2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This might be a first, but it actually looks like your question is better suited for Mathoverflow.net! :) $\endgroup$ – Deepak Feb 27 '16 at 12:52
  • $\begingroup$ I guess this answers that thought: mathoverflow.net/questions/232482/… $\endgroup$ – John Nicholson Mar 1 '16 at 13:29
  • 1
    $\begingroup$ 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. $\endgroup$ – Deepak Mar 1 '16 at 13:41
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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, $\frac{\psi_n }{p_n}<\frac{\log(n)}{\log(p_n)} <1$. $\endgroup$ – John Nicholson Jan 28 '18 at 3:05
  • $\begingroup$ $\psi_n < \frac{n\log^2(n)}{\log(p_n)}< \frac{p_n\log(n)}{\log(p_n)}$ ? $\endgroup$ – John Nicholson Feb 6 '18 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.