In above sequence of right angles I want to find the smallest integer n such that $OA_n>30$. I can do this by calculate $OA_2$ in the triangle $OA_1A_2$. In the same way, calculate $OA_3$ and so on until the solution is found. But I am sure there is another way to find based on Pythagorean triplets and Fibonacci numbers (?). I have also found the sequence of the squares of the hypotenuses:
$5,14,30,55,91,140,204\ldots$
Thanks in advance for any help.
The sequence of squares of the hypotenuses is $5,14,30,55,91,140,164,...$ These are fairly obviously successive partial sums of $1^2+2^2+3^2+4^2+5^2+6^2+7^2+...$. You are looking for the point when this sum exceeds $30^2$.
There is a formula for the sum of successive squares: $$1^2+2^2+3^2+4^2+5^2+...+k^2 = \frac{k(k+1)(2k+1)}{6} = \frac{2k^3+3k^2+k}{6}$$
So you want to solve: $$ \frac{2k^3+3k^2+k}{6} > 30^2$$ $$ 2k^3+3k^2+k > 6\times 30^2$$ This is tricky to solve exactly, but you can get a rough answer by ignoring the smaller terms: $$ 2k^3 \approx 6\times 30^2$$ $$ k \approx \sqrt[3]{3\times 30^2}$$ $$ k \approx 13.9$$ So then just try $k=13$ in the formula. This gives $819$, so is too small. Trying $k=14$ does give a number larger than $900$.
Therefore you want the last triangle side to be $14$, and $A_{14}$ is the first hypotenuse greater than $30$.