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+1 − pk
< ln²pk − ln pk
for k > 4;
(C)
pk+1 − pk
< ln²pk − ln pk − 1
for k > 9.
This is why we might say that Firoozbakht conjecture is stronger than Cramér's conjecture
pk+1 − pk = 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+1 − pk
< ln²pk − ln pk
for k > 4.
Proof.
Let g = pk+1 − pk.
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 = pk,
x + 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+1 − pk
< ln²pk − ln pk
for all primes pk ≥ 11,
i.e. for k > 4.
See also:
• Firoozbakht conjecture
• Verification for primes up to one million (106)
• Verification for primes up to four quintillion (4×1018)
• Doing math with JavaScript
|