I never solved this, I cheated. Here's the notes I made while failing.
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if the one before and the one after an adapter can connect, that adapter can be dropped
find all these possible drops
number of possible combinations of these drops
https://en.wikipedia.org/wiki/Binomial_coefficient maybe?
https://en.wikipedia.org/wiki/Combination "Number of k-combinations for all k"
(2^n)-1
Applying this blindly doesn't work because some combinations are invalid (multiple adapters next to each other cannot all be removed, because that leads to greater than 3 overall). I don't want to have to check every possible combination, that would be slow.
Some adapters can only be skipped in the case others are not. Could I check for these and count them as one?
3s cannot be changed. Runs of 1s can be changed.