An Egyptian fraction represents a given rational number as a sum of distinct unit fractions. If a rational number ${\displaystyle x/y}$  is a sum of unit fractions with odd denominators,

${\displaystyle {\frac {x}{y}}=\sum {\frac {1}{2a_{i}+1}},}$ 

then ${\displaystyle y}$  must be odd. Conversely, every fraction ${\displaystyle x/y}$  with ${\displaystyle y}$  odd can be represented as a sum of distinct odd unit fractions. One method of finding such a representation replaces ${\displaystyle x/y}$  by ${\displaystyle Ax/Ay}$  where ${\displaystyle A=35\cdot 3^{i}}$  for a sufficiently large ${\displaystyle i}$ , and then expands ${\displaystyle Ax}$  as a sum of distinct divisors of ${\displaystyle Ay}$ .^[1]

However, a simpler greedy algorithm has successfully found Egyptian fractions in which all denominators are odd for all instances ${\displaystyle x/y}$  (with odd ${\displaystyle y}$ ) on which it has been tested: let ${\displaystyle u}$  be the least odd number that is greater than or equal to ${\displaystyle y/x}$ , include the fraction ${\displaystyle 1/u}$  in the expansion, and continue in the same way (avoiding repeated uses of the same unit fraction) with the remaining fraction ${\displaystyle x/y-1/u}$ . This method is called the odd greedy algorithm and the expansions it creates are called odd greedy expansions.

Stein, Selfridge, Graham, and others have posed the question of whether the odd greedy algorithm terminates with a finite expansion for every ${\displaystyle x/y}$  with ${\displaystyle y}$  odd.^[2] As of 2021^[update], this question remains open.

Let ${\displaystyle x/y}$  = 4/23.

23/4 = 53/4; the next larger odd number is 7. So the first step expands

161/5 = 321/5; the next larger odd number is 33. So the next step expands

5313/4 = 13281/4; the next larger odd number is 1329. So the third step expands

Since the final term in this expansion is a unit fraction, the process terminates with this expansion as its result.

It is possible for the odd greedy algorithm to produce expansions that are shorter than the usual greedy expansion, with smaller denominators.^[3] For instance,

A similar phenomenon occurs with other numbers, such as 5/5809 (an example found independently by K. S. Brown and David Bailey) which has a 27-term expansion. Although the denominators of this expansion are difficult to compute due to their enormous size, the numerator sequence may be found relatively efficiently using modular arithmetic. Nowakowski (1999) describes several additional examples of this type found by Broadhurst, and notes that K. S. Brown has described methods for finding fractions with arbitrarily long expansions.

The odd greedy algorithm cannot terminate when given a fraction with an even denominator, because these fractions do not have finite representations with odd denominators. Therefore, in this case, it produces an infinite series expansion of its input. For instance Sylvester's sequence can be viewed as generated by the odd greedy expansion of 1/2.

• Breusch, R. (1954), "A special case of Egyptian fractions, solution to advanced problem 4512", American Mathematical Monthly, 61: 200–201, doi:10.2307/2307234, JSTOR 2307234
• Guy, Richard K. (1981), Unsolved Problems in Number Theory, Springer-Verlag, p. 88, ISBN 0-387-90593-6
• Guy, Richard K. (1998), "Nothing's new in number theory?", American Mathematical Monthly, 105 (10): 951–954, doi:10.2307/2589289, JSTOR 2589289
• Klee, Victor; Wagon, Stan (1991), Unsolved Problems in Elementary Geometry and Number Theory, Dolciani Mathematical Expositions, Mathematical Association of America
• Nowakowski, Richard (1999), "Unsolved problems, 1969–1999", American Mathematical Monthly, 106 (10): 959–962, doi:10.2307/2589753, JSTOR 2589753
• Stewart, B. M. (1954), "Sums of distinct divisors", American Journal of Mathematics, 76 (4): 779–785, doi:10.2307/2372651, JSTOR 2372651, MR 0064800
• Wagon, Stan (1991), Mathematica in Action, W.H. Freeman, pp. 275–277, ISBN 0-7167-2202-X

• MathPages - Odd-Greedy Unit Fraction Expansions, K. S. Brown