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I have been looking at the literature on sieve theory which proves theorems similar to the following:

For all $x > x_0$ the interval $[x - x^{\theta}, x]$ contains prime numbers.

For example, I understand the best result in this line of work is $\theta = 0.525$ from Baker, Harman, and Pintz (2000) "The Difference Between Consecutive Primes, II".

That paper, and the other references for similar theorems I have found, leave out the precise value of $x_0$ for which this holds. The Baker et al. paper, for example, says "with enough effort, the value of $x_0$ could be determined effectively". I take this to mean that, if one were to keep careful track of bounds in the argument presented in that paper, one would obtain an explicit value for $x_0$ in the above theorem.

My main question is: What would be the approximate order of magnitude of this $x_0$ value, for this and similar, weaker theorems?

Of course, once one has the theorem for a particular value of $x_0$ one may lower that bound computationally by finding primes that cover lower intervals. In theory, one may do this for as long as one pleases, or until one runs up against an interval that has no prime, at which point the lowest possible value of $x_0$ for which the theorem holds is found. I am interested in what the machinery of the proof itself gives in terms of a bound.

As a related request, I would be interested in seeing any references which give an explicit $x_0$, even for a different value of $\theta$ or potentially for a different problem altogether, as long as it has this property of being "a bound that is possible to lower by checking primalities of numbers".

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    $\begingroup$ For all $x\ge1$, the interval $[x,2x]$ contains a prime. Maybe mathoverflow.net/questions/44466/… gives you some answers. $\endgroup$ Commented Aug 24, 2021 at 4:54
  • $\begingroup$ @GerryMyerson I am aware of Bertrand's postulate, in fact it partially inspired this question. Going off the wikipedia proof, you could say $x_0 = 468$ for that result. I was wondering if better asymptotics need larger $x_0$. $\endgroup$ Commented Aug 24, 2021 at 5:26
  • $\begingroup$ Huh? For Bertrand, you can say $x_0=1$, indeed, I did say exactly that. Was there nothing of interest in the link I gave? "I am aware of Bertrand's postulate" – I apologize for my inadequate mind-reading skills. $\endgroup$ Commented Aug 24, 2021 at 5:33
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    $\begingroup$ Indeed $x_0 = 1$ for Bertrand, I was merely saying that the $x_0$ given by the machinery of the standard proof is 468. One then explicitly checks primes to get the number down to 1. Your link is very helpful, in particular, Tao's link in the comments. I am reading through it now, but it definitely has a problem of the form I want. $\endgroup$ Commented Aug 24, 2021 at 5:44

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