Iterations of $x^2 + y^2$

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We construct a sequence $S$ of distinct positive integers as follows

1) the sequence $S$ starts as $1,2,3$

2) If $x,y$ are in the sequence , then $x^2 + y^2 $ is also in the sequence.

3) the sequence is strictly increasing.

4) the sequence is completely determined by the above.

The questions are unsurprisingly the following :

How dense are the integers in this set that belong in the interval $[a,b]$ ;

A) compared to the Sum of 2 squares in that interval.

B) compared to the integers in that interval.

C) such that They are prime compared to the primes of the form $v^2 + u^2$.

??


Edit

An example of a Number that is not in the sequence $S$ is $4^2 + 1^2 = 17$ because $4$ is neccessary to Sum to 17 ( 17 is a prime hence a Sum of 2 squares in only one way ) and $4$ is not in the set $S$. $4$ is not even a Sum of two (nonzero) squares !

I considered estimating the density as follows :

$$z = z_1^2 + z_2^2$$

Where $z_1,z_2$ have the probability of Being a sum of 2 squares equal to ( about ) $ ln(z_1)^{-1/2} * ln(z_2)^{-1/2}$. See the Landau-Ramanujan results :

https://en.m.wikipedia.org/wiki/Landau–Ramanujan_constant

With Some " handwaving " we simplify ( informal ) to probability $ln(z)^{-1}$.

However $z_1 = z_3^2 + z_4^2, z_2 = z_5^2 + z_6^2$ When both $z_1,z_2$ are large. And those new variables Also have to be the Sum of 2 squares ! Thus we now estimate the probabilty as $ln(z)^{-1} * ln(\sqrt z)^{-4} = 1/16 * ln(z)^{-5}$ These iterations and estimates go on until $z_k$ becomes small. Kinda.

Finally I think the counting function approximation

$$ \frac{b-a}{ b^{ln(2)}} $$

( counting the numbers in $S$ that are in the interval $[a,b]$ approximately , inspired by the informal idea above )

Works pretty well ?

Just my guess