How many irreducibles in $\mathbb Z[\sqrt - 5]$? Finding and counting irreducibles in non-UFD's seems harder to me than finding and counting primes in UFD's.
The number of primes in the UFD $\mathbb Z[\sqrt -1]$ (the gaussian integers) is about $\pi(n)$ where $\pi$ is the usual prime counting function and $n$ is the norm.
In other words, the number of gaussian primes with norm at most $n$ is almost equal to $\pi(n)$.

To see this notice primes that are the Sum of 2 squares are $ 1 \mod 4$. Let the counting function of such primes be $\pi_2(n)$. It is Well known that $\pi_2(n) $ ~ $ \pi(n)/2$. 
The number of gaussian primes is ( ignore unit multiples ) thus $2  \pi_2(n) $ ~ $\pi(n)$.
A similar thing happens for Eisenstein integers.


How many irreducibles are there in the integral domain $\mathbb Z[\sqrt - 5]$ ?

Is it asymptotic to
$$ T_5 \pi(n) $$
Where $n$ is again the norm and $T_5$ is rational ?
Does a non-UFD imply more irreducibles than a UFD or less ... Or neither ?
Are the number of irreducibles in $\mathbb Z[\sqrt - p]$ for $p$ a prime always asymptotic of the form 
$$ T_p \pi(n) $$
Where again $T_p$ is rational.
Is $T_p$ the class number ??? If so why ??
 A: The order of magnitude for the number of irreducibles in $\mathbb{Z}[\sqrt{-5}]$ with norm up to $x$ is larger, it is: 
$$\frac{x}{\log x} \log \log x$$
(I say "order of magnitude" as I am not sure about the constant; yet it is asymptotic to this with some constant.)
Yet you are right that in the UFD case it is always asymptotic to 
$$\frac{x}{\log x}$$
More generally, the count of irreducibles (disregarding multiplication by units) in the maximal order of a number field is of order of magnitude: 
$$\frac{x}{\log x} (\log \log x)^c$$
where $c$ depends on the class group (not only the class number). 
Specifically it is $c+1$ is the Davenport constant of the class group. In case the group is cyclic it happens to equal the order, that is the class number, but in general it will be smaller. 
(Note that $\mathbb{Z}[\sqrt{-p}]$ might not always be the maximal order though. But similar results exists for any order.)
To appreciate where this might come from, note that what is more uniform is the number of prime ideals. One has a total of about 
$$\frac{x}{\log x }$$
and they are uniformly distributed over the classes. 
If you have, say, class number $2$ then you can take two nonprincipal prime ideals $P_1, P_2$ say of norm $\le \sqrt{x}$ and their product will be a principal ideal that is irreducible as a principal ideal and thus generated by an irreducible.
Or differently, the number of irreducibles of norm $\le x$ is at least the number of products of two nonprincipal ideals such that the norm is $\le x$.
And the count of those  prime ideals is up to a constant like that of the prime numbers. Thus that count should correspond to the count of numbers that are the product of two prime numbers. The count of those is also (up to a constant) 
$$\frac{x}{\log x} \log \log x$$
See the Wikipedia page on almost primes. 
If you have a more complex class group than you can have more complex products of nonprincipal prime ideal that yield an irreducible, and this further increases the order of magnitude, just like the number that are a product of at most $k$ primes is 
$$\frac{x}{\log x} (\log \log x)^{k-1}$$
The above mentioned Davenport constant corresponds exactly to the maximum length of such a product of prime ideals. Thus the power of the $\log \log$ is one less than this constant.
A book on algebraic number theory that deals with such things is Narkiewicz's "Elementary and Analytic Theory of Algebraic Numbers" (at least in the 2nd and 3rd edition, but likely you'd have a harder time to find the 1st edition than one of those anyway).
There is varied more specialized literature too. 
