A lucky nest to call home ....

We can write the expression as $S = \displaystyle\sqrt{9 - \dfrac{A}{3}}$ ,

where $A = \displaystyle\sqrt{13 + \sqrt{13 - \sqrt{13 + \sqrt{13 - \sqrt{13 + .....}}}}}$ .

Now $(A^{2} - 13)^{2} = 13 - A$ , which by observation has $A = 4$ as a solution. (There is another positive solution, namely $A = \frac{\sqrt{53} - 1}{2}$ , but since this value is less than $\sqrt{13}$ it can be discarded.)

So $S = \displaystyle\sqrt{9 - \dfrac{4}{3}} = \displaystyle\sqrt{\dfrac{23}{3}}$ ,

and thus $a + b = 23 + 3 = \boxed{26}$ .

$S=\sqrt{9-\frac{1}{3}\sqrt{13+\sqrt{13-\sqrt{13+\sqrt{13+\ldots}}}}}$

Let $\sqrt{13+\sqrt{13-\sqrt{13+\sqrt{13+\ldots}}}}=x$

also, let $\sqrt{13-\sqrt{13+\sqrt{13+\ldots}}}=y$

then, we see that

$\sqrt{13+y}=x~~~~~(1)$

and

$\sqrt{13-x}=y~~~~~(2)$

squaring both $(1)$ and $(2)$ and subtract to get:

$y+x=x^2-y^2=(x+y)(x-y)\\1=x-y\implies{y=x-1}$

Substitution $y=x-1$ to $(1)$ , we have $x^2-x-12=0$ . So, we have $x=4$ and $x=-3$ . But, since $S$ is positive, then $x$ should be positive, so $x=-3$ is discarded.

Finally, $S=\sqrt{9-\frac{1}{3}\times{4}}=\sqrt{\dfrac{23}{3}}$

and, the required answer is $~~23+3=\boxed{26}$