjecture. – 2017-01-01
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    Yes, I agree that if we knew $c < N$ then we could prove that conjecture and we do not know if that conjecture is true... – 2017-01-01
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    Right. So, we aren't going to establish the existence of $c$ here. I expect that any tight bound on $E$ will depend on some hard analytic number theory. After all, if there were only finitely many primes then, for sufficiently large $N$, $E$ could not exist. – 2017-01-01
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    Just to say: to prove that $E$ exists, invoke the Chinese Remainder Theorem. if the primes in your list are $3=p_1,p_2,\cdots, p_k$ then simultaneously solve $E-1\equiv 1\pmod 2, E-1\equiv -1\pmod {p_i}$. That gives us a class $\pmod {2\times \prod p_i}$. – 2017-01-01
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    Thanks, I will have a look at that theorem, though this math is probably over my current knowledge level. – 2017-01-01
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    @MartinVis The chinese remainder theorem is not very difficult to learn and understand. It says : If we have pairwise coprime integers $n_1,\cdots,n_k$ and arbitary integers $a_1,\cdots,a_k$ and if we denote $N:=n_1\cdot\ \cdots\ \cdot n_k$, then there exists a unique integer $s$ with $0\le s\le N-1$ with $s\equiv a_j\mod n_j$ for $j=1,\cdots,k$ – 2017-01-03
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    Unfortunately, in general, nothing can be said about the magnitude of the number $s$ – 2017-01-03

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