Why does one of the units, namely $50 - 7\sqrt{51}$, in $\mathbb{Z}[\sqrt{51}]$ have a decimal expansion that is the Catalan numbers? $50 - 7\sqrt{51}$ is one of the units in $\mathbb{Z}[\sqrt{51}]$. In fact, corresponding to $n = 51$, $x = 50$ and $y = 7$ is the smallest integer solution to $x^2 - ny^2 = 1$.
The Catalan numbers have a great deal of combinatorial significance, and they appear in the decimal expansion of  $50 - 7\sqrt{51} \approx 0.010001000200050014004201320429.$
Why is this the case?

wikipedia names $$  c(x) = \sum_0^\infty \; C(n) \; x^n $$
and says that it satisfies
$$ c(x) = 1 + x \, c^2(x)  $$
Define $$   A = c \left( \frac{1}{10000} \right) $$
Then we have
$$  A = 1 + \frac{1}{10000} \; A^2 $$
so
$$  A^2 - 10000A + 10000 = 0. $$ So
$$  A = \frac{10000 \pm \sqrt{100000000 - 40000}}{2} =  \frac{10000 \pm 100 \sqrt{10000 - 4}}{2}$$
$$   A = \frac{10000 \pm 200 \sqrt{2500 - 1}}{2}= 5000 \pm 100 \sqrt {2499} $$
Note $$ 2499 = 51 \cdot 49  $$
Actually, later they say
$$ c(x) = \frac{1 - \sqrt{1 - 4 x}}{2x} = \frac{2}{1 + \sqrt{1 - 4x}} $$
which confirms that we should take 
$$   A = 5000 - 100 \sqrt {2499} = 100 \left( 50 - 7 \sqrt{51} \right)$$
