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.



Infinitely many $k$ such that $[a_k,a_{k+1}]>ck^2$
In fact there are sequences $\{a\_k\}$ of pairwise distinct positive integers
such that $[a\_k, a\_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$.


We first exhibit a sequence with
$[a\_k, a\_{k+1}] \ll k^{3/2} \log^3 k$ for all $k>1$, which already disproves
the conjecture that $[a\_k, a\_{k+1}] \gg k^2$ infinitely often.
The sequence will consist of all numbers of the form $p\_m p\_n$
with $m$ odd and $n$ even, listed in the following order:
$$
\begin{array}{cccccccc}
17\cdot 3 & \rightarrow & 17\cdot 7 & \rightarrow & 17\cdot 13 & \rightarrow & 17\cdot 19 &
\cr
\uparrow & & & & & & \downarrow &
\cr
11\cdot 3 & \leftarrow & 11\cdot 7 & \leftarrow & 11\cdot 13 & & 11\cdot 19 &
\cr
& & & & \uparrow & & \downarrow &
\cr
5\cdot 3 & \rightarrow & 5\cdot 7 & & 5\cdot 13 & & 5\cdot 19 &
\cr
\uparrow & & \downarrow & & \uparrow & & \downarrow &
\cr
2\cdot 3 & & 2\cdot 7 & \rightarrow & 2\cdot 13 & & 2\cdot 19 & \rightarrow
\end{array}
$$
Then each $a\_k$ is $p\_m p\_n$ with $m,n \ll \sqrt k$, so
$p\_m, p\_n \ll \sqrt k \log k$; and each $[a\_k,a\_{k+1}]$ is the product of
three such primes, so $\ll k^{3/2} \log^3 k$, as claimed.


Likewise, for each $M>1$ we can use products of $M$ primes
to obtain a sequence $\{a\_k\}$ of pairwise distinct positive integers
such that $[a\_k, a\_{k+1}] \ll k^{(M+1)/M} \log^M k \ll k^{1+\epsilon}$
for all $\epsilon > 1/M$. Finally, by concatenating ever-longer initial
segments of the sequences for $M=2$, $M=3$, $M=4$, etc. we construct
a sequence $\{a\_k\}$ of pairwise distinct positive integers
such that $[a\_k, a\_{k+1}] \ll k^{1+\epsilon}$ for all $\epsilon > 0$.


