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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.
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If all the coins are quantum-mechanical, you should never quit, nor if all the coins are logical (digits of pi). If the first coin is logical ("what laws of physics are true?", in the LHC dilemma), the following coins are quantum, and your utility is linear in squared amplitude of survival, again you should never quit. However, if your utility is logarithmic in squared amplitude (i.e., dying in half of your remaining branches seems equally bad no matter how many branches you have remaining), then you should quit if your first throw comes up heads.
I'm not getting the same result... let's see if I have this right.
If you quit if the first coin is heads: 50%*75% death rate from quitting on heads, 50%*50% death rate from tails
If you never quit: 50% death rate from eventually getting tails (minus epsilon from branches where you never get tails)
These deathrates are fixed rather than a distribution, so switching to a logarithm isn't going to change which of them is larger.
I don't think the formula you link to is appropriate for this problem... it's dominated by the log(2^-n) factor, which fails to account for 50% of your possible branches being immune to death by tails. Similarly, your term for quitting damage fails to account for some of your branches already being dead when you quit. I propose this formula as more applicable.