Inspired By Garrett Clarke

We are given $a_{9}=0$ . It is really easy to find $a_{8}=1, a_{7}=2, a_{6}=3$ . But from now I would suggest you to make some observations. If you are searching for any $n$ which is decreasing, Then note that for some positive integer $p$ . We have $x>a_{n+1}$ . where $x$ is the integers between $p^{2}$ and $(p+1)^{2}$ and mathematically representing $2p > n$ .

Now substitute the value of $n=5$ .

$2p>3\\ p>1.5\\ Let \quad p=2$

This means that $a_{5}$ lies between $2^{2}$ and $3^{2}$ . Also $3^{2}-2^{2}=5$ , according to the sequence equation we see $a_{6}$ should be the difference of ${ a }_{ 5 }-{\left\lfloor \sqrt { { a }_{ 5 } } \right\rfloor }^{2}$ . This gives $a_{5}=7=3^{2}-2=(p+1)^{2}-2$ .

By similar method of substitution you will get $a_{1}=13053767$ .