We make the following assumptions:

1) That M > N
2) If there is an even number of pirates left the senior pirate votes twice once in the regular vote and once to break a tie if needed
3) The coins cannot be broken down into smaller pieces of gold
4) There is zero cost for the hangings (For example the cost of the rope is zero)
5) Each pirate acts strictly in his best financial interest.

The solution is the senior pirate distributes one coin to each priate, except the second most senior pirate gets no coins and he keeps M+2-N coins for himself.

We start with N=2. In this secnario the senior pirate keeps all the coins and uses his tie vote to outvote the junior pirate
N = 3 The senior pirate only needs the vote of one pirate. So he pays the cheapest cost by giving one coin to the most junior and keeping the rest etc.