The numerator really does drop

Greedy isn't the shortest

Which n need all four?

Fibonacci and Sylvester

[Demo] Mei's sketch shows the greedy algorithm needs at most four terms for 

. Is the "numerator goes down" argument complete, and is anything known about which 

The greedy algorithm for Egyptian fractions takes the largest unit fraction that fits, subtracts it, and repeats. For 

. So the numerator goes down by at least one every step: 

Greedy is not always the shortest though. For 

[Demo] The greedy algorithm never needs more than four terms for 4/n

[Demo] Mei's sketch shows the greedy algorithm needs at most four terms for 4/n. Is the "numerator goes down" argument complete, and is anything known about which n need all four?