# A Conjecture Sharper than Cramér's and Firoozbakht's

Notation: $\lfloor\cdot\rfloor$ is floor function; and $\pi(x)$ is the prime-counting function up to $x$. $g_k := p_{k+1} - p_k$ .

OEIS sequence A267549 is "Primes prime(k) such that floor( (prime(k)/k)^2 ) <= prime(k+1) - prime(k)" https://oeis.org/A267549, or stated differently if $p_k \neq 3, 5, 7, 13, 23, 113$ and $2 \le p_k < 10^{s}$, with $s$ currently at $12$, then $g_k < \lfloor\left(\frac{p_k}{k}\right)^2\rfloor$ holds. Define $$a_k := \left(\frac{p_k}{k}\right)^2.$$ It is conjectured that $$a_k > g_k, \text{ for all }p_k > 113.\qquad \qquad(1)$$

Pierre Dusart proved in 2010:

$$\frac{x}{\log{(x)} - 1} < \pi(x) \text{ for }x \ge 5393.\qquad \qquad(2)$$

By rearranging, taking $x = p_k$, and noting $\pi(p_k) = k$ we get

$$\frac{p_k}{k} < \log{(p_k)} - 1 < \log{(p_k)} \text{ for }p_k \ge 5393.\qquad \qquad(3)$$

Clearly, the square of (3) is $$\left(\frac{p_k}{k}\right)^2 < (\log{(p_k)} - 1)^2 < (\log{(p_k)})^2\text{ for }p_k \ge 5393.\qquad \qquad(4)$$ One can also take (3) multiply by $\frac{p_k}{k}$ and have $$\left(\frac{p_k}{k}\right)^2 < \frac{p_k}{k}(\log{(p_k)} - 1) < \frac{p_k}{k}(\log{(p_k)}).\qquad \qquad(5)$$

Which is to say (1) is a stronger upper bound than both Cramér's conjecture, right-side of (4), and Firoozbakht's conjecture, right-side of (5) is the same as the right-side of (*).

Note for the 64-th record maximal gap $$p_k = 1693182318746371, k = 49749629143526, g_k = p_{k+1} - p_k = 1132$$

$$a_k = (1693182318746371/49749629143526)^2 = 1158.3178633021$$

$$a_k - g_k = 1158.3178633021 - 1132 = 26.3178633021$$

$$\frac{g_k}{a_k} = 1132 / 1158.3178633021 = 0.977279239 < 1.$$ One can deduce that the gaps get arbitrarily smaller in proportion to the primes: the quotient $$\lim_{k\to\infty}\frac{g_k}{p_k}=0.\qquad \qquad(6)$$ In 1931, E. Westzynthius proved that prime gaps grow more than logarithmically. That is, $$\limsup_{k\to\infty}\frac{g_k}{\log p_k}=\infty.\qquad \qquad(7)$$

We expect that there is constant $C$ such that, $$\limsup_{k\to\infty}\frac{g_k}{C (\log p_k)^2}=1.\qquad \qquad(8)$$

This, (6) and (7), explains why Charles R Greathouse IV wrote the commented on OEIS sequence A267549:

$\qquad$"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."

So, if there is constant $G$ such that:

$$\limsup_{k\to\infty}\frac{g_k}{G a_k} = 1$$ then: If $G < 1$ then (1) holds. If $G > 1$ then (1) fails. As for $G = 1$, a bound above must be found so that $\frac{g_k}{a_k} < 1$ holds.

But we do not know $G$ value or hints of it past $4*10^{18}$. Or do we? What information have I left off? What is needed to prove $113$ is the largest counterexample of (1) or the value of the counterexample?

(*) See Section 4, page 4, (12) of arXiv:1506.03042 (Journal of Integer Sequences, 18, 2015, article 15.11.2, Alexei Kourbatov) Upper bounds for prime gaps related to Firoozbakht's conjecture.

• This might be a first, but it actually looks like your question is better suited for Mathoverflow.net! :) – Deepak Feb 27 '16 at 12:52
• I guess this answers that thought: mathoverflow.net/questions/232482/… – John Nicholson Mar 1 '16 at 13:29
• Oh dear. You might have done better to make the full post on MO rather than link to the post here. But I hope you get a good answer to your question either way. – Deepak Mar 1 '16 at 13:41

In 2016, R. Farhadian conjectured that if $p_n$ be the $n$-th prime and $\psi_n= \left(\frac{p_{n+1}}{p_n}\right)^n$, then

$$p_n^{\psi_n} < n^{p_n}, \quad \text{ for all } n>4$$

The Farhadian's conjecture is more stronger than some other conjectures as Cramer's, Firoozbakht's, Granville's, Nicholson's conjectures [1, 2].

Furthermore, we know that the sequence $\psi_n$ is directly related to the Firoozbakht's and Nicholson's conjectures, since the Firoozbakht's conjecture states that $$\psi_n < p_n, \quad \text{ for all } n\geq 1$$ and the Nicholson's conjecture states that $$\psi_n < n\log(n), \quad \text{ for all } n>4$$ Therefore, we believe that studying the sequence $\psi_n$ is useful for examining these conjectures.

[1] C. Rivera, ed., Conjecture 78: p(n)^(p(n+1)/p(n))^n<=n^p(n), 2016, Available at http://www.primepuzzles.net/conjectures/.

[2] R. Farhadian, R. Jakimczuk, On a New Conjecture of Prime Numbers, 12 (2017), 559 - 564.

• While interesting, I must note that this answer is for another conjecture of mine. That being said, taking the ln and rearranging: $\frac{\psi_n }{p_n}<\frac{\log(n)}{\log(p_n)}$, for all n<4 –The Firoozbakht's conjecture as stated has the constant 1 on the right-side. Clearly, $\frac{\psi_n }{p_n}<\frac{\log(n)}{\log(p_n)} <1$. – John Nicholson Jan 28 '18 at 3:05
• $\psi_n < \frac{n\log^2(n)}{\log(p_n)}< \frac{p_n\log(n)}{\log(p_n)}$ ? – John Nicholson Feb 6 '18 at 0:15