Let's say I have $M$ gallons of water left in a tank.
I know that it will take me some amount of time to remove half the water. Call this $t$. I know it will take me $2t$ to remove half of what's left after that. Then $4t$ for half of what's left after that, and so on.
But this should be more of a continuous function as opposed to a discrete thing.
How would I create such a function?
You take $T$ second for taking half of the stored amount (allow me to call it $T$ in caps, the discrete time).
Lets call $x$ the amount you take out after $T$ seconds.
Hence
$$ x(0)=0\\ x(T)=M\frac{1}{2}\\ x(2T)=M\frac{1}{2}\frac{1}{2}=M\frac{1}{2^2}\\ x(2^2T)=M\frac{1}{2}\frac{1}{2^2}=M\frac{1}{2^3}\\ x(2^3T)=M\frac{1}{2}\frac{1}{2^3}=M\frac{1}{2^4}\\ $$
Hence you realize magicly (by induction) that each decrement immediately after the time $2^nT$ is:
$$ x(2^nT)=M\frac{1}{2^{n+1}} $$
Now allow me to have this change, $2^nT=t$:
$$ x(t)=M\frac{T}{22^nT}=M\frac{T}{2t} $$
Hence at some T, I could have some specific amount of water as a approximation, which is different from the given examples, but reasonable. If i take $t=1, 2, 4, 8$ i will recover the given values. If i take $t=1.5$, i will have:
$$ x(1.5)=M\frac{T}{3} $$
Which is a rough approximate, not linear, in this specific point case, an inverse average. Let it be.
I dont think, as is, this problem is convenient to make further continuous, since at the limit, you require an infinite amount of time to remove an infinitesimal amount of water.
Please note, this problem is not an exponential decay. Under that case, you take always a same amount of time $t$ for halving your container, in which case the limit is easier to express. Could you confirm, we are not under that case?.