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A182514
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Primes prime(n) such that (prime(n+1)/prime(n))^n > n.
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6
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OFFSET
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1,1
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COMMENTS
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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
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
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REFERENCES
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Farhadian, R. (2017). On a New Inequality Related to Consecutive Primes. OECONOMICA, vol 13, pp. 236-242.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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) 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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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