I'll guess that in your analysis, given the base case of D and E's game being a tie vote on a (D=100, E=0) split, results in a (C=0, D=0, E=100) split for three pirates since E can blackmail C into giving up all the coins in exchange for staying alive? D may vote arbitrarily on a (C=0, D=100, E=0) split, so C must consider E to have the deciding vote.

If so, that means four pirates would yield (B=0, C=100, D=0, E=0) or (B=0, C=0, D=100, E=0) in a tie. E expects 100 coins in the three-pirate game and so wouldn't be a safe choice of blackmailer, but C and D expect zero coins in a three-pirate game so B could choose between them arbitrarily. B can't give fewer than 100 coins to either C or D because they will punish that behavior with a deciding vote for death, and B knows this. It's potentially unintuitive for C because C's expected value in a three-pirate game is 0 but if C commits to voting against B for anything less than 100 coins, and B knows this, then B is forced to give either 0 or 100 coins to C. The remaining coins must go to D.

In the case of five pirates C and D except more than zero coins on average if A dies because B may choose arbitrarily between C or D as blackmailer. B and E expect zero coins from the four-pirate game. A must maximize the chance that two or more pirates will vote for A's split. C and D have an expected value of 50 coins from the four-pirate game if they assume B will choose randomly, and so a (A=0, B=0, C=50, D=50, E=0) split is no better than B's expected offer for C and D and any fewer than 50 coins for C or D will certainly make them vote against A. I think A should offer (A=0, B=n, C=0, D=0, E=100-n) where n is mutually acceptable to B and E.

Because B and E have no relative advantage in a four-pirate game (both expect zero coins) they don't have leverage against each other in the five-pirate game. If B had a non-zero probability of being killed in a four-pirate game then A should offer E more coins than B at a ratio corresponding to that risk. As it is, I think B and E would accept a fair split of n=50, but I may be overlooking some potential for E to blackmail B.