Since you are trying to switch on battery power, you have battery power available. So use it.

Your design brief contains two completely incompatible requirements; powered from the signal, and small size. To get significant power from the signal, you need to collect significant signal; and that means a large antenna. 

The classic AM band "crystal set" circuitry you are studying needs about a hundred foot wire antenna to collect enough power to run sensitive headphones. At which point you might as well run the wire to the transmitter!

At higher frequencies you can reduce the antenna size to some extent but if you need the antenna to be small you MUST expect the received power to be very low, and need amplification.

Even your transistor switch is an amplifier, so admit it; there will be amplification. Then focus on the least power you need to power that amplifier - off the battery. One way might be to use a very low power micro like the MSP430 (under 1 microamp in the right usage pattern) to turn the receiver on very briefly every few seconds, to see if a signal is being transmitted.

What are the components needed for this receiver?
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?



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 $0.5 - \sqrt{0.249}$.
