==== TABLE OF CONTENTS ====
Short-Egyptian-fractions//counting.tex
Short-Egyptian-fractions//descent.tex
Short-Egyptian-fractions//divisors.tex
Short-Egyptian-fractions//elementary.tex
Short-Egyptian-fractions//introduction.tex
Short-Egyptian-fractions//main.tex
Short-Egyptian-fractions//prescribed.tex
Short-Egyptian-fractions//random.tex
Short-Egyptian-fractions//residues.tex
Short-Egyptian-fractions//unicode-mappings.tex
Short-Egyptian-fractions//references.bib
===========================
==== BEGIN Short-Egyptian-fractions//counting.tex ====
\section{Counting representations of one}\label{sec:counting}
We prove Corollary~\ref{cor:counting} using the uniform upper bound
in Theorem~\ref{thm:main}, together with the prime number theorem and
elementary identities. Divisor-rich denominators and divisor-indexed
splitting are also used by Konyagin and Elsholtz
\cite{Konyagin2014,Elsholtz2016}. The first step is
to obtain a short distinct expansion of one with a denominator having
many divisors. Lemma~\ref{lem:distinct} concerns totals below one; at
total one we need a different repetition-removal argument, preserving
an odd divisor of a denominator.
\begin{lemma}\label{lem:marked-cleanup}
Let $Q>1$ be odd. Suppose that a finite list of positive integers
$m_1,\ldots,m_t$ satisfies $\sum_{i=1}^t1/m_i=1$ and contains a
multiple of $Q$. Then $1$ has a representation by at most $t$ distinct
positive unit fractions, still with a denominator divisible by $Q$.
\end{lemma}
\begin{proof}
Whenever a denominator $m$ occurs twice, replace those two terms by
\[
\frac2m=\frac1{m/2}\quad(m\text{ even}),
\qquad
\frac2m=\frac1{(m+1)/2}+\frac1{m(m+1)/2}
\quad(m\text{ odd}).
\]
The sum stays equal to one and the length does not increase.
There remains a denominator divisible by $Q$: if no such denominator
is removed it persists, while if $Q\mid m$, the even replacement
$m/2$ is a multiple of $Q$ because $Q$ is odd, and the larger odd
replacement is a multiple of $m$.
This invariant excludes a denominator one, since that would force the
singleton list $(1)$. It also excludes a repeated denominator two,
since two halves exhaust the sum and neither denominator is divisible
by $Q$. Thus an odd repeated denominator is at least three. In an odd
replacement the smaller new denominator is strictly below $m$ and the
larger is strictly above $m$. The sorted denominator tuple therefore
strictly decreases lexicographically: all entries below $(m+1)/2$
are unchanged, and one extra copy of $(m+1)/2$ is inserted.
At a fixed length such a descent in positive integer tuples terminates.
Indeed, the first coordinate can decrease only finitely often, and
after it stabilizes the same argument applies successively to the other
coordinates. Even replacements strictly reduce length, so only finitely
many occur. The procedure consequently terminates with distinct
denominators and preserves the required multiple of $Q$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:counting}]
\emph{A short initial expansion.}
Let $r$ be a sufficiently large integer and let $Q$ be the product of
the first $r$ odd primes. The prime number theorem, already used in
Section~\ref{sec:residues}, gives
\[
\log Q=O(r\log r),\qquad \log\log Q=O(\log r).
\]
Apply Theorem~\ref{thm:main} to $(Q-1)/Q$ and append $1/Q$.
Lemma~\ref{lem:marked-cleanup} produces a distinct expansion of one
whose denominator set $S$ has size $s=O(\log r)$ and contains a
multiple $n$ of $Q$. Fix this one set $S$ and this one $n$.
Writing $\tau(n)$ for the number of positive divisors of $n$, we have
$\tau(n)\ge 2^r$.
\emph{Branching over divisors.}
For every positive proper divisor $d$ of $n$, the identity
\begin{equation}\label{eq:divisor-split}
\frac1n=\frac1{n+d}+\frac1{n+n^2/d}
\end{equation}
gives two distinct integer denominators, with
\[
n1$,
so it is not the singleton $\{1\}$; the operation applies to all
the expansions just constructed.
Choose an absolute constant $B\ge1$ such that $\ell\le B\log r$ for
all sufficiently large $r$. For each sufficiently large integer $k$,
take $r=\lfloor\exp(k/(2B))\rfloor$. Then $\ell\le k$, and applying
\eqref{eq:padding-split} exactly $k-\ell$ times to each of our
common-length expansions gives
\[
F(k)\ge 2^r-2s+1\ge 2^{r-1}.
\]
Since $r\ge\tfrac12\exp(k/(2B))$ for large $k$, this proves
$\log\log F(k)\ge k/(2B)+O(1)$, hence the desired lower bound
for every sufficiently large integer $k$.
\emph{The upper bound.}
Fix a tuple counted by $F(k)$ and put $P_0=1$ and
$P_i=n_1\cdots n_i$. Before the $i$th term, the positive remainder
\[
R_i=1-\sum_{je^S\), and the first random
list otherwise. Its entries may coincide; their indices are always
retained. Having defined \(G_{j+1}\), let
\begin{equation}\label{eq:good-backwards}
\begin{aligned}
G_j=G_{j+1}\ \cup\ \bigl\{u\in\N:\ &1\le u\le X_j,\\
&h_{j,t_{j,i}}(u)\in G_{j+1}\cup\{0\}
\text{ for some }i\in I_j\bigr\}.
