A245396 Largest prime not exceeding prime(n)^(1 + 1/n).

3, 5, 7, 11, 17, 19, 23, 23, 31, 37, 41, 47, 53, 53, 59, 67, 73, 73, 83, 83, 89, 89, 97, 107, 113, 113, 113, 113, 127, 131, 139, 151, 157, 157, 167, 173, 179, 181, 181, 193, 199, 199, 211, 211, 211, 223, 233, 241, 251, 251, 257, 263, 263, 277, 283, 283, 293, 293, 293, 307, 307, 317, 331, 337

Offset: 1

Author

M. F. Hasler, Nov 03 2014

Keywords

• nonn

Comments

Firoozbakht's conjecture, prime(n+1) < prime(n)^(1 + 1/n), is equivalent to a(n) > prime(n). See also A182134.

Here prime(n) = A000040(n). The conjecture is also equivalent to a(n) - prime(n) >= A001223(n), the n-th gap between primes. See also A246778(n) = floor(prime(n)^(1 + 1/n)) - prime(n).

It is also conjectured that the equality a(n) - prime(n) = A001223(n) holds only for n in the set {1, 2, 3, 4, 8}, see A246782. a(n) is also largest prime less than prime(n)^(1 + 1/n), since prime(n)^(1 + 1/n) is never prime. - Farideh Firoozbakht, Nov 03 2014

a(n) = A007917(A249669(n)) = A244365(n,A182134(n)) = A006530(A245722(n)). - Reinhard Zumkeller, Nov 18 2014

Links

• 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
• Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT]
• Wikipedia, Firoozbakht's conjecture

Crossrefs

Cf. A000040, A001223, A007917, A182134, A246778, A249669.

Cf. A244365.

Programs

Haskell

a245396 n = a244365 n (a182134 n)  -- Reinhard Zumkeller, Nov 16 2014

Maple

seq(prevprime(ceil(ithprime(n)^(1+1/n))),n=1..100); # Robert Israel, Nov 03 2014

Mathematica

Table[NextPrime[Prime[n]^(1 + 1/n), -1], {n, 64}] (* Farideh Firoozbakht, Nov 03 2014 *)

PARI

a(n)=precprime(prime(n)^(1+1/n))

PARI

a(n)=precprime(sqrtnint(prime(n)^(n+1),n)) \\ Charles R Greathouse IV, Oct 29 2018

Formula

A245396 = A007917 o A249669, i.e., a(n) = A007917(A249669(n)). Although one could say "less than" in the definition of this sequence, one cannot use A151799 in this formula because for n = 2 and n = 4, one has a(n) = A249669(n).