### Firoozbakht Conjecture vs. Cramer Conjecture

Doing Math with JavaScript | Firoozbakht conjecture

The Firoozbakht conjecture implies very tight upper bounds for prime gaps:
(A)     pk+1  <  (pk)1+1/k   (an equivalent form of the conjecture);
(B)     pk+1pk  <  ln²pk − ln pk     for k > 4;
(C)     pk+1pk  <  ln²pk − ln pk − 1     for k > 9.
This is why we might say that Firoozbakht conjecture is stronger than Cramér's conjecture pk+1pk = O(ln²pk).

Inequalities (B), (C) follow from (A). Inequality (B) was first stated in 2002 by Luis Rodriguez on the PrimePuzzles website, and later mentioned (with a slightly different right-hand side, ln²pk − ln pk + 1) by N.K.Sinha (2010, arXiv:1010.1399) and Zhi-Wei Sun (2012-2013, arXiv:1207.7059). Below we give a short proof that (A) ⇒ (B).

For a proof that (A) ⇒ (C), see arXiv:1506.03042.

Theorem:   If pk is the k-th prime and pk+1 < (pk)1+1/k, then

pk+1pk  <  ln²pk − ln pk     for k > 4.

Proof.   Let g = pk+1pk. We have the inequalities:
(1)     x + ln²x ln x − 1  <  x ln x (1 + 1 ln x + 2 ln²x)     for  x ≥ 8676.
(2)     x ln x (1 + 1 ln x + 2 ln²x)  ≤  π(x)     for  x ≥ 88783
(Theorem 6.9, arXiv:1002.0442); π(x) is the prime-counting function.
(3)     π(x)  <  ln x ln(x + g) − ln x     for prime x
(Firoozbakht conjecture (A);  x = pkx + g = pk+1,  π(x) = k).
(4)     g x + g  <  ln(x + g) − ln x     for any positive x and g.

Inequalities (1), (2), (3) form a chain, so the left-hand side of (1) is less than the right-hand side of (3):
(5)     x + ln²x ln x − 1  <  ln x ln(x + g) − ln x     for  x = pk ≥ 88783.

Equivalently,
(6)     (ln(x + g) − ln x)(x + ln²x)  <  ln²x − ln x.

Using (4) we replace the left-hand side of (6) by a smaller quantity:
(7)     g x + g (x + ln²x)  <  ln²x − ln x.

Rearranging (7) we indeed find that g < ln²x − ln x for large prime x = pk:

g (x + ln²x)  <  (x + g)(ln²x − ln x)
g (x + ln²x − ln²x + ln x)  <  x(ln²x − ln x)
g  <  x x + ln x (ln²x − ln x)  <  ln²x − ln x.
The above is based on the assumption that x = pk ≥ 88783; see (2). Separately, we verify by direct computation that the theorem is true for 11 ≤ pk ≤ 88783 (but false for pk = 7). This completes the proof; thus, if the Firoozbakht conjecture is true, we have
g  =  pk+1pk  <  ln²pk − ln pk     for all primes pk ≥ 11,   i.e. for k > 4.