\end{aligned}
\end{equation}
These sets have the asserted expansions. For membership inherited
directly from \(G_{j+1}\), divide the known expansion by the integer
\(C_j/C_{j+1}\). For membership obtained from a residue, put
\[
z=\left\lceil\frac{C_jt_{j,i}}u\right\rceil,\qquad
h=uz-C_jt_{j,i}.
\]
Then
\begin{equation}\label{eq:rescaled-descent}
\frac{u}{MC_j}
=\frac1{(M/t_{j,i})z}
+\frac1{zC_j/C_{j+1}}\frac{h}{MC_{j+1}}.
\end{equation}
Both \(M/t_{j,i}\) and \(zC_j/C_{j+1}\) are positive integers.
This includes the possible transition from \(C_j=C\) to
\(C_{j+1}=1\), where the second scaling factor is \(zC\).
If \(h=0\), the second term is absent. Otherwise it uses the known
expansion for \(h\in G_{j+1}\). Consequently every member of \(G_j\)
has an expansion of length at most
\begin{equation}\label{eq:descent-length}
B_0\log S+d-j.
\end{equation}
It remains to prove that almost all integers up to \(X_0\) belong to
\(G_0\). We have built expansions whenever a residue reaches a
previously treated numerator; the following count controls all other
numerators.
\paragraph{Counting unsuccessful numerators.}
Define
\[
H_j=\{1,\ldots,\lfloor X_j\rfloor\}\setminus G_j,
\qquad
\delta_j=\frac{|H_j|}{X_j}.
\]
Thus \(0\le\delta_j\le1\) and \(\delta_d=0\).
Fix \(j\sqrt Y\), order its prime factors with multiplicity and let
\(d\) be the longest initial product at most \(\sqrt Y\), allowing
\(d=1\). Let \(p\) be the next prime. Then
\[
d\le\sqrt Y\sqrt Y\) forces \(d\) to
be a large product of small primes. The total reciprocal weight of
such prefixes is small, so \eqref{eq:divisor-multiples} limits their
occurrence in every shifted interval.
If \(\log p\ge v^{15/16}\), the exponential factor in
\eqref{eq:divisor-prefix} is at most \(e^{rBv^{1/16}}\).
For \(n\le\sqrt Y\), instead take \(d=n\) and use
\(d_X(n)\le\tau(d)\). By \eqref{eq:divisor-multiples} and
\eqref{eq:divisor-harmonic}, these two cases together contribute at most
\begin{align}
2Ye^{rBv^{1/16}}\sum_{d\le\sqrt Y}\frac{\tau(d)^r}{d}
&\le Y\exp\!\left(
O_{D,r}(S^{1/16}\log\log S+\log S)\right).
\label{eq:divisor-large-contribution}
\end{align}
For the remaining integers, \(\log p\tfrac12\log Y-\log p
\ge v/4-v^{15/16}>v/5
\end{equation}
once \(S\) is sufficiently large.
This large prefix makes integers with small next prime rare.
We estimate the ensuing smooth-prefix sums by Rankin's exponential
weighting method; see \cite[p.~414, (1.3)]{HildebrandTenenbaum1993}.
If \(p1\) gives
\[
\sum_{\substack{d>e^{v/5}\\
\ell\mid d,\ \ell\ {\rm prime}\Rightarrow\elle^{v/5}\\
\ell\mid d,\ \ell\ {\rm prime}\Rightarrow\ell\le e^{2t}}}
\frac{\tau(d)^r}{d}.
\]
The last sum is at most
\[
e^{-\lambda v/5}
\prod_{\substack{\ell\le e^{2t}\\\ell\ {\rm prime}}}
\left(1+\sum_{j\ge1}(j+1)^r\ell^{-j(1-\lambda)}\right).
\]
By \eqref{eq:divisor-local-factor} and
\eqref{eq:prime-reciprocals}, the logarithm of this product is
bounded by
\[
C_r\sum_{\substack{\ell\le e^{2t}\\\ell\ {\rm prime}}}
\ell^{-1+\lambda}
\ll_r e^{2t\lambda}\log(2t)
=Q^{1/5}O_r(\log(2t)).
\]
The contribution of the interval is consequently at most
\begin{equation}\label{eq:divisor-band}
2Y\exp\left(rBQ-\frac{Q\log Q}{50}
+O_r(Q^{1/5}\log(2t))\right).
\end{equation}
Both positive terms in this exponent are uniformly negligible compared
with \(Q\log Q\). Indeed, \eqref{eq:divisor-B} and
\(Q\ge v^{1/16}\) give
\[
\frac{rB}{\log Q}
\ll_{D,r}\frac{\log\log S}{\log S}\longrightarrow0,
\qquad
\frac{Q^{1/5}\log(2t)}{Q\log Q}
\ll v^{-1/20}\longrightarrow0.
\]
Thus \eqref{eq:divisor-band} is at most
\(2Y\exp(-Q\log Q/100)\le2Y\), after increasing \(S_0(D,r)\).
Summing the intervals costs \(O_D(Y\log S)\).
Combining this bound with \eqref{eq:divisor-large-contribution}
and \eqref{eq:divisor-small-contribution} gives
\[
\sum_{1\le h\le Y}d_X(N+h)^r
\le Y\exp\!\left(
O_{D,r}(S^{1/16}\log\log S+\log S)\right)
+2Ye^{-v/100}+O_D(Y\log S).
\]
The logarithmic loss is \(o(S^{1/4})\), proving
\eqref{eq:divisor-moment}. Every enlargement of the threshold depended
only on the fixed \(D,r\); this completes the required uniformity check.
