A182514 Primes prime(n) such that (prime(n+1)/prime(n))^n > n.

2, 3, 7, 113, 1327, 1693182318746371

Offset: 1

Author

Thomas Ordowski, May 04 2012

Keywords

• nonn

Comments

The Firoozbakht conjecture: (prime(n+1))^(1/(n+1)) < prime(n)^(1/n), or prime(n+1) < prime(n)^(1+1/n), prime(n+1)/prime(n) < prime(n)^(1/n), (prime(n+1)/prime(n))^n < prime(n).

Using the Mathematica program shown below, I have found no further terms below 2^27. I conjecture that this sequence is finite and that the terms stated are the only members. - Robert G. Wilson v, May 06 2012 [Warning: this conjecture may be false! - N. J. A. Sloane, Apr 25 2014]

I conjecture the contrary: the sequence is infinite. Note that 10^13 < a(6) <= 1693182318746371. - Charles R Greathouse IV, May 14 2012

[Stronger than Firoozbakht] conjecture: All (prime(n+1)/prime(n))^n values, with n >= 5, are less than n*log(n). - John W. Nicholson, Dec 02 2013, Oct 19 2016

The Firoozbakht conjecture can be rewritten as (log(prime(n+1)) / log(prime(n)))^n < (1+1/n)^n. This suggests the [weaker than Firoozbakht] conjecture: (log(prime(n+1))/log(prime(n)))^n < e. - Daniel Forgues, Apr 26 2014

All a(n) <= a(6) are in A002386, A205827, and A111870.

The inequality in the definition is equivalent to the inequality prime(n+1)-prime(n) > log(n)*log(prime(n)) for sufficiently large n. - Thomas Ordowski, Mar 16 2015

Prime indices, A000720(a(n)) = 1, 2, 4, 30, 217, 49749629143526. - John W. Nicholson, Oct 25 2016

Examples

			7 is in the list because, being the 4th prime, and 11 the fifth prime, we verify that (11/7)^4 = 6.09787588507... which is greater than 4.
11 is not on the list because (13/11)^5 = 2.30543740804... and that is less than 5.
		

References

• Farhadian, R. (2017). On a New Inequality Related to Consecutive Primes. OECONOMICA, vol 13, pp. 236-242.

Links

• Reza Farhadian, A New Conjecture On the primes, Preprint, 2016.
• R. Farhadian, and R. Jakimczuk, On a New Conjecture of Prime Numbers Int. Math. Forum, vol. 12, 2017, pp. 559-564.
• Luan Alberto Ferreira, Some consequences of the Firoozbakht's conjecture, arXiv:1604.03496v2 [math.NT], 2016.
• Luan Alberto Ferreira, Hugo Luiz Mariano, Prime gaps and the Firoozbakht Conjecture, São Paulo J. Math. Sci. (2018), 1-11.
• A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
• A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2.
• Carlos Rivera, Conjecture 78. P_n^((P_n+1/P_n)^n) <= n^P_n, 2016.
• Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT], 2010.
• Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.
• Wikipedia, Firoozbakht’s conjecture

Crossrefs

Cf. A111870.

Programs

Mathematica

Prime[Select[Range[1000], (Prime[# + 1]/Prime[#])^# > # &]] (* Alonso del Arte, May 04 2012 *)
firoozQ[n_, p_, q_] := n * Log[q] > Log[n] + n * Log[p]; k = 1; p = 2; q = 3; While[ k < 2^27, If[ firoozQ[k, p, q], Print[{k, p}]]; k++; p = q; q = NextPrime@ q] (* Robert G. Wilson v, May 06 2012 *)

PARI

n=1;p=2;forprime(q=3,1e6,if((q/p*1.)^n++>n, print1(p", "));p=q) \\ Charles R Greathouse IV, May 14 2012

PARI

for(n=1,75,if((A000101[n]/A002386[n]*1.)^A005669[n]>=A005669[n], print1(A002386[n],", "))) \\ Each sequence is read in as a vector as to overcome PARI's primelimit \\ John W. Nicholson, Dec 01 2013

PARI

q=3;n=2; forprime(p=5, 10^9,result=(p/q)^n/(n*log(n));if(result>1, print(q," ",p, " ", n, " ", result));n++;q=p) \\ for stronger than Firoozbakht conjecture \\ John W. Nicholson, Mar 16 2015, Oct 19 2016

Extensions

a(6) from John W. Nicholson, Dec 01 2013