1

Not so much an question as adding another level to the same question as Ratio of logarithmic primes. (See answers, same as here.)

The Firoozbakht's conjecture (1982) is equal to:

$$(p_{n+1})^{n} < (p_n)^{n+1}.$$

Then the natural log is: $$n \ln(p_{n+1}) < (n+1)\ln(p_n).$$

Now, $$\ln(p_n) \leq \ln(n) + \ln(\ln(n)) + 1\text{, for $n \geq 2$. (*)}$$ With $n = n+1$ we have: $$\ln(p_{n+1}) \leq \ln(n+1) + \ln(\ln(n+1)) + 1, \text{for $n \geq 2$.}$$

And, because $$p_n \geq n*\ln(n)\text{, for $n \geq 2$; (**)}$$

the natural log of $p_n$ is: $$\ln(n) + \ln(\ln(n)) \leq \ln(p_n), \text{for $n \geq 2$}.$$

With $n = n+1$ we have: $$\ln(n+1) + \ln(\ln(n+1)) \leq \ln(p_{n+1}), \text{for $n \geq 2$}.$$

So, if $$n (\ln(n+1) + \ln(\ln(n+1))) < n \ln(p_{n+1}) < n(\ln(n+1) + \ln(\ln(n+1)) + 1) < (n+1)(\ln(n) + \ln(\ln(n))) < (n+1)\ln(p_n) < (n+1)(\ln(n) + \ln(\ln(n)) + 1)$$ holds then the conjecture is true for all terms. If only the outer terms only hold then the primes with largest gap maximal primes will hold true. Primes with smaller prime gaps require sharper bounds on the inner terms.

With the outer terms, dividing by $n(\ln(n) + \ln(\ln(n)))$ we have: $$\frac{\ln(n+1) + \ln(\ln(n+1))}{\ln(n) + \ln(\ln(n)) + 1} < \frac{n+1}{n} \text{, for $n \geq 2$}.$$

This inequality is true because the left-side increases slower than the right-side.

With the inner terms, dividing by $n(\ln(n) + \ln(\ln(n)))$ we have: $$\frac{\ln(n+1) + \ln(\ln(n+1)) + 1}{\ln(n) + \ln(\ln(n))} < \frac{n+1}{n} \text{, for $n \geq 2$}.$$

This inequality is false for every $n$ value tested.

Is there a problem with any of these statements?

References:

$(*)$ Proved in Dusart 2010, ESTIMATES OF SOME FUNCTIONS OVER PRIMES WITHOUT R.H.,section 4. Useful Bounds)

$(**)$ Used in Dusart 1999, THE k th PRIME IS GREATER THAN k(ln k + ln ln k − 1) FOR k ≥ 2 Lemma 1. p. 413

Community
  • 1
user160140
  • 147
  • 1
    You last inequality is false for all $n$ that I've checked it for. For instance, if $n=10$, then the lhs evaluates to 1.362... while the rhs is 11/10=1.1. – Matthew Conroy Apr 11 '14 at 19:56
  • Moving the "+ 1" from the numerator to the denominator seems to help to computations, but that is the for the widest spread or the for the largest primes. – user160140 Apr 11 '14 at 21:30
  • Now edited as to remove the bad statement. – user160140 Apr 11 '14 at 23:39

2 Answers2

4

This approach will not work since Dusart's bounds are not strong enough to prove that inequality. His bounds have error $n/\log^k n$ but you need closer to $\log^2 n$ to be able to prove Firoozbakht's conjecture.

It's easy to see that your inequality will fail for large $n$. Multiply by $n\log n+n\log\log n$ (which is positive in this range) to get $$ n\log(n+1)+n\log(\log(n+1))+n\stackrel{?}{<}n\log n+n\log\log n+\log n+\log\log n $$ which is $$ n<n(\log(n+1)-\log n)+n(\log(\log(n+1))-\log\log n)+n\stackrel{?}{<}\log n+\log\log n $$ but clearly $\log n+\log\log n<n$ for large $n$.

As an aside, work by Maier, Granville, and others since Firoozbakht made her conjecture has shown that Firoozbakht's conjecture is likely false, though the first counterexample is probably very large.

Charles
  • 32,122
  • It looks like you are restating my comments with the inner terms and proving them. Thanks. What about the outer terms? – user160140 Apr 29 '14 at 22:12
  • 1
    @user160140: The calculations are not hard, and fail for the same reason indicated above: the error bounds are much too large (on the order of $n^1\log^in$ when you need them to be on the order of $n^0\log^jn$). – Charles Apr 29 '14 at 22:13
0

Firoozbakh's conjecture wants to show that a particular sequence defined on prime numbers is descending. It is not always easy to prove that a sequence is a monotonic sequence. I think it's better to prove stronger results. For example, you can study the Farhadian's conjecture. If Farhadian's conjecture is proved, then the Firoozbakht's conjecture is also true.

user747431
  • 1