Remarks on the ``greedy odd'' Egyptian fraction algorithm (Q2735218)

Fibonacci proved that the greedy algorithm gives a (finite) representation of a reduced fraction \(0< a/b<1\) as a sum of distinct reciprocals. A variant of it, the ``greedy odd'' algorithm, is defined as follows: Define \(x_1\) to be the least odd positive integer with \(1/x_1 \leq a/b\). If \(a/b-1/x_1 >0\), then inductively define \(x_2\) to be the least odd positive integer so that \(1/x_2 \leq a/b - 1/x_1\). NEWLINENEWLINENEWLINEIt is an open question whether this algorithm terminates for all odd denominators \(b\). See \textit{R. K. Guy}, Unsolved problems in number theory, 2nd ed., Section D11 (1994; Zbl 0805.11001) and \textit{P. Erdős} and\textit{R. L. Graham}, Old and new problems and results in combinatorial number theory (1980; Zbl 0434.10001). Erdős and Graham attribute this problem to Sherman K.~Stein. It is known, due to the works of \textit{R. Breusch} [Solution to problem 4512, Am. Math. Mon. 61, 200-201 (1954)] and \textit{B. M. Stewart} [Sums of distinct divisors, Am. J. Math. 76, 779-785 (1954; Zbl 0056.27004)], that there exists a representation (in fact there exist infinitely many) of finite length by means of reciprocals of distinct odd positive integers. NEWLINENEWLINENEWLINEThe author proves some elementary results on the greedy odd algorithm. He proves that for \(s \in \mathbb{N}\) there are infinitely many fractions requiring exactly \(s\) steps. He also describes an infinite class of fractions where the numerators \(a_i\) of \(a_i/b_i=a/b-1/x_1 - \cdots -1/x_i\), where \((a_i,b_i)=1\), increase in the first steps. For example for \(2/(360k-101)\) one has \(a_1=3, a_2=4, a_3=5, a_4=6\).