A144104 Primes p such that log(nextPrime(p))/log(p) is smaller for larger primes.

2, 3, 7, 13, 23, 31, 47, 53, 113, 139, 199, 211, 293, 317, 523, 1327, 1669, 1951, 2179, 2477, 2971, 3271, 4297, 4831, 5591, 5749, 5953, 6491, 6917, 7253, 8467, 9551, 9973, 10799, 11743, 12163, 12853, 15683, 16141, 19609, 31397, 34061, 35617, 35677

Offset: 1

Author

T. D. Noe, Sep 11 2008

Keywords

• nonn

Comments

log(nextPrime(p))/log(p) is another measure of the (relative) gap between consecutive primes. See A144105 for the primes at the upper end of the gaps.

The statement log(prime(k+1))/log(prime(k)) < 1 + 1/k, for k >= 1, is a rewrite of the Firoozbakht conjecture. - John W. Nicholson, Dec 06 2013

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

prime(n+1) < prime(n)^(1+1/n), which can be rewritten as: (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n. This suggests a weaker conjecture: (log(prime(n+1))/log(prime(n)))^n < e. - Daniel Forgues, Apr 28 2014

Examples

			Examples for (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n < e:
(log(3)/log(2))^1 = 1.58... < (1+1/1)^1 = 2 < e;
(log(1361)/log(1327))^217 = 2.14... < (1+1/217)^217 = 2.712... < e;
(log(8501)/log(8467))^1059 = 1.59... < (1+1/1059)^1059 = 2.716... < e;
(log(35729)/log(35677))^3795 = 1.69... < (1+1/3795)^3795 = 2.717... < e. - _Daniel Forgues_, Apr 28 2014
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..176
• A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
• Wikipedia, Firoozbakht's conjecture

Programs

Mathematica

nn=10^5; ps=N[Log[Prime[Range[nn]]]]; ps=Rest[ps]/Most[ps]; k=1; t={}; While[k