\end{proof}
==== END Short-Egyptian-fractions//divisors.tex ====
==== BEGIN Short-Egyptian-fractions//elementary.tex ====
\section{From a dense set to every numerator}\label{sec:elementary}
The main construction produces short expansions for most numerators over a
carefully chosen denominator. We first state exactly what is needed from
that construction, and then show why it suffices for every numerator.
\begin{proposition}[A dense set of short expansions]\label{prop:density}
There are absolute constants \(D_M>1\), \(L>0\), and \(S_0>1\) with the
following property. Put \(D_X=D_M+2\) and \(D_C=4D_X\).
For every real \(S\ge S_0\) and every integer \(C\) satisfying
\[
e^{D_CS}\le C\le e^{2D_CS},
\]
there are an integer \(M\) and a set
\(G\subseteq\{1,\ldots,\lfloor X\rfloor\}\), where \(X=e^{D_XS}\), such that
\[
e^S0\) and \(k\), there are only finitely many nondecreasing
lists of \(k\) denominators with reciprocal sum \(x\).
Indeed, the first denominator is at most \(k/x\).
Once it is chosen, induction applies to the remaining positive sum and
the remaining \(k-1\) terms; when no terms remain the sum must be zero.
If a denominator occurs twice, apply one of the identities
\[
\frac2{2p}=\frac1{p+1}+\frac1{p(p+1)},
\qquad
\frac2{2p+1}=\frac1{p+1}+\frac1{(p+1)(2p+1)}.
\]
Because the total is less than \(1\), a denominator \(1\), or a repeated
denominator \(2\), is impossible. Thus \(p\ge2\) in the first identity
and \(p\ge1\) in the second. Each replacement preserves the number of
terms and increases the sum of their denominators: the increases are
\((p-1)^2\) and \(2p^2\), respectively. Finiteness of the set of lists
therefore forces termination, at which point all denominators are
distinct. Positivity and the unchanged total exclude a denominator
\(1\) throughout.
\end{proof}
\begin{lemma}[Preparing the denominator]\label{lem:greedy}
Let \(1\le ab\) be real.
After at most
\[
1+\left\lceil\log_2\left(\frac{\log T}{\log 2}\right)\right\rceil
\]
greedy steps, either \(a/b\) has been expressed as a sum of unit fractions,
or its positive remainder can be written as \(A/C\), with integers
\[
1\le A\le a,\qquad T\le C1/T\). This is impossible when
\(2^{j-1}\log2\ge\log T\), proving the stated bound.
Every current positive value is less than \(1\), so \(z\ge2\).
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}, assuming
Proposition~\ref{prop:density}]
For the upper bound, fix \(1\le aX/4\) for large \(S\),
some split has both entries in \(G\).
Their expansions together express
\[
\frac{n}{MC}=\frac{gA}{C}
\]
using at most \(2L\log S\) unit fractions. Multiplying every denominator
by the integer \(g\) gives an expansion of \(A/C\) of the same length.
Adjoin the greedy prefix and apply Lemma~\ref{lem:distinct} to the total
\(a/b<1\). The resulting distinct expansion has \(O(\log S)\) terms,
uniformly in \(a\). No reduction of \(a/b\), or restriction on the final
denominator sizes, has been used.
For the lower bound, consider an expansion of \((b-1)/b\) with \(k\)
terms and append \(1/b\). Sort the resulting \(s=k+1\) denominators as
\(d_1\le\cdots\le d_s\); repetitions are harmless here.
Put \(L_0=1\) and \(L_j=d_1\cdots d_j\).
For each \(1\le j\le s\), the amount remaining after the first \(j-1\)
terms is positive and has denominator dividing \(L_{j-1}\).
Consequently
\[
\frac1{L_{j-1}}
\le 1-\sum_{i0$ and $b_0$ such that, for every
integer $b\ge b_0$,
\[
c_1\log\log b\le N(b)\le c_2\log\log b.
\]
In particular, the upper bound holds for every integer numerator
$1\le a0$ and every sufficiently large $b$, every $a/b\in(0,1)$
has a distinct expansion of length at most
$(1+\varepsilon)\log b/\log\log b$, with all denominators at most
$4b(\log b)^2\log\log b$ \cite[Theorem, p.~151]{TenenbaumYokota1990}.
Their length estimate serves this simultaneous constraint.
Here we optimize only the order of the length and leave the
denominator sizes unrestricted.
The uniform theorem also determines the double-logarithmic order of
the number of representations of one. For each positive integer $k$, put
\[
F(k)=\#\left\{(n_1,\ldots,n_k):
1\le n_1<\cdots0$ and $k_0$ such that, for every
integer $k\ge k_0$,
\[
ck\le\log\log F(k)\le Ck.
\]
\end{corollary}
Erd\H{o}s and Graham asked for good estimates for this counting function
\cite[p.~32]{ErdosGraham1980}. Konyagin stated a lower bound with
double logarithm of order $k/\log k$ \cite[Theorem~1]{Konyagin2014};
Elsholtz obtained this lower order even when every denominator is
congruent to $1$ or $-1$ modulo a fixed squarefree integer $P>1$:
the bound holds for every sufficiently large $k$, with $k$ required
to be odd when $P$ is even, and its constant may depend on $P$
\cite[Theorem~1.1]{Elsholtz2016}. Elsholtz and Planitzer proved an upper
bound with double logarithm $O(k)$, allowing repetitions as well
\cite[Corollary~3(2)]{ElsholtzPlanitzer2021}.
Corollary~\ref{cor:counting} determines the double-logarithmic order for
unrestricted distinct denominators.
We can also ask which exact denominators occur among these expansions.
