From $0 = 0.5 - 0.5 = 0.5 - \sqrt{0.25}$, we can adjust the subtrahend slightly to obtain


$$0.5 - \sqrt{0.249} = 0.001\ 001\ 002\ 005\ 014\ 042\ldots$$


where the decimal representation contains the first few Catalan numbers: $1, 1, 2, 5, 14, 42 
\ldots$


We can see even more Catalan numbers (albeit spaced apart with more $0$s) by using more $9$s. 


(For example, check the decimal representation of $0.5 - \sqrt{0.24999999999}$.)


My question is not how to show this formally; it is a straightforward problem to show how one derives, e.g., the decimal representation given above. (I'll include a derivation below for anyone who doesn't want to think this through her/himself.)


Instead, my **question** is: why would it make sense, *intuitively*, for the Catalan numbers to show up in these decimal representations?




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**Derivation:**


Recall that the generating function for the Catalan numbers $c(x)$ satisfies $c(x) = 1 + xc(x)^2$. Rearranging, we find that $c(x) = \frac{2}{1 + \sqrt{1 - 4x}}.$ Then


$$\sum\_{n = 0}^{\infty}\frac{C\_n}{10^{3n + 3}} = \frac{1}{1000}\sum\_{n=0}^{\infty}\frac{C\_n}{1000^{n}} = x \sum\_{n=0}^{\infty}C\_n x^n = xc(x) = \frac{2x}{1 + \sqrt{1 - 4x}},$$


where we have simplified our computations by letting $x = \frac{1}{1000}$.


Evaluating at this value of $x$ yields $0.5 - \sqrt{0.249}$.



Intuition Behind a Decimal Representation with Catalan Numbers
To make it less surprising is to fade some of the magic! But, ok. 


First let us say that any real sequence $a\_1,a\_2,\cdots$ has an ordinary generating function (ogf) $f(x)=\sum a\_ix^i$ which may be a formal series with radius of convergence $0$ (and still be useful) BUT if the $a\_i$ are positive integers and $f(x)$ converges at $\frac{1}{b^k}$ then $f(\frac{1}{b^k})$ is a number whose base $b$ expansion is the sequence $a\_i$ buffered by $0$'s until they start to bump into each other. I'll stick to $b=10$ and I'll use $(10^{-j})f(10^{-k})$ if there is an $a\_0$ term I want to shift past the decimal point. 


So the question might be which series have a nice ogf? That the Catalan numbers do is very nice.


From $\frac{1}{(1-x)^k}=\sum \binom{k+i}{k}x^i$ one obtains


$\frac{1}{10\cdot 0.9998}=0.10002000400080016003200640128025605121024$


$\frac{1}{10(0.999)^2}=0.10002000300040005000600070008000900100011$


$\frac{1}{10(0.999)^3}=0.10003000600100015002100280036004500550066$


Since $\binom{i}{2}+\binom{i+1}{2}=i^2$, 


we can use $(\frac{1}{10}+\frac{1}{100000})\frac{1}{(1-x)^3}$ at $x=0.0001$ to get


$\frac{1.0001}{10(0.999)^3}=\frac{100010000000}{999700029999}=0.100040009001600250036004900640081010001210$


But it is more productive to use $\binom{i}{1}+2\binom{i}{2}=i^2$ for 


$\frac{1}{10(1-x)^2}+\frac{1}{5000(1-x)^3}.$ Another function which coincides with the previous one at $x=0.0001.$ So the same rational but obtained another way. This second approach makes it clear how to get any polynomial sequence $a\_i=p(i).$


This already seems less fun. So thinking of a dramatic last target, the Fibonacci numbers remind us that anything given by a recurrence relation (linear, with constant coefficients...) has a nice ogf.


At $x=\frac{1}{100}$, 


$\frac{1}{10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$


A cute point is that all my examples (none of which are as nice as yours) lead to a rational number so as the $a\_i$ increase in size and overlap they result in an eventually repeating decimal. For example the base $10$ "Fibonacci" rational I gave has period $496620$ while
$\frac{10}{89}=0.\overline{11235955056179775280898876404494382022471910}$


