Ok, so if i understand correctly, if i do not do a analogRead() it won't use up any amps? (Except for keeping the cap filled).
Yes and no.
There is some (considerable) confusion here.
The 50 nA figure CrossRoads quotes is not related to charging the ADC capacitor (because firstly, 1V across 10k is 100 *micro*amps, not *nano*amps), but the maximum notional input leakage current of the ATmega input itself.
Other than this, the ADC only consumes power to charge that integrator capacitor, each time to perform a read. And since it is actual charge (nanoCoulombs or such) that is required, the value of the series resistor is irrelevant, only changing how long it takes for that charge to be transferred.
The real trick of this arrangement, is that the input leakage current of the ATmega is only meaningful whilst it is powered up. Remove the supply voltage, and all inputs are pulled down to 0.5V by the protection diodes.