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? 


Solution

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):

***

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

(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))˛

hence, by (3), we obtain

pn+1 - pn < C( log( pn))˛

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:

(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

***

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.

***

 


Records   |  Conjectures  |  Problems  |  Puzzles