Problems & Puzzles:
Conjectures
Conjecture
30. The
Firoozbakht
Conjecture
Faride Firoozbakht,
from the University of Isfahan, Iran, conjectures that:
(p_{n})^{1/n}
decreases increasing n, in which p_{n} 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)
p_{n+1}  p_{n} =
O((ln(p_{n}))^2):
***
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."
***
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.
***
Said El Aidi wrote:
Suppose true the Firoozbakht's
conjecture, then
(p_{n}_{+1})^{1/n}^{+1
}<
(p_{n})^{1/n}
for all n
hence
p_{n}_{+1
}<
(p_{n})^{n+1}^{/n}
=
p_{n}
(p_{n})^{1/n}
=
p_{n}
exp(log
(p_{n})/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
(p_{n})/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
p_{n}_{+1
}<
p_{n}(
1 + C(log(
p_{n})
/n) )
and so
p_{n}_{+1
}
p_{n}
< C
p_{n}
log(
p_{n})
/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 =
p_{n}_{
}and have
n =
π(
p_{n}
) >
p_{n}
/log(
p_{n}),
p_{n}
/n <
log(
p_{n})
,
p_{n}
log(
p_{n}
)/n < (log(
p_{n}))²
hence, by (3), we obtain
p_{n}_{+1
}
p_{n}
< C(
log(
p_{n}))²
which is The
Cramer's Conjecture.
***
Felice Russo has sent (25/3/2002) the following argument in
favor the Faride's Conjecture:
The Faride's
conjecture is equal to:
(p_{n}_{+1})^{1/n}^{+1
}<
(p_{n})^{1/n}
Then
n*ln(p_{n}_{+1})^{
}<
(n+1)*ln(p_{n})
But according to the Prime
Theorem: p_{n}
= 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
***
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.
***
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 kth 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).
***
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.
***
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)."
***
On April 3,
2017, Reza Farhadian wrote:
Recently, in 2017, Rafael Jakimczuk showed that for almost all pairs of
consecutive primes p_(n+1) and p_n, the inequality log(p_(n+1) )log(p_n
)<log(p_n )/n is hold. Now if we take exponential of this inequality,
then we have
p_(n+1)<(p_n)^((n+1)/n)
That this is the same Firoozbakht’s conjecture. Hence, the Firoozbakht’s
conjecture is true for almost all pairs of consecutive primes. For
Jakimczuk’s paper, see
Rafael Jakimczuk ,The Difference between log pn+1 and log pn,
International Mathematical Forum, Vol. 12, 2017, no. 6, 293 – 302. See
http://www.mhikari.com/imf/imf2017/582017/718.html
***
On Dec 25, 2017 Reza Farhadian wrote:
"Already (Refer to previous post at the date February
27, 2017), I
thought I was able to prove the Firoozbakht's conjecture by my
article at the link http://vixra.org/abs/1702.0318. But
recently I realized that this is not true. My article was published
without arbitration, so it was not valid, hence I removed the
article from the link."
***
