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Statement

For a positive rational $r$ and a positive integer $t$, let $\mathcal H_t(r)$ be the set of $t$-tuples $(x_1,\ldots,x_t)$ of integers with $x_1>\cdots>x_t\ge1$ and $\sum_{i=1}^t1/x_i=r$, and let

$$M_t(r)=\inf\{x_1:(x_1,\ldots,x_t)\in\mathcal H_t(r)\}$$

be the least largest denominator over all $t$-term Egyptian fraction representations of $r$ ($M_t(r)=\infty$ if there is none; display (3), p. 2). Let $t_0(r)$ be the fewest terms an Egyptian fraction representation of $r$ can have, with the convention $t_0(1)=3$ (p. 2: $1$ has a one-term representation and none with two terms).

Theorem 2 (p. 2). "For all positive rational numbers $r$ and all integers $t\ge t_0(r)$, we have

$$M_t(r)=\frac{t}{1-e^{-r}}+O_r\Bigl(\frac{t\log\log3t}{\log3t}\Bigr),$$

which is best possible."

The specialization to Problem 285: its $f(k)$, the least possible $n_k$ over representations $1=1/n_1+\cdots+1/n_k$ with $n_1<\cdots<n_k$, is $M_k(1)$ (the same tuples, listed in the opposite order), and $1/(1-e^{-1})=e/(e-1)$, so

$$f(k)=\frac{e}{e-1}k+O\Bigl(\frac{k\log\log3k}{\log3k}\Bigr)=(1+o(1))\frac{e}{e-1}k \qquad(k\ge3).$$

"Best possible" refers to both terms: the deduction on p. 5 (display (9)) gives $x_1\ge t/(1-e^{-r})+\delta(r)\,t\log\log t/\log t$ for every $t$-term representation and large $t$, so the error term cannot be lowered in order.

Source. G. Martin, Denser Egyptian fractions, arXiv:math/9811112v1 (18 November 1998), Theorem 2 on p. 2, read on the page image and in the text layer of that preprint. The journal version, Acta Arith. 95 (2000), no. 3, 231--260 (DOI 10.4064/aa-95-3-231-260), was not consulted; its numbering and pagination were not compared.

Read depth. Claims checked: the statement, the definitions of $\mathcal H_t(r)$, $M_t(r)$ and $t_0(r)$ (p. 2) were read clause by clause on the page image. The reduction of Theorem 2 to Propositions 5 and 6 (pp. 4--5) was read for structure; the proofs of the propositions (Sections 3--5, pp. 7--19) were not read.

Proof pointer

Section 2 reduces Theorems 1 and 2 to two propositions (p. 4). Proposition 5: for a closed interval $I\subset(0,\infty)$ there is $T(I)$ such that for all integers $t>T(I)$ and all $r=a/b\in I$ with $P^*(b)<t\log^{-22}t$ ($P^*$ the largest prime power divisor) there is a set $E$ of $t$ distinct positive integers with $\sum_{n\in E}1/n=r$ and $\max E<t/(1-e^{-r})+O_I(t\log\log t/\log t)$. Proposition 6: there is $\delta(r)>0$ such that for large $x$ every set $E\subseteq[1,x]$ with $\sum_{n\in E}1/n=r$ has $|E|\le(1-e^{-r})x-\delta(r)\,x\log\log x/\log x$. The deduction (p. 5): Proposition 5 with $I=\{r\}$ gives the upper bound for large $t$, extended to all $t\ge t_0(r)$ by enlarging the implied constant; Proposition 6 applied to a representation with largest element $x_1$ gives $t\le(1-e^{-r})x_1-\delta(r)x_1\log\log x_1/\log x_1$, hence the lower bound (9). Proposition 5 is itself reduced (pp. 5--6) to Proposition 7 (a set $\mathcal R\subset[x/2e^r,x]$ of prescribed size whose reciprocal sum leaves a remainder $a/b$ with $1/\log x<a/b<1$ and $P^*(b)\le x^{1/5}$) and Proposition 8 (representing such a remainder by $2\pi^*(y)$ integers in $[1,2y^4]$). Proposition 6 is proved in Section 3 (p. 8), Proposition 7 in Section 4 (p. 14) and Proposition 8 in Section 5 (p. 18); the method classifies denominators by the size of their largest prime power ("very large", "large" and "small" prime powers).

Dependencies

Same-paper Propositions 5--8 and Lemmas 9--17; the paper says its methods combine Croot's techniques (the paper's [2]) with the author's earlier paper [8] (Dense Egyptian fractions, Trans. Amer. Math. Soc., arXiv:math/9804045).

Bears on

• Problem 285: with $r=1$ it gives $f(k)=(1+o(1))\frac{e}{e-1}k$, the asymptotic the problem asks for; display (9) (p. 5) gives $f(k)\ge\frac{e}{e-1}k+\delta(1)\,k\log\log k/\log k$ for large $k$, so $f(k)-\frac{e}{e-1}k$ has exact order $k\log\log k/\log k$.
• Problem 286: with $r=1$, every large $k$ has a $k$-term representation of $1$ with all denominators in $[2,M_k(1)]$, an interval of width below $(e-1)k$ since $e/(e-1)<e-1$; the deduction is the corpus's own, not the paper's.
• Problem 292: context only; the Erdős--Graham density question on the possible largest denominators is Theorem 4, and Theorem 3 treats the second-largest and later positions.