For each integer $k\ge1$, let $D_k$ be the set of integers $m\ge2$
for which there are positive integers $n_1<\cdots0$ \cite[Theorem~1.1]{vanDoornTang2026}. They also show
that an exact prescribed denominator can be
retained while increasing the length of a distinct expansion
\cite[Lemma~2.1]{vanDoornTang2026} and anticipate the connection
between uniform short expansions and a double-exponential lower bound
for $v(k)$ \cite[Section~3]{vanDoornTang2026}. We give the required
marker-avoidance argument below, thereby deriving the qualitative order
from Theorem~\ref{thm:main}. A direct construction additionally gives
the numerical lower slope in Corollary~\ref{cor:prescribed}, without
quantifying the constant $c_2$ in the uniform theorem. The elementary
upper bound is convenient rather than best known: sharper public upper
bounds follow from the counting estimates of Elsholtz and Planitzer
\cite[Corollary~3(2)]{ElsholtzPlanitzer2021}.
The two consequences impose different requirements on a construction.
To obtain many expansions, it suffices to retain a denominator divisible
by an integer with many prime factors. To include a prescribed integer
$m$ in $D_k$, the exact term $1/m$ must survive. The latter requirement
governs both the initial construction and the operation that increases
its length.
\subsection*{The proof in outline}
Put $S=\log b$. After $O(\log S)$ greedy steps, either the expansion
is complete or its remainder has the form $A/C$, where $A\le b$ and
$C$ lies in a prescribed exponential range in $S$.
We then construct an auxiliary integer $M$ with $e^S0$ and $0\le R<2m$ such that
\[
1=\frac1m+\sum_{i=1}^{j}\frac1{n_i}+\frac Rq,
\qquad q=m\prod_{i=1}^{j}n_i,
\qquad
j<3+\log_2\log_2 T.
\]
Writing $q_0=m$ and $q_i=m\prod_{h=1}^{i}n_h$ for $1\le i\le j$,
we have $n_i\le q_{i-1}+1$ for every $1\le i\le j$.
If $R>0$, then
\[
q\ge T,\qquad
\frac Rq<\frac{2m}{T}\le\frac1m,
\qquad \frac Rq<\frac1{n_i}\quad(1\le i\le j).
\]
In either case $q0$ and $q_in_{r-2}\ge2$ and
in particular $p$ is odd. All denominators other than $p$ and $v$
are below $p$; write the sum of their reciprocals as $A/B$ with
$p\nmid B$. The two exceptional choices give respectively
\[
1-\frac AB
=\frac1p+\frac1{p+1}
=\frac{2p+1}{p(p+1)},
\qquad\text{or}\qquad
1-\frac AB
=\frac1p+\frac1{p(p+1)}
=\frac{p+2}{p(p+1)}.
\]
After clearing denominators and reducing modulo $p$, these equations
give respectively $0\equiv B\pmod p$ and $0\equiv2B\pmod p$,
both impossible. Thus in every valid case an unmarked term can be
replaced by two distinct terms, preserving $m$ and increasing the
length by exactly one.
\end{proof}
\section{A quantitative prescribed-denominator bound}
\label{sec:prescribed-quantitative}
The uniform theorem gives the short marked expansion above with an
unspecified absolute length constant. We now prove the following
numerical estimate by constructing the tail directly.
\begin{proposition}\label{prop:marked-length}
For every $\varepsilon>0$ there is an integer $m_\varepsilon$ such
that every integer $m\ge m_\varepsilon$ occurs as an exact denominator
in a representation of $1$ by distinct positive unit fractions with
at most
\[
\left(\frac{257}{\log 2}+\varepsilon\right)\log\log m
\]
terms.
\end{proposition}
The construction has two ingredients. A common integer $K_m$ supplies
short representations of small integers by positive rationals whose
numerators divide $K_m$. The unreduced denominator of the greedy
remainder has divisors at every multiplicative scale. Combining these
properties gives a short expansion of the tail: divisors of the greedy
denominator control how many pieces are needed, and the rational
representations supplied by $K_m$ bound the number of unit fractions
used for each piece.
\subsection{A common supply of rational divisors}
We construct a moderately sized integer whose divisors, after division by
positive integers, represent every integer in a prescribed initial interval
with a bounded number of summands. The denominators of these rational
summands need not agree. This freedom lets us first represent a positive
multiple of a prime and then divide by its integer multiplier.
\begin{lemma}\label{lem:rational-divisor-supply}
For every sufficiently large integer $m$, put
\[
L=\log\log m,\qquad Y=m^4,\qquad
\ell(v)=\left\lceil\frac{4\log v}{\log 2}\right\rceil
\quad(v\geq2).
\]
Let $P(v)$ be the product of the first $\ell(v)$ primes, and define
\[
K_m=P(Y)^2\,\lfloor L\rfloor!.
\]
Then $K_m>m$ and
\begin{equation}\label{eq:supply-size}
\log K_m\leq
\left(\frac{32}{\log 2}+o(1)\right)\log m\log\log m
\qquad(m\longrightarrow\infty).
\end{equation}
Moreover, every integer $1\leq s\leq Y$ has a representation
\begin{equation}\label{eq:rational-supply}
s=\sum_{i=1}^{h}\frac{e_i}{t_i},\qquad
h\leq16,\quad e_i\mid K_m,\quad e_i,t_i\in\mathbb Z_{>0}.
\end{equation}
Repeated summands are allowed.
\end{lemma}
\begin{proof}
The prime number theorem \cite[p.~305, (1.1)]{Selberg1949} gives
\[
\log P(v)=(1+o(1))\ell(v)\log\ell(v)
\ll\log v\log\log v.
