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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...

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Your proto-Euclidean algorithm is basically equivalent to the Greedy algorithm for finding the alternating Egyptian fraction representation of a rational. For instance, in your example, if we expand the nested parentheses we get: $$\frac{22}{51} = \frac{1}{2}(1-\frac{1}{7}(1-\frac{1}{25}(1-\frac{1}{51}))) = \frac{1}{2}-\frac{1}{14}+\frac{1}{350}-\frac{1}{17850}$$

UPDATE: This is almost the Fibonacci-Sylvester Algorithm for finding Egyptian Fractions. The difference being that alternating signs between the fractions that the proto-Euclidean algorithm creates. I'm not sure how that affects the rate of convergence and such. You could probably eliminate the sign changes by choosing the signs on the $r_i$ and/or adding/subtracting 1 from each of them. (This is what I had to do for a similar project but can't remember which turned out to give the right answer.) Some heuristics on the F-S method can be found here.

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Your proto-Euclidean algorithm is basically equivalent to the Greedy algorithm for finding the alternating Egyptian fraction representation of a rational. For instance, in your example, if we expand the nested parentheses we get: $$\frac{22}{51} = \frac{1}{2}(1-\frac{1}{7}(1-\frac{1}{25}(1-\frac{1}{51}))) = \frac{1}{2}-\frac{1}{14}+\frac{1}{350}-\frac{1}{17850}$$