Problem Statement

An   Egyptian fraction   is the sum of distinct unit fractions such as: Each fraction in the expression has a numerator equal to   1   (unity)   and a denominator that is a positive integer,   and all the denominators are distinct   (i.e., no repetitions).
Fibonacci's   Greedy algorithm for Egyptian fractions   expands the fraction

x y

{\displaystyle {\tfrac {x}{y}}}

to be represented by repeatedly performing the replacement

(simplifying the 2nd term in this replacement as necessary, and where

⌈ x ⌉

{\displaystyle \lceil x\rceil }

is the   ceiling   function).

For this task,   Proper and improper fractions   must be able to be expressed.

Proper  fractions   are of the form

a b

{\displaystyle {\tfrac {a}{b}}}

where

a

{\displaystyle a}

and

b

{\displaystyle b}

are positive integers, such that

a < b

{\displaystyle a<b}

,     and improper fractions are of the form

a b

{\displaystyle {\tfrac {a}{b}}}

where

a

{\displaystyle a}

and

b

{\displaystyle b}

are positive integers, such that   a ≥ b.

(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.) For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].

Let's start with the solution:

frac[p, q] :=
{
    a = makeArray[[0]]
    if p > q
    {
        a.push[floor[p / q]]
        p = p mod q
    }
    while p > 1
    {
        d = ceil[q / p]
        a.push[1/d]
        [p, q] = [-q mod p, d q]
    }
    if p == 1
        a.push[1/q]
    a
}

showApproximations[false]

egypt[p, q] := join[" + ", frac[p, q]]

rosetta[] :=
{
    lMax = 0
    longest = 0

    dMax = 0
    biggest = 0

    for n = 1 to 99
        for d = n+1 to 99
        {
            egypt = frac[n, d]
            if length[egypt] > lMax
            {
                lMax = length[egypt]
                longest = n/d
            }
            d2 = denominator[last[egypt, 1]@0]
            if d2 > dMax
            {
                dMax = d2
                biggest = n/d
            }
        }

    println["The fraction with the largest number of terms is $longest"]
    println["The fraction with the largest denominator is $biggest"]
}