Function or bounds for the number of solutions of $\sum\limits_{i=0}^k \frac{1}{x_k} = 1$

Hi, is there any result known for the number of different solutions of 1 = $\sum\limits_{k=0}^n \frac{1}{x_k}$ in dependency of the length $n$ of this partition? All I know, up to now, is that there are for every $n$ only finite many different solutions and the maximal $x_k$ is given by the $k$th element of the sylvester-sequenze. Father there is a result by Hofmeister/Stoll about the number of solutions up to lenght $n$ for $\frac{a}{b} = \sum\limits_{k=0}^m \frac{1}{x_k}$ with $m < n$ and $a \neq b$. So unfortunally this is not usable in my case too.

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Christian Elsholtz seems to have a preprint (with coauthors) called "Egyptian fractions with restrictions": math.tugraz.at/~elsholtz/WWW/papers/papers.html I believe that this paper will contain the state of the art on the problem you asked. Christian might be willing to send you a copy. – Greg Martin Mar 8 '11 at 0:39
Johannes: If you don't register an account, you won't be able to edit your question, and you won't be able to accept an answer. – S. Carnahan Mar 15 '11 at 16:35

An upper bound of about $c_0^{2^k}$ follows by elementary induction, (from your comment $x_k$ is bounded by the Sylvester sequence). Here $c_0 \approx 1.264$ is $\lim u_n^{\frac{1}{2^n}}$, ($u_n$ the $n$-th term of the Sylvester sequence).

An upper bound of $c_0^{(1+\epsilon) 2^{k-1}}$ is in a paper by C. Sándor. Sándor also gives a lower bound: for $k \geq 3$: $exp(c \frac{k^3}{\log k })$, for some positive constant $c$. His paper is: Periodica Mathematica Hungarica Volume 47, Numbers 1-2, 215-219. On the number of solutions of the Diophantine equation $\sum_{i=1}^n\frac{1}{x_i}=1$ http://www.springerlink.com/content/tn50293862257447/

An upper bound of $c_0^{(\frac{5}{24} +\epsilon) 2^{k-1}}$ was proved by T Browning and myself in our paper "On sums of unit fractions", which is to appear in the Illinois J of Mathematics. It is online here http://www.maths.bris.ac.uk/~matdb/preprints/es.pdf This paper gives an essentially best possible answer for a general fraction $\frac{m}{n}$ and $k=2$, then a nontrivial upper bound for $k=3$, and as a corollary lifts these results to general $k$. The case $\frac{m}{n}=1$ then follows.

All in all, there is a large gap between upper and lower bound for your original question.

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The first $8$ terms (under the assumption $x_1\le\cdots\le x_n$) are given at http://oeis.org/A002966 (without the restriction, it's http://oeis.org/A002967).

There is no information there about exact formulas or asymptotics, which suggests to me that nothing is known (but I haven't checked the Elsholtz paper that Greg Martin mentions).

There is some discussion of the problem at D11 in Guy, Unsolved Problems In Number Theory, especially page 257 (of the 3rd edition), but the main piece of information there is the attribution to Erdos of the question of asymptotic behavior.

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There are definitely not asymptotics known, but there are upper and lower bounds (that are not particularly close together). If I recall correctly, both are doubly exponential in n. – Greg Martin Mar 11 '11 at 22:05
@Greg, thanks. @Johannes, are you still here? Any reaction to what Greg and I have posted? – Gerry Myerson Mar 12 '11 at 3:54

One correction: the result of Hofmeister/Stoll is for $m \leq n$, about the number of all solutions up to length $n$.

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