2
$\begingroup$

The idea of this post arises when I've considered simple variants of the known as Firoozbakht's conjecture (see this corresponding Wikipedia or [1]), and comparisons by trial and error with means (if you are interested feel free to do yourself comparisons motivated from the Wikipedia Generalized mean).

I've considered the variant, that is a comparison, $$\frac{n+p_{n+1}}{n+1}>\lambda(n)\cdot p_{n}^{\frac{1}{n}}+\mu(n)\tag{1}$$ where $p_k$ denotes the $k$-th prime number and $\lambda(n)$ and $\mu(n)$ are simple functions. As I've said in the introductory paragraph I got my comparison by trial and error using a Pari/GP program, from these calculations I've considered $\lambda(n)=\log n$ and $\mu(n)=\log\log n$, and my proposal (below in my Question) will be for integers $n\geq 2700$.

The identity $(1)$ evokes a comparison for means because the quantity $\frac{n\cdot 1+p_{n+1}}{n+1}$ is the arithmetic mean of $n$ ones $1$'s and $p_{n+1}$ while that $p_{n}^{1/n}$ is the geometric mean of $n-1$ ones $1$'s and the prime $p_n$. The functions $\log n$ and $\log\log n$ are simple functions that additionally belong to a class of functions studied in Rafael Jakimczuk, Functions of Slow Increase and Integer Sequences, Journal of Integer Sequences, Article 10.1.1, Vol. 13 (2010).

Question. Prove or refute the following conjecture as a variant of Firoozbakht's conjecture:

For each integer $n\geq 2700$ the following inequality $$\frac{n+p_{n+1}}{n+1}>p_{n}^{\frac{1}{n}}(\log n)+\log\log n\tag{2}$$ holds.

Many thanks.

My proposal $(2)$ is true for integers $2700\leq n\leq 50000$. You can see it from the web Sage Cell Server using this Pari/GP program (choose GP as language)

for(n=2700, 50000, if((n+prime(n+1))/(n+1)<(log(n))*prime(n)^(1/n)+log(log(n)),print(n)))

(you need to wait a minute to see that there aren't counterexamples as outputs of this program), or this other output, it is a line written in Pari/GP, showing the differences of LHS and RHS

for(n=2700, 50000, if((n+prime(n+1))/(n+1)>(log(n))*prime(n)^(1/n)+log(log(n)),print((n+prime(n+1))/(n+1)-(log(n))*prime(n)^(1/n)-log(log(n)))))

As was said feel free to do yourself variants in your home if you can to create a more interesting variant of Firoozbakht's conjecture using generalized means.

References:

[1] Conjecture 30. The Firoozbakht Conjecture, Retrieved 22 August 2012 in Carlos Rivera's web The Prime Puzzles & Problems Connection.

Improve this question
$\endgroup$
6
  • $\begingroup$ I think that the idea is thus if it is possible to prove or refute the inequality $(2)$ for some $n_0>2700$ using the explicit inequalities for primes, since I cann't find a counterexample. Feel free to add comments/critics about this question, I'm waiting your feedback. Thank you very much. $\endgroup$
    – user142929
    Dec 18, 2019 at 21:19
  • $\begingroup$ Also is welcome comments about if my inequality $(2)$ is interesting or if there are similar inequalities in the literature. $\endgroup$
    – user142929
    Dec 18, 2019 at 23:40
  • $\begingroup$ Due the form of the functions $\lambda(n)$ and $\mu(n)$ that I've choosen by experimenting, the inequality $(2)$ can be written as $$e^{e^{A(n)}}>n^{n^{G(n)}},\text{ for integers } n\geq 2700$$ where the exponent $A(n)$ denotes the arithmetic mean $A(\underbrace{1,\ldots,1}_{n\text{ ones}},p_{n+1})=\frac{n+p_{n+1}}{n+1}$ and $G(n)$ denotes the geometric mean $G(\underbrace{1,\ldots,1}_{n-1\text{ ones}},p_{n})=\sqrt[n]{p_{n}}$. This is just to emphasize that my inequality is a comparison of means, in the same way that Firoozbakht's conjecture can be interpreted in terms of geometric means $\endgroup$
    – user142929
    Dec 19, 2019 at 13:21
  • 1
    $\begingroup$ Replacing $p_n$ with $n\log n$ in (2) yields a false inequality for all large $n$. I'm guessing that some improved form of the prime number theorem may imply that (2) is false. $\endgroup$
    – Joel Moreira
    Dec 19, 2019 at 16:15
  • 1
    $\begingroup$ Hm, apparently $p_n/(n\log n)$ decays to $1$ rather slowly (it behaves like $1+\frac{\log\log n}{\log n}$), so I take back my guess that an improvement of PNT will show that (2) is false. $\endgroup$
    – Joel Moreira
    Dec 19, 2019 at 16:46

1 Answer 1

Reset to default
1
$\begingroup$

Yes, (2) holds for all large enough $n$. According to Wikipedia (see also this question) $\frac{p_n}n=\log n+\log\log n-1+\frac{\log\log n}{\log n}(1+o(1))$, so $$\frac{n+p_{n+1}}{n+1}=\log(n+1)+\log\log(n+1)+\frac{\log\log n}{\log n}(1+o(1))$$ Since $\log\log(n+1)\approx\log\log n+\frac1{n\log n}$, so that $\log\log(n+1)-\log\log n=o\left(\frac{\log\log n}{\log n}\right)$, (2) now follows from:

Lemma : $\log(n+1)-p_n^{1/n}\log n=o\left(\frac{\log\log n}{\log n}\right)$.

Proof: Using the prime number theorem to write $p_n=n\log n\cdot(1+o(1))$ we have $p_n^{1/n}=\exp\left(\frac1n(\log n+\log\log n+o(1))\right)=1+o\left(\frac{\log\log n}{(\log n)^2}\right)$. On the other hand, $\log(n+1)=\log n+o\left(\frac{\log\log n}{\log n}\right)$ and the lemma follows.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Many thanks I'm going to study your nice proof. As I've said if some user wants to try a more suitable variant than mine (a variant of Firoozbakht's conjecture, using combinations of different generalized means) that he/she feels free to do it. $\endgroup$
    – user142929
    Dec 19, 2019 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.