A249669 - OEIS

A249669

a(n) = floor(prime(n)^(1+1/n)).

4, 5, 8, 11, 17, 19, 25, 27, 32, 40, 42, 49, 54, 56, 60, 67, 74, 76, 83, 87, 89, 96, 100, 107, 116, 120, 122, 126, 128, 132, 148, 152, 159, 160, 171, 173, 179, 186, 190, 196, 203, 204, 215, 217, 221, 223, 236, 249, 253, 255, 259, 265, 267, 278, 284, 290, 296, 298, 304, 308, 310, 321

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Firoozbakht's conjecture (prime(n)^(1/n) is a decreasing function), is equivalent to say that prime(n+1) <= a(n). (One has equality for n=2 and n=4.) See also A182134 and A245396.

This is not A059921 o A000040, i.e., a(n) != A059921(prime(n)), since the base is prime(n) but the exponent is n.

A245396(n) = A007917(a(n)). - Reinhard Zumkeller, Nov 16 2014

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

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, arXiv:1506.03042 [math.NT], 2015 and J. Int. Seq. 18 (2015) 15.11.2

Wikipedia, Firoozbakht's conjecture

FORMULA

a(n) = prime(n) + (log(prime(n)))^2 - log(prime(n)) + O(1), see arXiv:1506.03042, Theorem 5. - Alexei Kourbatov, Nov 26 2015

MAPLE

seq(floor(ithprime(n)^(1+1/n)), n=1..100); # Robert Israel, Nov 26 2015

PROG