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 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? 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) 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:
*** 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 hence where log is the natural logarithm (base e), and exp is
the exponential function. On the other hand, if 0 < x < 1, exp(x) < 1 + Cx, (2) for a convenient positive constant C. and so RS showed in 1962 that π(x)> x/log(x) for all x > 10 where hence, by (3), we obtain 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
}< Then n*ln(p_{n}_{+1})^{
}< 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*(ln(n+1)
+ lnln(n+1))^{
}< [ln(n+1)
+ lnln(n+1)]/[ln(n)
+ lnln(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):
***





