Let sequence $49,4489,444889,...$ is formed when in between digits $4$ and $9$ is the number $48$. For any number larger than $444889$ continue to put $48$.
Example: $49,4489,444889,44448889,...$ Prove that all numbers in the sequence are perfect squares.
I think of finding generating function of the sequence and then the closed form of a sequence.
How to evaluate generation function (in $\sum\limits$ notation)?
A number $444\ldots 4$ consisting of $n$ fours is equal to $(10^n-1)\cdot\frac{4}{9}$, see why for yourself.
Then the number consisting of $n/2$ fours, then $n/2-1$ eights and then a nine is equal to:
$$\frac{4}{9}(10^n-1)+\frac{4}{9}(10^{n/2}-1)+1=\frac{4}{9}(10^n+10^{n/2}+\frac{1}{4})$$
which you can easily see to be equal to:
$$(\ \frac{2}{3}(10^{n/2}+\frac{1}{2})\ )^2$$
which is the square of:
$$\frac{1}{3}(1+2\cdot10^{n/2}) $$
The thing in the parenthesis is most certainly divisible by $3$, so the number we started off with was a perfect square.