Short Egyptian fractions and exact denominators

Source and verification

This article explains family 025 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Distinct unit fractions

An Egyptian-fraction expansion expresses a positive rational as a finite sum of distinct unit fractions. For $1\le a<b$, let $N(a,b)$ be the least possible length in

$$\begin{array}{c}\frac{a}{b}=\frac{1}{n_1}+\cdots +\frac{1}{n_k},\\ 2\le n_1<\cdots <n_k.\end{array}$$

Set $N(b)={\max }_{1\le a<b}N(a,b)$. The input need not be reduced, and there is no upper bound on the denominators used.

For example,

$$\frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20}.$$

This is an expansion, not a claim about minimality. The greedy algorithm repeatedly chooses the largest unit fraction no greater than the current remainder. It proves existence, but its elementary length bounds do not establish the optimal uniform order below.

Claimed shortest-length order

The manuscript asserts absolute constants $c_1,c_2>0$ such that, for all sufficiently large $b$,

$$c_1\log \log b\le N(b)\le c_2\log \log b.$$

The upper bound holds for every numerator. The matching lower order is classical; it does not hold separately for every numerator, since $N(1,b)=1$. (OpenAI, 2026)

🏷️ A one-term reduction by divisors

The source introduces a common auxiliary denominator $M$. At one stage it represents $u/(MQ)$, where $Q$ is either a fixed denominator factor or one. Choose a divisor $t\mid M$ and set

$$z=\lceil \frac{Qt}{u}\rceil ,h=uz-Qt.$$

Then $0\le h<u$, and the exact identity is

$$\frac{u}{MQ}=\frac{1}{(M/t)z}+\frac{1}{z}\frac{h}{MQ}.$$

To check it, combine the right side over denominator $MQz$: its numerator is $Qt+h=uz$.

One unit fraction therefore replaces the numerator $u$ by a smaller residue $h$. A divisor $t$ is useful when $-Qt\mathrm{mod}u$ is particularly small. The remainder’s expansion can then be scaled by $z$.

This identity is elementary. The main research problem is constructing one $M$ with enough useful divisors for enough numerators at every stage, while keeping the total number of stages small.

🏷️ Dense families and exceptional sets

Write $S=\log b$. A short greedy prefix either finishes the expansion or leaves a remainder $A/C$ in a controlled denominator range. The source then chooses $M=\exp (O(S))$ and builds, through $O(\log S)$ descending numerator ranges, a dense family of short expansions.

At large ranges, products of distinct primes give residue distribution. At smaller ranges, auxiliary random products provide the needed distribution. At the bottom, a uniform reduction and binary expansion finish the remaining numerators.

The obstruction is accumulation of exceptions. Even if most $h$‘s have good expansions at one stage, many $u$‘s could funnel into a small bad set. For fixed $t,h$, the identity implies

$$u\mid Qt+h.$$

Thus the number of predecessors of a bad numerator is controlled by a divisor-counting problem. A uniform moment estimate for small divisors in shifted intervals, followed by Hölder’s inequality, gives a recurrence keeping exceptional proportions small throughout the descent.

The construction fixes its constants, then a sufficiently large moment order, and only then a sufficiently large $S$. Treating a moment estimate with an uncontrolled dependence on these choices would not justify the iteration.

🏷️ From density to every numerator

A dense set of integers in an interval can have a sumset covering a shorter central interval. Indeed, if $T$ and $s-T$ occupy more than half of a common interval, they must intersect; an intersection gives $s=u+v$ with $u,v\in T$.

The source selects a suitable multiple $gAM$ in such a central range and writes it as two good numerators. Adding their expansions over $MC$ and dividing by $g$ gives

$$\frac{u+v}{gMC}=\frac{gAM}{gMC}=\frac{A}{C}.$$

This is why a dense-family theorem can yield a result for every original numerator. The scaling and the central-range placement are part of the argument, rather than consequences of density alone.

Intermediate lists may repeat denominators. The manuscript includes a separate lemma removing repetitions when the total is below one; the counting application at total one needs its own treatment. Distinctness cannot be silently assumed after adding two expansions.

🏷️ Exact lengths and prescribed denominators

Let $F(k)$ count expansions of one into exactly $k$ distinct unit fractions. The source deduces

$$\log \log F(k)=\Theta (k).$$

A denominator rich in divisors creates many splitting choices. An injective padding operation then reaches every sufficiently large exact length without identifying different expansions.

For a stronger marker problem, let $v(k)$ be the least integer $m\ge 2$ absent from every exact $k$-term expansion of one. The source claims, eventually,

$$\begin{array}{c}\exp (\exp (k/600))\\ \le v(k)\\ \le 1+k^{2^{k-1}}.\end{array}$$

The construction must preserve the exact term $1/m$. Keeping a denominator merely divisible by $m$ would not suffice. A marker-avoiding greedy prefix and a separate padding lemma retain that term. The paper further gives lower asymptotic slope $(\log 2)/257$ for $\log \log v(k)/k$.

These conclusions optimize length while leaving denominator size unrestricted.

🏷️ Key steps and proof outlines

Preparing a denominator costs only doubly logarithmic length

For a remainder $x=A/C<1$, a greedy step uses $z=\lceil C/A\rceil$ and leaves $(Az-C)/(Cz)$. Its numerator decreases, and its denominator is at most $C^2$.

