I don't understand exactly what your sequence is, but if $v$ is your final value you can apply
$$h(v) = 1 - e^{-\alpha v}, \alpha \gt 0$$
to $v$ and get roughly what you've described, i.e. it takes $0$ to $0$, increases as $v$ increases, and goes to $1$ as $v$ gets infinitely large. $\alpha$ is a tunable parameter that controls how quickly $h$ increases. Since you're creating a visualization I'd say just start with $\alpha = 1$ and play with it to get the result you like best.
BTW you could also use any of
$$h(v) = 1 - e^{-\alpha v^2}$$
$$h(v) = 1 - e^{-\alpha v^3}$$
$$h(v) = 1 - e^{-\alpha v^{1/2}}$$
which will have the same properties but give better discrimination of either larger or smaller values. You can see that generically these look like
$$h(v) = 1 - e^{-\alpha v^\beta}, \alpha \gt 0, \beta \gt 0$$
so you've now got two parameter to tune your images.
I got curious to see how these looked and created a Desmos graph. You can go here to see it. If you look on the left hand side, towards the bottom, there are two sliders there labelled $a$ and $b$ (corresponding to $\alpha$ and $\beta$ here). You can see how different values of them affect your mapping.