An inequality that involves consecutive primes, prime gaps and roots of prime numbers as a weak form of Firoozbakht's conjecture

52 Views Asked by At

In this post for integers $n\geq 1$ we denote the $n$-th prime number as $p_n$.

When we consider that $k>1$ runs over integers, from the theory of the Stolarsky mean we can deduce that as $k\to \infty$ the inequality $$\frac{n(p_{n+1}-p_n)}{\log p_n}<\left(\frac{k\left(\sqrt[k]{p_{n+1}}-\sqrt[k]{p_{n}}\,\right)}{p_{n+1}-p_n}\right)^{-k/(k-1)}\tag{1}$$

becomes in the known as Firoozbakht's conjecture, because as $k\to \infty$ the RHS of previous inequality tends to the logarithmic mean of consecutive primes numbers $p_{n+1}$ and $p_{n}$.

Wikipedia has an article for Stolarsky mean in which I was inspired (that is the specialization for $p=1/k$ from the quotient that defines the Stolarsky mean and after I do a comparison with the logarithmic mean of the section Special cases). For the mentioned conjecture ([1]) Wikipedia has the article Firoozbakht's conjecture that refers in the first paragraph the history of this conjecture.

Question. I would like to know what is (approximately) the largest integer $k>1$ for which we can to prove that $(1)$ is true $\forall n\geq 1$ or $\forall n\geq n_0$ with $n_0$ a suitable choice ($n_0=n_0(k)$ maybe depending on $k$) from your calculations. Many thanks.

I know how grows the logarithmic mean of consecutive prime numbers by the squeeze theorem (I mean the inequality between the logarithmic mean, and the geometric and arithmetic means) and invoking the prime number theorem, but I believe that it will not be enough to approach our Question, and I know that the questions about prime gaps $g_n:=p_{n+1}-p_n$ are very difficult.

References:

[1] Conjecture 30 The Firoozbakht Conjecture, Carlos Rivera's The Prime Puzzles & Problems Connection, (2012).

[2] Kenneth B. Stolarsky, Generalizations of the logarithmic mean, Mathematics Magazine. 48, (1975), 87–92.