Question about the gaps between prime numbers

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The discussion centers on the relationships between consecutive prime numbers, specifically whether there exists a prime number \( p_n \) such that \( p_{n+1} > p_n^2 \) and \( p_{n+1} > 2p_n \). Participants reference Bertrand's Postulate, which provides insights into prime gaps, and Firoozbakht's conjecture, noting that while related, it does not directly address the \( p_{n+1} > p_n^2 \) condition. The conversation concludes that the existence of such primes remains unproven and invites further exploration of related mathematical postulates.

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DuckAmuck
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Is there any prime number pn, such that it has a relationship with the next prime number pn+1
p_{n+1} > p_{n}^2
If not, is there any proof saying a prime like this does not exist?

I have the exact same question about this relation:
p_{n+1} > 2p_{n}
 
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DuckAmuck said:
Is there any prime number pn, such that it has a relationship with the next prime number pn+1
p_{n+1} > p_{n}^2
If not, is there any proof saying a prime like this does not exist?

I have the exact same question about this relation:
p_{n+1} > 2p_{n}
https://en.wikipedia.org/wiki/Prime_gap

There is also a proof for arbitrary gaps, but see the section "upper bounds".
 
Interesting.Bertrand's Postulate answers the second part of my question. :)

I see Firoozbakht's conjecture, which is similar to my first part, but it's not quite the same thing as
p_{n+1} > p_{n}^2

I wonder if this can be proved or disproved from other postulates...
 
DuckAmuck said:
Interesting.Bertrand's Postulate answers the second part of my question. :)

I see Firoozbakht's conjecture, which is similar to my first part, but it's not quite the same thing as
p_{n+1} > p_{n}^2

I wonder if this can be proved or disproved from other postulates...

This also follows very easily from Bertrand's postulate.
 
DuckAmuck said:
but it's not quite the same thing as
p_{n+1} > p_{n}^2

I wonder if this can be proved or disproved from other postulates...

Compare ##2p_n## to ##p^2_n## .
 
micromass said:
This also follows very easily from Bertrand's postulate.

Yeah it does. Wow. I'm dumb. :p
 

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