Proof outline. The remainder satisfies

$$0\le x-\frac{1}{\lceil 1/x\rceil }\le \frac{x^2}{1+x}.$$

After the first positive step it is below $1/2$, and after $j$ positive steps it is at most $2^{-2^{j-1}}$. Before the denominator reaches a target $T$, a positive integer numerator forces the remainder to exceed $1/T$. Thus after $O(\log \log T)$ steps either the process terminates or the denominator first enters $[T,T^2)$. Taking $T$ to be a fixed power of $b$ costs $O(\log \log b)$ terms and places the remainder in the range needed by the divisor construction. (OpenAI, 2026)

Count predecessors of bad residues, not just their proportions

The source constructs lists of divisors of one auxiliary denominator $M$. At descent level $j$, most numerators $u\le X_j$ have many list indices $i$ for which the remainder $h=uz-C_j{t}_{j,i}$ lies below $X_{j+1}$. A good $h$ gives a one-term extension of its already constructed expansion. Lists may contain repeated values, so the index $i$ is part of the data.

Proof outline. Let $H_j$ be the bad numerators and ${\delta }_j=|H_j|/X_j$. If $u$ is bad but not in the small Fourier-exception set, all its many short remainders must lie in $H_{j+1}$. For a fixed pair $(h,i)$, the identity forces $u\mid C_j{t}_{j,i}+h$. Thus its number of predecessors is at most the truncated divisor count $d_{X_j}(C_j{t}_{j,i}+h)$. Count the edges of this bipartite relation from both sides. A uniform $r$th-moment estimate for the divisor counts and Hölder’s inequality bound the sum over the small bad set by its size to the power $\alpha =1-1/r$. The resulting recurrence is

$${\delta }_j\le ϵ+A{\delta }_{j+1}^{\alpha },ϵ=e^{-c_{\ast }m},A=e^{2S^{1/4}},S=\log b.$$

The bottom level has ${\delta }_d=0$, and $d=O(\log S)$. Choose the fixed moment order $r$ sufficiently large that ${\alpha }^d\ge S^{-1/4}$. Using $(x+y{)}^{\alpha }\le x^{\alpha }+y^{\alpha }$ to iterate gives

$${\delta }_0\le d\exp \left(2r{S}^{1/4}-c_{\ast }m{S}^{-1/4}\right).$$

Here $m\ge S/(2\log S)$, so the negative term has order at least $S^{3/4}/\log S$ and dominates. Consequently the good set eventually occupies at least seven eighths of its interval. This calculation is the mechanism preventing a small bad set from accumulating too many predecessors over the descent.

Two good numerators recover every prescribed numerator

Let $G\subseteq [1,X]\cap \mathbb{Z}$ be the good set, with at most $X/8$ missing integers. For the prepared remainder $A/C$, the construction ensures $AM<X$. Put $g=\lfloor X/(AM)\rfloor$ and $n=gAM$.

Proof outline. Since $\lfloor t\rfloor \ge t/2$ for $t\ge 1$, $X/2\le n\le X$. Among the $n-1$ positive decompositions $n=u+(n-u)$, at most twice the number of missing integers have a bad summand, hence at most $X/4$. For large $X$, some decomposition has both summands in $G$. Add their short expansions over $MC$, then multiply every denominator by $g$. Their sum is $(u+n-u)/(gMC)=A/C$. Adjoin the initial greedy prefix. This upgrades a dense-family construction to a uniform result without assuming the original numerator itself was good.

Removing repeated denominators preserves the length

For a total strictly below one, repeated terms can be replaced using

$$\begin{array}{rl}\frac{2}{2p}& =\frac{1}{p+1}+\frac{1}{p(p+1)},\\ \frac{2}{2p+1}& =\frac{1}{p+1}+\frac{1}{(p+1)(2p+1)}.\end{array}$$

Proof outline. A repeated denominator two is impossible when the total is below one. In every allowed replacement the number of terms is unchanged and the sum of their denominators increases strictly. For fixed positive total and fixed length, there are only finitely many nondecreasing denominator lists: the first denominator is bounded by length divided by total, and induction bounds each subsequent choice. Hence the replacements terminate with distinct denominators. This supplies the distinctness omitted by the addition-and-scaling argument; denominator sizes need not remain controlled.

Divisor branching and reversible padding give many exact-length expansions

For a denominator $n$ with many divisors, each proper divisor $d$ supplies

$$\frac{1}{n}=\frac{1}{n+d}+\frac{1}{n+n^2/d}.$$

Proof outline. The two new denominators lie on opposite sides of $2n$. Each unchanged denominator forbids at most two choices of $d$. Starting from the source’s short expansion containing a multiple of a product of $r$ odd primes leaves roughly $2^r$ distinct choices, all at the same length $O(\log r)$. The smaller new denominator recovers $d$, so they really are different expansions. To reach a larger exact length, repeatedly split the largest denominator $v$ into $v+1$ and $v(v+1)$. These become the two largest denominators, so the operation is reversible and preserves the count.

For a prescribed exact denominator, the largest-term split may remove the marker. The source instead splits a nonmarked term. If splitting the second-largest $s$ would hit the marked largest value, a factorization $s=ab$ permits the alternative split into $s+a$ and $b(s+a)$; the inequalities between these and $s+1,s(s+1)$ avoid the collision. The remaining prime-denominator case is excluded arithmetically in the marker-padding lemma. This is why preserving a multiple of a marker and preserving the marker itself require different constructions.

🏷️ Connections

Uniform residue avoidance also requires estimates valid for every prescribed input, rather than a typical one. Totient counting illustrates a different situation where representation multiplicities must be controlled before counting distinct objects.

🏷️ Manuscripts and verification scope

• Short Egyptian fractions — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects shortest-length order, exact-length counts, and the prescribed-denominator results. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. Short Egyptian fractions.