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?

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$50 - 7\sqrt{51}$ 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?

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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 $$ A = 5000 - 100 \sqrt {2499} = 100 \left( 50 - 7 \sqrt{51} \right)$$