\]
Since $\ell(Y)=16\log m/\log 2+O(1)$, and
$\log(\lfloor L\rfloor!)=O(L\log L)$, this proves
\eqref{eq:supply-size}. Also $P(Y)\geq2^{\ell(Y)}\geq Y^4$,
so $K_m>m$ for large $m$.
We next prove an assertion whose threshold is independent of $m$.
Fix a sufficiently large odd integer $u$, and let $A$ be the set of positive
divisors of $P(u)$. Write $N=|A|=2^{\ell(u)}\geq u^4$.
We estimate exceptional primes by counting congruent pairs, the
mechanism underlying Gallagher's larger sieve \cite{Gallagher1971}.
The coarse count below suffices.
For each prime $p\leq u$, let $A_p$ be the image of $A$ in
$\mathbb F_p$. Call $p$ exceptional when $|A_p|\leq p^{3/4}$.
If $c_a$ is the number of members of $A$ in residue class $a$, then
Cauchy--Schwarz shows that the number of ordered pairs of distinct members
of $A$ congruent modulo an exceptional prime is at least
\[
\sum_{a\in\mathbb F_p}c_a^2-N
\geq\frac{N^2}{|A_p|}-N
\geq\frac{N^2}{2u^{3/4}}.
\]
For any two distinct members $a,b\in A$, the nonzero integer $|a-b|0$ such that every sufficiently large odd integer $u$ has
at least $cu^2/(\log u)^3$ ordered representations as a sum of three
primes; see the statement in \cite[(1)--(2)]{Kumchev1997}
and the exposition in \cite[Theorem~1 and Section~3.1]{KumchevTolev2005}.
To make the uniformity in odd $u$ explicit, its singular series is
\[
\mathfrak S(u)=
\prod_{p\mid u}\left(1-\frac{1}{(p-1)^2}\right)
\prod_{p\nmid u}\left(1+\frac{1}{(p-1)^3}\right).
\]
For odd $u$ its factor at $2$ is $2$, and
\[
\mathfrak S(u)\geq
2\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)>0.
\]
The infinite product is positive because the sum of the subtracted
quantities converges. Thus the asymptotic formula supplies an absolute
lower bound with an absolute threshold, with no dependence on the
factorization of $u$.
At most $3uB(u)$ ordered prime triples summing to $u$ contain an
exceptional prime: choose its position, its value, and one further entry;
the last entry is then fixed. By \eqref{eq:supply-exceptional},
\[
3uB(u)\ll u^{7/4}\log u\log\log u
=o\left(\frac{u^2}{(\log u)^3}\right).
\]
Consequently every sufficiently large odd $u$ is the sum of three primes
that are not exceptional for this same set $A$.
It remains to represent each such prime using divisors of $P(u)^2$.
Fix one of them, say $p$, and put $H=A_p$ and $h=|H|>p^{3/4}$.
Let $U$ be the product of two independent uniformly chosen members of
$H$, and write $\mathrm e_p(z)=\exp(2\pi i z/p)$.
The following classical bilinear estimate is recorded in
\cite[Proposition~1.1]{GlibichukKonyagin2007}; we give its short proof
and the five-product consequence needed here.
For every nonzero $r\in\mathbb F_p$, orthogonality and
Cauchy--Schwarz give
\begin{align*}
\left|\sum_{a,b\in H}\mathrm e_p(rab)\right|^2
&\leq h\sum_{a\in\mathbb F_p}
\left|\sum_{b\in H}\mathrm e_p(rab)\right|^2\\
&=ph^2.
\end{align*}
After division by $h^4$, this says
\[
\left|\mathbb E\,\mathrm e_p(rU)\right|
\leq\frac{\sqrt p}{h}0.
\end{align*}
Here the lower bound follows by bounding the absolute value of the sum
of the nonzero Fourier coefficients.
Choose pairs of residues witnessing this positive probability and lift
each residue to any positive member of $A$ representing it. The five
integer products $e_1,\ldots,e_5$ are positive divisors of $P(u)^2$,
and their sum is divisible by $p$. Hence
\[
t=\frac{e_1+\cdots+e_5}{p}\in\mathbb Z_{>0},\qquad
p=\sum_{i=1}^{5}\frac{e_i}{t}.
\]
The three chosen primes thus represent $u$ in fifteen summands of the
required kind.
For a sufficiently large integer $s$, use $u=s$ if $s$ is odd and
$u=s-1$ otherwise. In the even case append the summand $1=1/1$.
This uses at most sixteen summands. If $s\leq Y$, monotonicity of
$\ell$ implies $P(u)\mid P(Y)$, so every numerator constructed above
divides $K_m$.
Finally, all preceding thresholds concern $s$ alone and are absolute.
Choose an integer $S_0$ above them. For all sufficiently large $m$ we
have $\lfloor L\rfloor\geq S_0$. Each remaining integer
$1\leq s\leq S_0$ then divides $\lfloor L\rfloor!$ and hence $K_m$,
so it has the one-summand representation $s=s/1$. This proves the
assertion simultaneously for every $1\leq s\leq Y$.
\end{proof}
\subsection{Grouping the remaining numerator}
The denominator retained by the greedy procedure has divisors at every
multiplicative scale. We first record this property, then use it to reduce
the remaining numerator to a short list of integers within the range of
the rational-divisor supply.
\begin{lemma}\label{lem:marked-divisor-density}
Let $m\ge2$ be an integer. Suppose that $q$ is obtained from $m$ by a
finite sequence of replacements $q\mapsto nq$, where at each replacement
$n$ is a positive integer satisfying $n\le q+1$. Then, for every real
number $w$ with $1\le w\le q$, there is a positive divisor $d$ of $q$ such
that
\[
\frac wm\le d\le w.
