Problems & Puzzles: Conjectures

Conjecture 30. The Firoozbakht Conjecture

Faride Firoozbakht, from the University of Isfahan, Iran, conjectures that: 

(pn)1/n decreases increasing n, in which pn is the nth prime number.

Firoozbakht, claims that she has "verified (his conjecture) with the table of "Maximal gaps between consecutive primes <=4.444 * 10^12" and that proving it "...This can also give simple proofs to many theorems related to Prime numbers..."  

Firoozbakht comments that she has been told by Huxley that her Conjecture is a hard one to prove but hasn't been posted before.

Question:

1) Of, course that I will ask you if you want or to prove or to disconfirm it?
2) Can you obtain new/interesting results about prime numbers, supposing true this conjecture? 


Boris Bukh, Luis Rodríguez and Said El Aidi have sent some developments over the Firoozbakht's conjectural statement showing, nearly the same way, that it is another form to express the Cramer unproven but well known conjecture (1934) pn+1 - pn = O((ln(pn))^2):

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Boris Bukh:

The Faride's conjecture can be written this way:

pn^1/n > pn+1 ^(1/n+1)

Then 

pn > pn+1 ^(n/(n+1))

pn+1 < pn ^((n+1)/n)

pn+1 < pn ^(1+1/n)

pn+1 < pn*pn^1/n

pn+1 < pn + pn*(pn^1/n -1)

dn = pn+1 - pn  < pn*(pn^1/n -1)

Now, the Taylor's expansion for a^x is:

a^x = 1 +x.lna/1! + ((x.lna)^2)/2! + ...

Then 

pn*(pn^1/n -1) = pn*(ln(pn)/n + ...)

pn*(pn^1/n -1) < pn*(ln(pn)/n)

dn < pn*(lnpn/n) = (pn/n)*ln(pn)

By the prime number theorem we know that n < pn/ln(pn), then ln(pn) < pn/n

dn < (ln(pn))^2 or dn = O((ln(pn))^2)

According to Bukh this last has been posed as a problem to solve, in "R. Graham, D. Knuth, O. Patashnik "Concrete Mathematics". Problem 4.69". 1994. In the "solutions" section to this problem has been written the following comment:

"Plausibility of this conjecture was shown by Cramer [166] from probabilistic arguments, and was confirmed by computational experiments. Brent [38] has shown that p_(n+1)-p_n=<602 for p_(n+1)<=2.86*10^12."

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Luis Rodríguez development goes like this:

Firoozbakht's conjecture means:

[p(n)]^1/n > [p(n+1)]^1/(n+1)

[p(n)]^(n+1) > [p(n+1)]^n

[p(n)]^(n+1) = p(n)*p(n)^n

p(n) > [p(n+1)/p(n)]^n

Dp = p(n+1)- p(n)

p(n+1)=p(n) + Dp

p(n) > [(p(n) + Dp)/p(n)]^n

p(n) > [1 + Dp/p(n)]^n

Log(p(n))/n > Log[ 1 + Dp/(p(n)]

Lim  Log[1 + Dp/(p(n)] = Dp/p(n), because Lim log(1+x) = x  when x -->0 

But always  Log (1 + x )  > x , and the inequality remains:

Log(p(n))/n > Dp/p(n)    ------>    Dp  <  p(n)Log(p(n)) / n

According to Dusart(1998),  n  >  p(n) / [ Log (p(n)) – 1 ] , replacing  n in prevoious formula:

Dp < Log(p(n))*(Log(p(n)) - 1)

This represents is an improvement over Cramer because Cramer stated that:

Dp< [Log(p(n))]^2

Then the  Firoozbakht's conjecture means:

Dp < Log(p(n))*[log(p(n)) - 1]

This inequality passes the Nyman's difference(1999):

Dp (for p(n)=1693182318746371) = 1132, the closest test to Cramer's conjecture.

 

Conclusions:

1.- Having the Firoozbakht' conjecture the same characteristic than the Cramer's one, there is few hope to be solved, as the Cramer's one. The best approach to this issue is for now too coarse: Dp < [p*Log(p))^(1/2), by Cramer himself. See Ribenboim pag. 253.

2. Being the  Firoozbakht's conjecture tighter than the Cramer's one, the first should be applied in every case in which the second is applied but with better (smaller) results, by example in the Fortune's conjecture.

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Said El Aidi wrote:

Suppose true the Firoozbakht's conjecture, then

(pn+1)1/n+1 < (pn)1/n     for all n

hence

pn+1 < (pn)n+1/n = pn (pn)1/n = pn exp(log (pn)/n)           (1)

where log is the natural logarithm (base e), and exp is the exponential function.

