

A249669


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


8



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
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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.


LINKS



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

(PARI) a(n)=prime(n)^(1+1/n)\1
(Magma) [Floor(NthPrime(n)^(1+1/n)): n in [1..70]]; // Vincenzo Librandi, Nov 04 2014
(Haskell)
a249669 n = floor $ fromIntegral (a000040 n) ** (1 + recip (fromIntegral n))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