\]
\end{lemma}
\begin{proof}
For the initial value $q=m$, the divisor $1$ works throughout $[1,m]$.
Suppose the property holds for $q$, and put $q'=nq$. If $1\le w\le q$,
use a divisor of $q$, which is also a divisor of $q'$. If $n\le w\le nq$,
apply the property at $w/n$ and multiply the resulting divisor by $n$.
These two cases cover $[1,nq]$ unless $qm^4$ decreases the remainder by a factor
strictly greater than $m^2$. There are therefore at most
$\lceil\log X/(2\log m)\rceil$ such steps, followed by at most one
final group. The number $G$ of groups satisfies
\[
G\le\left\lceil\frac{\log X}{2\log m}\right\rceil+1
\le\frac{\log X}{2\log m}+2.
\]
This bound also covers $X=1$ and a zero remainder before the final group.
For each group, use the assumed representation
$s_i=\sum_{a=1}^{b_i}e_{i,a}/t_{i,a}$, with $b_i\le B$.
Dividing $X=\sum_i d_i s_i$ by $qK_m$ gives
\[
\frac{X}{qK_m}
=\sum_i\sum_{a=1}^{b_i}
\frac{1}{(q/d_i)(K_m/e_{i,a})t_{i,a}}.
\]
Each displayed denominator is a positive integer. This uses the two
divisibilities $d_i\mid q$ and $e_{i,a}\mid K_m$; no divisibility between
$t_{i,a}$ and $e_{i,a}$, or coprimality assumption, is needed. The total
number of terms is at most $BG$.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:marked-length}]
Fix a sufficiently large integer $m$, put $L=\log\log m$, and take
$K_m$ from Lemma~\ref{lem:rational-divisor-supply}. Apply
Lemma~\ref{lem:marked-greedy-prefix} with $T=2mK_m$, which is an
integer at least $2m^2$. It gives
\[
1=\frac1m+\sum_{i=1}^{j}\frac1{n_i}+\frac Rq,
\qquad 0\le R<2m.
\]
The prefix is distinct and avoids $m$. By \eqref{eq:supply-size},
\[
\log\log(2mK_m)\le L+O(\log L),
\qquad
j\le\frac{L}{\log2}+O(\log L)
=\left(\frac1{\log2}+o(1)\right)L.
\]
If $R=0$, the expansion is complete and already satisfies the asserted
length bound. Suppose henceforth that $R>0$.
The denominator $q$ is retained without cancellation, beginning at $m$
and using multipliers at most the preceding denominator plus one.
Lemma~\ref{lem:marked-divisor-density} therefore applies to this $q$.
Moreover $q\ge2mK_m$ and $R<2m$, so the integer $X=RK_m$ satisfies
$1\le X0$ proves the proposition.
\end{proof}
\subsection{Every sufficiently large prescribed length}
\begin{proof}[Proof of Corollary~\ref{cor:prescribed}]
Set $A=257/\log2$. Fix a real number $c$ with $00$ such that $c(A+\varepsilon)<1$.
Proposition~\ref{prop:marked-length} supplies an integer $M\ge4$ such
that every integer $m\ge M$ has a distinct marked expansion with
at most $(A+\varepsilon)\log\log m$ terms. The construction in
Section~\ref{sec:prescribed} also gives a finite distinct marked
expansion for each $2\le mu^{0.095}\).
Call a sample contributing to \(U_0\) a hit if its prime value divides
\(u\), and count hits with their sample multiplicities. If their number
is \(B_0\), then
\[
\gcd(U_0,u)\le \prod_{\text{hit samples }p}p
\le S^{(K+1)B_0}.
\]
Thus failure requires at least
\[
k_0=\left\lceil\frac{0.08V}{(K+1)\log S}\right\rceil
\]
hits. At most \(S\) primes in the sampling interval divide \(u\), so
each sample has hit probability at most
\(S/P\le S^{-(K-2)}\). There are at most \(m\le S\) independent samples
in \(U_0\). A union bound over sets of \(k_0\) positions yields
\begin{equation}\label{eq:random-gcd}
\Prob(qm-2s\), the event is empty. This argument allows prime powers
in \(u\) and repeated sampled primes.
Now fix an exposure with \(q\ge u^{0.9}\). Every possible value of
either fresh product is at most
\[
(2S^K)^s\le S^{(K+1)s}\le u^{0.7575}e^S\), Lemma~\ref{lem:deterministic} and Markov's
inequality show that the number of integers
\(u\in(X_{j+1},X_j]\) for which
\[
\left|2^{-m}\sum_{t\in T_0}\e(lCt/u)\right|>e^{-3\eta m}
\]
at some \(1\le l\le H\) is at most
\[
X_j H e^{6\eta m}e^{-0.01m}
\le X_j e^{-0.009m}.
\]
Conjugation changes the Fourier sign without changing the absolute
value. Lemma~\ref{lem:discrepancy}, with scale \(m\), therefore gives at
least \(\rho|T_0|/2\) entries with
\[
h_{j,t}(u)<\rho u\le X_{j+1}
\]
outside that exceptional set. This holds separately at every such
level and requires no probabilistic selection.
Next suppose \(je^m\), and every
\(u\in(X_{j+1},X_j]\) satisfies
\[
(1-\eta)m<\log u\le S,\qquad
w=\min(m,\log u)\ge(1-\eta)m.
\]
For sufficiently large \(S\), these integers satisfy the lower bound
in Lemma~\ref{lem:random}. Moreover
\(4\eta m\le0.005w\), so that Lemma applies to all \(1\le l\le H\).
