Is there a conjecture, statement, formula or anything that determines whether or not there is a prime between two numbers? I'm not looking for a computer algorithm. Any help would be appreciated.
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1Are you looking for something like this? – cansomeonehelpmeout Jan 25 '20 at 16:50
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Whether there exists a prime between any two positive integers depends on the size of the interval between them. See Bertrand's postulate and refinements thereto. – Keith Backman Jan 25 '20 at 16:52
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Thank you that is what I need. – Tobi Odeyemi Jan 25 '20 at 16:58
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we can actually iterate bertrand to show either there's a prime between $n$ and $\lceil{4n\over 3}\rceil$ or two between $\lfloor{2n\over 3}\rfloor$ and $2n$ – Jan 25 '20 at 22:29
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How can I do that? @Roddy MacPhee – Tobi Odeyemi Jan 26 '20 at 05:53
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it's literally use overlapping bounds ... – Jan 26 '20 at 10:14
3 Answers
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I highly doubt this is possible. Otherwise you could simply check any number of being prime, just by considering the number itself and the number +1 or something along these lines. But checking numbers being prime is hard...
Jonas Linssen
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Bertrand's postulate in loose terms says:
There is alway a prime between $n$ and $2n$
Once we know this we can iterate it by shrinking or enlarging $n$ we get statements like:
- There is always a prime between $\lfloor {2n\over 3}\rfloor$ and $\lceil{4n\over 3}\rceil$
- There is always a prime between $\lfloor {3n\over 4}\rfloor$ and $\lceil{3n\over 2}\rceil$
- $\ldots$
rounding is simply to make sure we land on integers. However there's overlap between our intervals, therefore we get :
- Either there's a prime in the insection of these intervals, with the original, or there are two primes in the complement to their intersection.
We can break this down to 5 intervals that don't intersect( or minimally) with implications on each other:
There is always either a prime between $n$ and $\lceil{4n\over 3}\rceil$ or one between $\lfloor {2n\over 3}\rfloor$ and $\lceil{3n\over 4}\rceil$ or $\lfloor {4n\over 3}\rfloor$ and $\lceil{3n\over 2}\rceil$ otherwise there's an interval where you can double a number without a prime.
There is always either a prime between $n$ and $\lceil{3n\over 2}\rceil$ or one between $\lceil{3n\over 2}\rceil$ and $2n$. otherwise there's an interval where you can double a number without a prime.
There is always either a prime between $\lfloor {3n\over 4}\rfloor$ and $\lceil{4n\over 3}\rceil$ or one between $\lfloor{2n\over 3}\rfloor$ and $\lceil {3n\over 4}\rceil$ . Otherwise there's an interval where you can double a number without a prime.
etc.
numeric example
if n=8
then
6 to 12 and 8 to 16 each have a prime, but it could be the same prime in each range because they overlap in between 8 and 12. If this weren't true ( 11 is prime, but lets say it isn't for a moment), we'd get that both of the ranges 6 to 8 and 12 to 16 would require primes for Bertrands postulate to hold. I used a pseudoexample here, because up to n=100 000 it's never true that center range doesn't have a prime.
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The best we can do is that for any $x \ge 89693$ there is always at least one prime between $x$ and $x + \dfrac{1}{\log^3 x}$. This is much stronger than Bertrand's postulate and was proved by Pierre Dusart.
Nilotpal Sinha
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