IRL_ABooklet_2023

Assume $a_1<a_2$ and let $r=\frac{a_2}{a_1}$. Then $r>1$ and $r$ is a rational number which can be written as $\frac{p}{q}$ with $p,q$ coprime. The sequence is then $$a_1,a_{1}r,a_1{r}^2,a_1{r}^3,a_1{r}^4,a_1{r}^5,a_1{r}^6.$$ The last term can only be an integer if $a_1$ is divisible by $q^6$. We may try values for $r$ that are equal to $(n+1)/n$ with $n\ge 2$. The first would be $r=3/2$, but $(3/2{)}^6=\frac{729}{64}>10$ and so $a_1{r}^6$ would have more digits than $a_1$. Therefore, $q>2$.

Next we try $r=4/3$. Then $a_1$ needs to be divisible by $3^6=729$. The smallest value for $a_1$ then is $2\cdot 3^6=1458$. Because $2\cdot 4^6=8192$, we indeed get seven four-digit numbers:

n	1	2	3	4	5	6	7
$a_n$	1458	1944	2592	3456	4608	6144	8192

In general, there will be a positive integer $a$ such that $a_1=a{q}^6$ and $a_7=a{p}^6$. Because $5^6$ is a five-digit number, neither $p$ nor $q$ may exceed 4. It now is easy to see that we found the only possible solution with $r>1$.