In block \(1\), Markov's inequality followed by the frequency union
bound shows that
\begin{align*}
&\Prob\left(
\max_{1\le l\le H}
\left|2^{-m}\sum_{t\in T_1}\e(lt/u)\right|>e^{-3\eta m}
\right)\\
&\hspace{2em}\le
\exp\bigl((10\eta-0.01(1-\eta))m\bigr)
=e^{-0.008999m}\le e^{-0.005m}.
\end{align*}
Let \(B_j\) count integers in this level that fail this Fourier bound.
Linearity of expectation and another application of Markov give
\[
\E B_j\le X_j e^{-0.005m},
\qquad
\Prob(B_j>X_j e^{-0.001m})\le e^{-0.004m}.
\]
As \(d=O(\log S)\), with probability \(1-o(1)\) these bounds hold at
every level under consideration. No independence between different
integers or levels is needed. Here \(C_j=1\), so
Lemma~\ref{lem:discrepancy} again gives the required
\(\rho|T_1|/2\) entries outside the exceptional sets.
Finally fix an integer \(S^{D_0}u^{-3\eta}
\right)\\
&\hspace{2em}\le u^{10\eta-0.01}
=u^{-0.009}\le u^{-0.005}
\end{align*}
in each block \(i\). On success, Lemma~\ref{lem:discrepancy}, with
scale \(\log u\), supplies a list entry for which
\[
u\lceil t/u\rceil-tS^{D_0}}u^{-5}=o(1).
\]
Thus every terminal integer has a suitable divisor, simultaneously,
with probability tending to one.
The sum of this failure probability and the middle-level failure
probability is \(o(1)\). Hence a realization satisfying both sets of
requirements exists. Fix it and the resulting \(M\).
Since \(e^{-0.009m}\le e^{-0.001m}\), the same
\(c_*=0.001\) is valid at every level. This proves both properties for
one common \(M\), for the fixed \(S\) and \(C\).
\end{proof}
==== END Short-Egyptian-fractions//random.tex ====
==== BEGIN Short-Egyptian-fractions//residues.tex ====
\section{Divisors with small residues}\label{sec:residues}
We now construct the auxiliary denominator.
Its useful divisors make most residues small at each of a sequence of
scales. At the smallest scales, several independent lists of divisors
ensure that every numerator has a suitable choice.
The underlying identity is elementary. For positive integers \(M,Q,u,t\)
with \(t\mid M\), set
\[
z=\left\lceil\frac{Qt}{u}\right\rceil,
\qquad h=uz-Qt.
\]
Then \(z\ge1\), \(0\le he^S\).
\subsection{The denominator and the residue statement}
Fix the absolute constants
\begin{equation}\label{eq:residue-constants}
\begin{gathered}
K=100,\quad R=1000,\quad D_0=100000,\quad \eta=10^{-4},\\
D_M=(2R+2)(K+2),\quad D_X=D_M+2,\quad D_C=4D_X.
\end{gathered}
\end{equation}
Here \(K\) sets the prime scale, \(R\) is the number of independent
blocks, \(D_0\) sets the binary cutoff, and \(\eta\) sets the rate of
descent. These numerical choices leave room in the estimates below.
Let \(S\) tend to infinity, put \(m=\lfloor S/\log S\rfloor\), and fix
an integer
\begin{equation}\label{eq:C-range}
e^{D_CS}\le C\le e^{2D_CS}.
\end{equation}
Define
\begin{equation}\label{eq:levels}
\begin{gathered}
\rho=e^{-\eta m},\qquad X_j=e^{D_XS}\rho^j,\\
d=\min\{j\ge0:X_j\le e^m\},\qquad
C_j=
\begin{cases}
C,&X_j>e^S,\\
1,&X_j\le e^S.
\end{cases}
\end{gathered}
\end{equation}
These cutoffs need not be integers. Every set or sum indexed by a
cutoff below consists of integers in the indicated range.
For \(0\le j\le d\), write
\begin{equation}\label{eq:least-residue}
h_{j,t}(u)=u\left\lceil\frac{C_jt}{u}\right\rceil-C_jt.
\end{equation}
In particular, \(h_{j,t}(u)\) is the least nonnegative residue of
\(-C_jt\) modulo \(u\).
\begin{lemma}\label{lem:residues}
For all sufficiently large \(S\) and every integer \(C\) satisfying
\eqref{eq:C-range}, there are an integer \(M\) and indexed lists
\(T_0,T_1,\ldots,T_R\), each with \(2^m\) entries, such that:
\begin{enumerate}
\item
\(e^Se^S\), and let
\(T=T_1\) otherwise. Apart from at most
\(X_j e^{-c_*m}\) integers \(u\in(X_{j+1},X_j]\), at least
\(\rho |T|/2\) entries \(t\in T\) satisfy
\[
h_{j,t}(u)\le X_{j+1},
\qquad c_*=0.001.
\]
\item
For every integer \(S^{D_0}S,
\]
whereas
\[
\log M
\le D_0\log S+\log2+(2R+1)m(K\log S+\log2)
\le D_MS
\]
eventually. The last inequality follows because
\((2R+1)Km\), so
\(\log X_{j+1}=\log X_j-\eta m\ge(1-\eta)\log X_j\).
Moreover \(C_d=1\), and \(C_{j+1}\mid C_j\).
