Let's put the number of coins at C and see what happens if we increase N. We'll number the pirates from meekest (1) to fiercest (N).
N = 1 1: C +
N = 2 1: 0 -
2: C *
N = 3 1: 1 +
2: 0 -
3: C-1 +
N = 4 1: 0 -
2: 1 +
3: 0 -
4: C-1 *
N = 5 1: 1 +
2: 0 -
3: 1 +
4: 0 -
5: C-2 +
N = 2C 1: 0 -
2: 1 +
3: 0 -
...
N-1: 0 -
N: 1 *
If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each N, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for N-1.
But now, we're at N = 2C + 1. Are we getting ready for carnage? Not quite:
N = 2C+1 1: 1 +
2: 0 -
3: 1 +
...
N-2: 1 +
N-1: 0 -
N: 0 +
Here, the fiercest pirate can escape alive, but without gold.
Okay, but now surely the sharks are getting fed? Let's see.
N = 2C+2 1: 0 -
2: 1 +
3: 0 -
...
N-2: 1 +
N-1: 0 -
N: 0 *
The fiercest pirate now broke the tie to stay alive. And now, finally, at N = 2C + 3, we reach our first unstable solution. The fiercest pirate can buy C votes, adds his own vote, but the opposition has C + 2 votes.
N = 2C+3 1: 1 +
2: 0 -
3: 1 +
...
N-3: 0 -
N-2: 0 -
N-1: 0 -
N: 0 + (shark bait)
But does that mean that from now on, all pirates get their feet wet? Not at all.
N = 2C+4 1: 1 +
2: 0 -
3: 1 +
...
N-4: 0 -
N-3: 0 -
N-2: 0 -
N-1: 0 +
N: 0 *
Now both the almost fiercest pirate is voting for his life, since the solution below is unstable. The fiercest pirate breaks the tie and the solution is stable.
There are two reasons for a pirate to vote in favour of a solution. The first is their live, the second one is gold.
N = 2C+5 1: 0 -
2: 1 +
3: 0 -
...
N-5: 0 +
N-4: 0 -
N-3: 0 -
N-2: 0 -
N-1: 0 -
N: 0 + (shark bait)
Here, pirates N-5 to N-1 vote against the solution, since the one below it is stable, is not worse than this one, and they like entertainment.
N = 2C+6 1: 0 -
2: 1 +
3: 0 -
...
N-6: 1 +
N-5: 0 -
N-4: 0 -
N-3: 0 -
N-2: 0 -
N-1: 0 +
N: 0 + (shark bait)
Here both pirates N and N-1 vote in favour, because in the solution for N-1, pirate N-1 is shark bait as well.
N = 2C+7 1: 0 -
2: 1 +
3: 0 -
...
N-7: 1 +
N-6: 0 -
N-5: 0 -
N-4: 0 -
N-3: 0 -
N-2: 0 +
N-1: 0 +
N: 0 + (shark bait)
N = 2C+8 1: 0 -
2: 1 +
3: 0 -
...
N-8: 1 +
N-7: 0 -
N-6: 0 -
N-5: 0 -
N-4: 0 -
N-3: 0 +
N-2: 0 +
N-1: 0 +
N: 0 *
And we've found a stable one again.
N = 2C+9 1: 1 +
2: 0 -
3: 1 +
...
N-9: 0 -
N-8: 0 -
N-7: 0 -
N-6: 0 -
N-5: 0 -
N-4: 0 -
N-3: 0 -
N-2: 0 -
N-1: 0 -
N: 0 + (shark bait)
As we can see, when N > 2C, we'll see more and more large runs of downvoting pirates, until the number of pirates upvoting to save their lives becomes equal to the number of downvoters and we find a stable solution again.
This yields the following formula for stable solutions:
N <= 2C ∨ N = 2C + 2^Z where Z ∈ {0, 1, 2, 3 ...}.
The procedure for doling out coins is to start with the lowest numbered pirate with the same parity as N, if N <= 2C.
If 2C + 2^(Z-1) < N <= 2C + 2^Z (where Z ∈ {0, 1, 2, 3 ...}), start with the pirate with the opposite parity of N + Z.