TITLE: Infinitely many $k$ such that $[a_k,a_{k+1}]>ck^2$
QUESTION [14 upvotes]: Let $a_n\in \mathbf{N}$ be an infinite sequence such that $\forall i\neq j, a_i\neq a_j$.
I have the following theorem:

For $0<c<\frac{3}{2}$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck$, where $[\cdots]$ denotes least common multiple.

Idea of proof: By contradiction.  Suppose there is an $M$ such that for all $k>M$, $[a_k,a_{k+1}]\leq ck$.
On the one hand,
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{ck}$
On the other hand,
$\displaystyle \frac{1}{[a_k,a_{k+1}]}$
$\displaystyle =\frac{(a_k,a_{k+1})}{a_ka_{k+1}}$
$\displaystyle =(a_k,a_{k+1})\frac{1}{a_k+a_{k+1}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle =\frac{1}{\frac{a_k}{(a_k,a_{k+1})}+\frac{a_{k+1}}{(a_k,a_{k+1})}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \leq \frac{1}{3}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\leq \frac{1}{3}\sum_{k=1}^{n}(\frac{1}{a_k}+\frac{1}{a_{k+1}})\leq \frac{2}{3}\sum_{k=1}^{n}\frac{1}{k}$
And I make the following conjecture:

$\exists c>0$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck^2$.

But I cannot prove or disprove it.

REPLY [10 votes]: In this paper
P. Erdős, R. Freud, and N. Hegyvári, Arithmetical properties of permutations of integers, Acta Mathematica Hungarica 41:1-2 (1983), pp 169-176.
http://renyi.mta.hu/~p_erdos/1983-02.pdf
the authors show that for permutations of {1,2,...,n} the answer is $\Theta(n^2/\log n)$.
See https://oeis.org/A064764 .
Also for infinite permutations they give an example where
${\rm lcm}(a_i, a_{i+1}) < i \exp (c \sqrt{\log i} \log\log i)$ for all $i$.
And I see that 
Chen, Y.-G.; Ji, C.-S.
The permutation of integers with small least common multiple of two subsequent terms. 
Acta Math. Hungar. 132 (2011), no. 4, 307–309 has improved on this still further, to  $i \exp (c' \sqrt{\log i\log\log i})$.