A267549 Primes prime(k) such that floor( (prime(k)/k)^2 ) <= prime(k+1) - prime(k).

3, 5, 7, 13, 23, 113

Offset

1,1

Comments

Prime index A000720 is: 2, 3, 4, 6, 9, and 30.

floor( (prime(k)/k)^2 ) is: 2, 2, 3, 4, 6, and 14.

Similarly, ceiling( (prime(k)/k)^2 ) > prime(k+1) - prime(k) holds for all prime(k) < 10^8 with the exception of prime(k) = 7. For prime(k) = 7, 4 = ceiling((prime(k) / k)^2) = prime(k+1) - prime(k).

Stronger than Firoozbakht's conjecture which states that the sequence prime(k)^(1/k) is strictly decreasing.

Conjecture: list is complete. If so, subsequence of A124147 and A174635.

Andrew Granville conjectures that lim sup (prime(n+1)-prime(n))/log(prime(n))^2 >= 2/e^gamma = 1.1229189.... If so (or at least if the lim sup is greater than 1) then this sequence is infinite. - Charles R Greathouse IV, Feb 18 2016

Links

Table of n, a(n) for n=1..6.

Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, arXiv preprint arXiv:1506.03042 [math.NT], 2015.

John Nicholson, A Conjecture Sharper than Cramér's and Firoozbakht's, Mathematics Stack Exchange.

Formula

Floor((A000040(k) / k)^2) <= A000040(k+1)-A000040(k), where k = A000720.

Floor(A001248(k) / A000290(k)) <= A001223(k), where k = A000720.

Example

For a(3) = 7, floor((7 / 4)^2) = 3 < 4 = 11 - 7. Note that all other a(n) use = instead of <.

Mathematica

Select[Prime@ Range[10^5], Floor[(#/PrimePi@ #)^2] <= NextPrime@ # - # &] (* Michael De Vlieger, Jan 21 2016 *)

Prog

(PARI) L=10^11; p=2; forprime(q=3, L, a=floor((p/primepi(p))^2.); if(a<=q-p, print1(p, ", ")); p=q)

Crossrefs

Cf. A001223, A001248, A124147.

Sequence in context: A126273 A007658 A275175 * A154321 A024724 A024946

Adjacent sequences: A267546 A267547 A267548 * A267550 A267551 A267552

Keyword

nonn,more

Author

John W. Nicholson, Jan 16 2016

Status

approved