Table~\ref{tab:regimes} summarizes the roles of the lists. The upper
two rows are classified by the level cutoff \(X_j\), so a high level
may include some numerators below \(e^S\). The binary step is carried
out in Section~\ref{sec:descent}.
\begin{table}[ht]
\centering
\small
\begin{tabular}{@{}p{0.40\linewidth}p{0.28\linewidth}p{0.23\linewidth}@{}}
\toprule
Range and denominator factor & Divisors used & Coverage\\
\midrule
High levels: \(X_j>e^S\), \(C_j=C\)
& Deterministic list \(T_0\) & Most numerators at each level\\[3pt]
Middle levels: \(e^me^S\), the large factor \(C\) makes the reciprocal
phases oscillate as \(u\) varies. The following elementary estimate
will make that observation uniform.
\begin{lemma}\label{lem:reciprocal-phase}
Fix \(B\ge4\). There are constants \(A_B,\delta_B>0\) such that,
for every sufficiently large \(U\), every real \(Z\) with
\(U^4\le |Z|\le U^B\), and every interval \(I\subset[U,2U]\),
\[
\left|\sum_{n\in I}\e(Z/n)\right|\le A_BU^{1-\delta_B}.
\]
\end{lemma}
\begin{proof}
We give the derivative argument, including the elementary estimates
it uses. This is the classical differencing method of van der Corput;
see \cite[Lemma~2.5]{GrahamKolesnik1991} for differencing and
\cite[Theorem~2.2]{GrahamKolesnik1991} for the second-derivative estimate.
First, if \(|a_n|\le1\) is supported on an interval of length at most
\(U\), then for \(2\le L\le U\),
\begin{equation}\label{eq:differencing}
U^{-2}\left|\sum_n a_n\right|^2
\ll L^{-1}
+\max_{1\le he^S\), and every integer
\(1\le\ell\le H:=\lfloor e^{4\eta m}\rfloor\), one has
\begin{equation}\label{eq:deterministic-mean}
\frac1X\sum_{X_{j+1}e^{S-\eta m}\ge e^{0.9S}\).
Expanding the square, the diagonal contributes at most \(2^{-m}\)
because the \(2^m\) subset products are distinct.
For an off-diagonal pair put \(Z=\ell C(t-t')\).
The difference \(t-t'\) is a nonzero integer, and
\eqref{eq:C-range} and \eqref{eq:product-size} give
\[
e^{D_CS}\le |Z|\le e^{(2D_C+K+2)S}.
\]
Split \((Y,X]\) into intersections with dyadic intervals \([U,2U]\).
We may take \(Y/2\le U\le X\), so for large \(S\),
\[
e^{0.8S}\le U\le e^{D_XS},\qquad
U^4\le |Z|\le U^{3D_C}.
\]
Lemma~\ref{lem:reciprocal-phase}, with the fixed value \(B=3D_C\),
bounds the sum over each piece by
\(O(Ue^{-0.8\delta_BS})\).
The sum of the dyadic \(U\)'s is \(O(X)\). Thus
\[
\left|\sum_{Y0\). Averaging over all off-diagonal pairs
cancels their number against \(|T_0|^2\). The left side of
\eqref{eq:deterministic-mean} is consequently at most
\[
2^{-m}+O(e^{-cS})\le e^{-0.01m}
\]
eventually, since \(S/m\to\infty\).
\end{proof}
The deterministic estimate controls the high levels for every
permitted \(C\). We next prove the random estimate for the lower
levels and for individual terminal numerators.
\input{random}
==== END Short-Egyptian-fractions//residues.tex ====
==== BEGIN Short-Egyptian-fractions//unicode-mappings.tex ====
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==== END Short-Egyptian-fractions//unicode-mappings.tex ====
==== BEGIN Short-Egyptian-fractions//references.bib ====
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year={1971}, pages={77--81}, doi={10.4064/aa-18-1-77-81},
url={https://matwbn.icm.edu.pl/ksiazki/aa/aa18/aa1819.pdf},
note = {\href{https://doi.org/10.4064/aa-18-1-77-81}{Publisher version}}
}
@article{KumchevTolev2005,
author={Kumchev, Angel V. and Tolev, Dimitar I.},
title={An invitation to additive prime number theory},
journal={Serdica Mathematical Journal},
volume={31}, year={2005}, pages={1--74},
url={https://tigerweb.towson.edu/akumchev/a17.pdf},
note = {\href{https://tigerweb.towson.edu/akumchev/a17.pdf}{Author version}}
}
@article{Martin2000,
author={Martin, Greg},
title={Denser {Egyptian} fractions},
journal={Acta Arithmetica},
volume={95}, number={3}, year={2000}, pages={231--260},
url={https://arxiv.org/abs/math/9811112},
note = {\href{https://arxiv.org/abs/math/9811112}{Author version}}
}
@article{Croot1999,
author={Croot, III, Ernest S.},
title={On some questions of {Erd{\H{o}}s} and {Graham} about {Egyptian} fractions},
journal={Mathematika},
volume={46}, number={2}, year={1999}, pages={359--372},
doi={10.1112/S0025579300007828},
url={https://ecroot.math.gatech.edu/erdos_graham.pdf},
note = {\href{https://doi.org/10.1112/S0025579300007828}{Publisher version}}
}
@incollection{GlibichukKonyagin2007,
author={Glibichuk, A. A. and Konyagin, S. V.},
title={Additive properties of product sets in fields of prime order},
booktitle={Additive Combinatorics},
series={CRM Proceedings and Lecture Notes},
volume={43}, year={2007}, pages={279--286},
publisher={American Mathematical Society},
url={https://arxiv.org/abs/math/0702729},
note = {\href{https://arxiv.org/abs/math/0702729}{Author version}}
}
==== END Short-Egyptian-fractions//references.bib ====