Thanks to the prime number theorem, we have

log (pn)/n < 1  for all n 1

On the other hand, if 0 < x < 1,

                                                           exp(x) < 1 + Cx,                                 (2)

for a convenient positive constant C.

From (1) and (2) it follows that

pn+1 < pn( 1 + C(log( pn) /n) )

and so

                                                   pn+1 - pn < C pn log( pn) /n                           (3)

RS showed in 1962 that

π(x)> x/log(x)  for all x > 10

where π(x) = the number of prime not exceeding x.

For n > 5, we put x =
pn and have

n = π( pn ) > pn /log( pn),        pn /n < log( pn) ,         pn log( pn )/n < (log( pn))²

hence, by (3), we obtain

pn+1 - pn < C( log( pn))²

which is The Cramer's Conjecture.

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Felice Russo has sent (25/3/2002) the following argument in favor the Faride's Conjecture:

The Faride's conjecture is equal to:

(pn+1)1/n+1 < (pn)1/n

Then

n*ln(pn+1) < (n+1)*ln(pn)

But according to the Prime Theorem: pn = n*ln(n) as n goes to infinite, then 

n*ln((n+1)*ln(n+1)) < (n+1)*ln(n*ln(n))

n*(ln(n+1) + lnln(n+1)) < (n+1)*(ln(n) + lnln(n))

[ln(n+1) + lnln(n+1)]/[ln(n) + lnln(n)] < (n+1)/n

But this "asymptotic" inequality is true because the function ln(n)+ln(ln(n)) increases slower than the n one

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Luis Rodríguez wrote (26/4/2002):

If Russo's "asymptotic inequality were true, then we would have a demonstration of Faride's Conjecture and also Cramer's.

The difficulty arises from the equation introduced by Felice: Lim p(n) = n Log(n). In reality is: Lim p(n)/nLog(n)= 1

Only when Lim f1(x) - f2(x) = 0 ,it is permited the replacement of f1(x) by f2(x), into an inequality.

Following a theorem of Dusart:

p(n) > n Log(n) + n(LogLog(n) - 1)

We cannot replace p(n) in Faride's inequelity

n Log(p(n+1)) < (n+1) Log(p(n))

by the lesser quantity: n Log(n), because the inequality can be put out of balance. Chiefly because this inequality is tightly adjusted.

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On June 10, 2015, Alexei Kourbatov wrote:

Dear Carlos and Luis,
The following arXiv preprint may be of interest to you -
a new proof that sharpens the prime gap bounds for Conjecture 30:
Upper bounds for prime gaps related to Firoozbakht's conjecture

Abstract:
We study two upper bounds for the prime gap g = p(k+1) − p(k) after the k-th prime p(k):
(A) p(k+1) < (p(k))^(1+1/k) and 
(B) p(k+1) − p(k) < log^2 p(k) − log p(k) − b  for k > 9. 
The upper bound (A) is equivalent to Firoozbakht’s conjecture. 
We prove that (A) implies (B) with b = 1; 
on the other hand, (B) with b = 1.17 implies (A).

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On February 27, 2017, Reza Farhadian wrote:

I think that conjecture 7 is true. Because we know that the Firoozbakht's conjecture (or conjecture 30) is stronger than conjecture 7.  On the other hand I proved that conjecture 30 is true as n goes to infinity (see http://vixra.org/abs/1702.0318). Therefore conjecture 7 must be true. 

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On March 15, 2017 Farideh's husband, Mr. Jahangeer Kholdi, wrote an email to inform about two papers from Ahmad Sabihi:

a) On the "On the Firoozbakht's Conjecture"

b) "On solutions of some of unsolved problems in number theory, specifically on the distribution of primes"

 See both papers in the links provided.

Regarding the first one:

"This paper proves Firoozbakht's conjecture using Rosser and Schoenfelds' inequality on the distribution of primes."

Regarding the second paper:

"We solve some famous conjectures on the distribution of primes. These conjectures are to be listed as Legendre's, Andrica's, Oppermann's, Brocard's, Cramer's, Shanks', and five Smarandache's conjectures. We make use of both Firoozbakht's conjecture (which recently proved by the author) and Kourbatov's theorem on the distribution of and gaps between consecutive primes. These latter conjecture and theorem play an essential role in our methods for proving these famous conjectures. In order to prove Shanks' conjecture, we make use of Panaitopol's asymptotic formula for pi(x) as well."

In his email, Mr. Jahangeer Kholdi adds:

"Moreover, I would like to let you know that, I think with the above results, Riemann's hypothesis
has also is proven by Jan Feliksiak in his book The symphony of primes, distributions of primes, and Riemannn's hypothesis . Xlibris, (2013)."

 

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