Egyptian fractions with very large denominators It is well-known that if we have a fraction $\frac ab$, with $a,b\in\mathbb N$ and $a<b$, if we apply the greedy algorithm in order to express it as a sum of unit fractions, then we may get fractions with very large denominators. For instance, if we apply the algorithm to $\frac5{31}$, what we get is$$\frac17+\frac1{55}+\frac1{3\,979}+\frac1{23\,744\,683}+\frac1{1\,127\,619\,917\,796\,295}$$and if we apply it to $\frac{1\,197}{2\,273}$, then the decimal representation of the last denominator has $14\,583$ digits. All this suggests that the following statement is true:

If $R>0$, then there are natural numbers $a$ and $b$ such that $a<b$ and that, if we apply the greedy algorithm to express $\frac ab$ as the sum of unit fractions, then one of the unit fractions $\frac1D$ that we get in that expression will be such that $\frac Db>R$.

My guess is that either this statement has already been proved or that it has already been stated as a conjecture. It that's so, can someone please provide a reference?
 A: There's a very simple answer to the original question. The greedy algorithm, applied to $a/b=2/(2n-1)$, gives $${2\over2n-1}={1\over n}+{1\over n(2n-1)}$$ so in the notation of the original question we have $${D\over b}={n(2n-1)\over2n-1}=n$$ which can be made arbitrarily large. 
A: I know this is an old question which already has an answer, but it is still interesting to try to maximize $D$, in terms of $a$ and $b$. By the greedy algorithm, we get $\frac{a}{b} = \frac{1}{\left \lceil \frac{b}{a} \right \rceil} + \frac{a \left \lceil \frac{b}{a} \right \rceil - b}{b \left \lceil \frac{b}{a} \right \rceil}$.
Since $a \left \lceil \frac{b}{a} \right \rceil - b$ (which is the numerator of the latter fraction) is strictly smaller than $a$, we know the greedy algorithm terminates in at most $a-1$ steps. If it takes exactly $a-1$ steps, then by induction we see $D = D_{a-1}$, where $D_0 = b$ and $D_i = D_{i-1} \left \lceil \frac{D_{i-1}}{a-i+1} \right \rceil$ for $1 \le i \le a-1$.
By Corollary 2 of this paper (which was referenced by Gerry Myerson in the comments), for every $a \in \mathbb{N}$ there are infinitely many $b > a$ such that the greedy algorithm applied to the fraction $\frac{a}{b}$ does indeed take $a-1$ steps to terminate. Assume $\frac{a}{b}$ is such a fraction with $a > 1$. Since $D_i = D_{i-1} \left \lceil \frac{D_{i-1}}{a-i+1} \right \rceil > \frac{D_{i-1}^2}{a}$, by induction this gives us $D = D_{a-1} > b\left(\frac{b}{a} \right)^{2^{a-1}-1}$. This means that if we fix $a$, then there is a constant $c = c(a) > 0$ such that there are infinitely many $b$ with $D > cb^{2^{a-1}}$. This is essentially best possible, since we always have $D_i \le D_{i-1}^2$ implying, once again by induction, $D \le b^{2^{a-1}}$ for all fractions $\frac{a}{b}$. To summarize:
Theorem.
For a fraction $\frac{a}{b}$ with $1 \le a < b$, let $D = D(a,b)$ be the largest denominator that occurs in the greedy egyptian expanion of $\frac{a}{b}$. Then $D \le b^{2^{a-1}}$. On the other hand, for every $a \in \mathbb{N}$ there is a positive constant $c = c(a)$ such that for infinitely many positive integers $b \in \mathbb{N}$ we have $D > cb^{2^{a-1}}$.
A: First, can you establish that the algorithm always halts for this problem? 
The reason I ask, is that several years ago, it was (and I believe still is) an open problem that if $\frac{a}{b}$ is a rational number with $b$ odd---

Can $\frac{a}{b}$ be expressed as $\sum \frac{1}{x_{i}}$, where the $x_{i}$ are chosen to be the least possible odd integers that leave a nonnegative remainder, in a finite number of terms?

This problem received a fair amount of attention from Richard K. Guy in his Unsolved Problems in Number Theory, 3rd Edition. In it, he highlights some of the efforts by certain individuals, such as Stan Wagon, who has tackled the fraction $\frac{3}{179}$, which produced 19 terms, the last of which has 439492 digits. This, Guy indicates was beaten by David Baily who considered $\frac{3}{2879}$ and used the greedy algorithm to expand it to the sum of 21 unit fractions, the last term of which has 3018195 digits. Later, it was shown that $\frac{5}{5809}$ has at least 22 terms with the last being in excess of 60,000,000 digits.   
Incidentally, although the 3rd edition of Guy's book is the most recent, it was however, published in 2005. Nevertheless, it seems that the problem is still open. See, for instance: 
https://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/203587#203587
I hope this puts a little bit of light on the problem. 
