Last night what I shared as $$\text{Rank}_i$$ is  `hot_s`.

$$
\text{Rank}_i \;=\; \sum_j \text{zap\_amount}_{ij}\,e^{\lambda\,\text{zap\_time}_{ij}}
\qquad\text{where}\qquad
\lambda=\frac{\ln 2}{4h}.
$$

We store `hot_t` and `hot_s`:

$$
\text{Rank}_i \;=\; e^{\lambda\,\text{hot\_t}_i}\,\text{hot\_s}_i
$$

with

$$
\text{hot\_t}_i \;=\; \max_j \text{zap\_time}_{ij},
\qquad
\text{hot\_s}_i \;=\; \sum_j \text{zap\_amount}_{ij}\,e^{\lambda(\text{zap\_time}_{ij}-\text{hot\_t}_i)}.
$$

Substituting those definitions directly:

$$
\text{Rank}_i
\;=\;
e^{\lambda\,\max_j \text{zap\_time}_{ij}}
\sum_j \text{zap\_amount}_{ij}\,
e^{\lambda\left(\text{zap\_time}_{ij}-\max_j \text{zap\_time}_{ij}\right)}.
$$

That’s not the full sort key. My bad.

I’ll try to write it in a way I understand tomorrow.

k00b

I think you'd have to update the rank of all the other items at the same time that item 

's rank relative to other items might actually go down, because its old rank gets down-weighted via the exponent.

 gets a zap of just 1 sat.  Its old rank gets down-weighted for time, by the increase from the 1 sat is negligible.)

The SN home page is an auction

Scoresby

I think you'd have to update the rank of all the other items at the same time that item $$i$$ gets zapped, otherwise item $$i$$'s rank relative to other items might actually go down, because its old rank gets down-weighted via the exponent.

(As an example, imagine item $$i$$ gets a zap of just 1 sat.  Its old rank gets down-weighted for time, by the increase from the 1 sat is negligible.)