Darkroom f-stop printing formalisation There is a common way to estimate your exposure while printing in a dark room and it's a stepped test. You cover the paper with some light-blocking sheet leaving only the narrow strip and expose it. Then you slide the sheet end expose the adjacent strip (and also all previous ones). To be most representative final exposure times of strips should be expressed as an exponential function like $t=t_0b^{ns}$. For your usual base of 2 the half-stop scale looks like $t=t_02^{1/2n}$. I'm struggling to express the time between sheet movements to meet this criteria. I was never strong in series to be honest.
 A: Let's start with a simple example.  Say your stepped test begins with an initial exposure time of $t_0 = 1$ second.  For additional steps in full stop increments, each step doubles the exposure time, so $$t_1 = 2, \\ t_2 = 4, \\ t_3 = 8, \\ t_4 = 16$$ seconds, and so forth.  In general, $$t_n = t_0 2^n,$$ where $t_0 = 1$.  The time increment between successive steps must therefore be
$$t_1 - t_0 = 1, \\ t_2 - t_1 = 2, \\ t_3 - t_2 = 4, \\ t_4 - t_3 = 8$$ and in general $$t_n - t_{n-1} = t_0 (2^n - 2^{n-1}) = t_0 2^{n-1}.$$
Now, we must be careful here:  the order in which these steps are taken is reversed.  In other words, your first strip is exposed for $t_4 - t_3 = 8$ seconds, then you pull back the cover sheet for the second step, which lasts for $t_3 - t_2 = 4$ seconds more; then for the third step, the sheet is pulled back again for $2$ seconds.  So the increment between successive steps should be longest at the beginning and get progressively faster until the final step.  This means you must choose the total number of steps to take before calculating the first step's exposure time.
So how would we do this for a more sophisticated example?  Say we want $n = 10$ steps in one-third-stop increments, with the shortest exposure time being $t_0 = 4$ seconds.  Then we calculate (all times in seconds)
$$\begin{align}
t_1 &= t_0 2^{1/3} \approx 5.03968 \approx 5, \\
t_2 &= t_0 2^{2/3} \approx 6.3496 \approx 6.3, \\
t_3 &= t_0 2^{3/3} = 8, \\
t_4 &= t_0 2^{4/3} \approx 10.0794 \approx 10.1, \\
t_5 &= t_0 2^{5/3} \approx 12.6992 \approx 12.7, \\
&\;\vdots \\
t_9 &= t_0 2^{9/3} = 32.
\end{align}$$
This gives the table of desired total exposure time for each step.  Then the time between increments is
$$\begin{align}
t_1 - t_0 &\approx 5 - 4 = 1, \\
t_2 - t_1 &\approx 6.3 - 5 = 1.3, \\
t_3 - t_2 &\approx 8 - 6.3 = 1.7, \\
t_4 - t_3 &\approx 10.1 - 8 = 2.1, \\
&\;\vdots \\
t_9 - t_8 &\approx 32 - 25.4 = 6.6,
\end{align}$$
and the first step is to expose the first strip for approximately $6.6$ seconds.
In general, then, the time differences for the $k^{\rm th}$ step in an $n$-step test, where there are $m$ steps in a full stop and the final step has an exposure time of $t_0$ seconds, is
$$\boxed{\delta_k = t_{n-k+1} - t_{n-k} = t_0 \left(2^{(n-k+1)/m} - 2^{(n-k)/m}\right)},$$
for $k \in \{1, 2, \ldots, n\}$.
This formula takes into account the need to reverse the steps.  As you can see, $\delta_1 = t_n - t_{n-1}$, which is the largest increment.  Also, don't forget to expose the entire sheet for an additional $t_0$ seconds, which is the desired minimum exposure time.
