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Let $p_n$ be the $n$th prime, let $L$ consist of the primes satisfying $p_{n+2} - 2p_{n+1} + p_{n} > 0$, and let $Q$ consist of the primes satisfying $p_{n+1}^2 < p_{n}p_{n+2}.$ Is $L=Q$?

Background: The arithmetic-geometric mean inequality implies $Q \subset L$, but the reverse containment, if true, may be harder to prove.

The inequalities have been verified equivalent by computer for the first $4,000,000$ primes (i.e., up to $67,867,967$). The number of these primes for which the inequalities hold is $1,941,180$, which is more than $48.5$ percent of the total. Further experimentation leads me to guess that a limiting ratio exists.

It's been checked that for $1<n<4,000,000$, when the two inequalities hold we also have $$p_{n+1}^2 - p_{n}p_{n+2} > p_{n+2} - 2p_{n+1} + p_{n}.$$

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    $\begingroup$ If you write $p_{n+2}=p_{n+1}+a$ and $p_{n+2}=p_{n+1}-b$, the product is $p_{n+1}^2+(a-b)p_{n+1}-ab$. For $L$ to succeed, but $Q$ to fail, you need $a>b$ hence $a-b\ge 1$ and so $ab\ge p_{n+1}$. This is a question about gaps between primes. Terry Tao's blog terrytao.wordpress.com/2014/08/21/… indicates that $a,b\ll p_{n+1}^{.525}$, but that this is considered a very weak upper bound. For a violation of your inequality, you would need 2 consecutive gaps almost as large as the maximum. This can probably be ruled out by some analytic # theory. $\endgroup$
    – Anthony Quas
    Commented May 21, 2015 at 20:55
  • $\begingroup$ There are least two cases of $p_{n+1} < a b$, namely $p_{4} = 7 < 4 \times 2$ and $p_{9} = 23 < 6 \times 4$. Of course these are not counterexamples to the OP: $p_4^2 - p_3 p_5 = -6$ and $p_9^2 - p_8 p_{10} = -22$. $\endgroup$
    – Robert Israel
    Commented May 21, 2015 at 21:38
  • $\begingroup$ I should have said $a-b\ge 2$, so that $ab\ge 2p_{n+1}$. $\endgroup$
    – Anthony Quas
    Commented May 21, 2015 at 22:18
  • $\begingroup$ It is believed that there are only finitely many gaps g (with adjacent prime p) such that g^2 > p, and the largest of these has p=113. My vote is that L is the same as Q, based on Anthony's analysis and Robert's results. $\endgroup$
    – The Masked Avenger
    Commented May 21, 2015 at 22:23

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I am using Anthony Quas's reformulation to restate the problem. Letting the $n$th prime gap be given by $d_n= p_{n+1} - p_n$, and given $n$ we relabel $a=d_n$ and $b=d_{n+1}$, we look at $L$ as the set of $p_{n+1}$ in which $n$ satisfies $b-a\gt 0$ and we look at $Q$ as the set of $p_{n+1}$ in which $0 \lt (b-a)p_{n+1} -ab$. As noted in the post, $p_{n+1}$ not in $L$ readily implies $p_{n+1}$ not in $Q$; it is natural to ask if $p_{n+1} \in L$ implies $p_{n+1} \in Q$. The question further notes that when $p_{n+1}$ is observed to be in $L$ one also has $ ab - p_{n+1}(b-a) \gt (b-a)$, which I think should be reversed as $7=p_4 \in L$ but $49 - 55 \lt 11 - 14 + 5$.

If the last inequality is reversed, it says $ab \lt ( p_{n+1} + 1)(b-a)$. If $p_{n+1} \in Q$, then clearly this last inequality holds. Finally as Anthony Quas observes, if $p_{n+1} \in L\setminus Q$ then $ab \geq (b-a)p_{n+1}$ and $b \gt a$, so one would have $ab \geq 2p_{n+1}$ if $n \gt 1$.

The formulation shows that the basic question is about consecutive prime gaps, and that $L$ is different from $Q$ only when a large gap $d_n$ is greater than the square root of an adjacent prime $p_n$. Such large gaps have not been observed for $n \gt 30$ (so $p_n \gt 113$), and the stronger inequality $ab \geq 2p_{n+1}$ is also not observed for $1 \lt n \leq 30$. The case $n=1$ is left to the reader, as is the conclusion that $L$ properly containing $Q$ would violate expectations and many conjectures in prime number theory.

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  • $\begingroup$ Indeed all known $n$ for which $d_nd_n \geq p_n$ occurs are $n=30,11,9,6,4$ and $2$, so there are very few opportunities for something like $ab \geq p_{n+1}$ to happen. $\endgroup$
    – The Masked Avenger
    Commented May 22, 2015 at 3:11

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