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Largest prime not exceeding prime(n)^(1 + 1/n).
6

%I #57 Oct 29 2018 05:39:44

%S 3,5,7,11,17,19,23,23,31,37,41,47,53,53,59,67,73,73,83,83,89,89,97,

%T 107,113,113,113,113,127,131,139,151,157,157,167,173,179,181,181,193,

%U 199,199,211,211,211,223,233,241,251,251,257,263,263,277,283,283,293,293,293,307,307,317,331,337

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

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

%C 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).

%C 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

%C a(n) = A007917(A249669(n)) = A244365(n,A182134(n)) = A006530(A245722(n)). - _Reinhard Zumkeller_, Nov 18 2014

%H A. Kourbatov, <a href="http://arxiv.org/abs/1503.01744">Verification of the Firoozbakht conjecture for primes up to four quintillion</a>, arXiv:1503.01744 [math.NT], 2015

%H A. Kourbatov, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kourbatov/kourb7.html">Upper bounds for prime gaps related to Firoozbakht's conjecture</a>, J. Int. Seq. 18 (2015) 15.11.2

%H Nilotpal Kanti Sinha, <a href="http://arxiv.org/abs/1010.1399">On a new property of primes that leads to a generalization of Cramer's conjecture</a>, arXiv:1010.1399 [math.NT]

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Firoozbakht%E2%80%99s_conjecture">Firoozbakht's conjecture</a>

%F 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).

%p seq(prevprime(ceil(ithprime(n)^(1+1/n))),n=1..100); # _Robert Israel_, Nov 03 2014

%t Table[NextPrime[Prime[n]^(1 + 1/n), -1], {n, 64}] (* _Farideh Firoozbakht_, Nov 03 2014 *)

%o (PARI) a(n)=precprime(prime(n)^(1+1/n))

%o (PARI) a(n)=precprime(sqrtnint(prime(n)^(n+1),n)) \\ _Charles R Greathouse IV_, Oct 29 2018

%o (Haskell)

%o a245396 n = a244365 n (a182134 n) -- _Reinhard Zumkeller_, Nov 16 2014

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

%Y Cf. A244365.

%K nonn

%O 1,1

%A _M. F. Hasler_, Nov 03 2014