UPDATE #2: Here's a more detailed explanation of the similarity. The Greedy/Fibonacci-Sylvester algorithm can be rephrased to look like a Euclidean-ish Algorithm. Here is the example above:$$ 51 = 3 \cdot 22 - 15$$$$51 \cdot 3 = 11 \cdot 15 - 12$$$$51 \cdot 3 \cdot 11 = 141\cdot 12-9$$$$51\cdot 3 \cdot 11 \cdot 141 = 26367 \cdot 9 - 0$$so the Greedy/F-S algorithm gives $$\frac{22}{51} = \frac{1}{3}+\frac{1}{11}+\frac{1}{141}+\frac{1}{26367}$$So the Greedy/F-S algorithm for $a/b$ at the $n$th step is doing a modified division algorithm with $bq_1q_2q_3\cdots q_{n-1}$ as the dividend and $r_{n-1}$ as the divisor (where $q_i$ is the $i$th quotient and $r_i$ is the $i$th remainder) and the Egyptian fraction is given by $\sum 1/q_i$. I say "modified division algorithm" because instead of the usual $b=aq+r$, the $+$ is replaced by a $-$. In your PEA (I think), you just kept the plus.
This is why I conjecture that the heuristics and such are the same. It seems like for every long $a$, $b$ pair in the Greedy/F-S algorithm, there should be an analogous long $a$, $b$ pair for the PEA. I don't have anything at this time other than a gut feeling to back me up. Maybe I'll try to construct an example...
