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\hypersetup{pdftitle={Small-index subsequences in finite cyclic groups},
 pdfauthor={Chao Liu},pdfsubject={Research manuscript},
 pdfkeywords={index-one, index-r, cyclic group, Jacobsthal function, zero-sum sequence}}
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\title[Small-index subsequences in cyclic groups]{Small-index subsequences in finite cyclic groups}
\author{Chao Liu}
\address{Department of Mathematics, University of Mississippi, USA}
\email{chaoliuac@gmail.com}
\date{}
\subjclass[2020]{11B30, 11B50, 11P70, 20K01}
\keywords{Cyclic group, index of a sequence, zero-sum sequence, subsequence sum}

\begin{document}
\begin{abstract}
For $r\ge1$, let $I_r(n)$ be the least length forcing a nonempty
zero-sum subsequence of index at most $r$ over the cyclic group $C_n$.
We prove $I_r(n)\le n^{1+o(1)}$ when
$r\ge C[\omega(n)\log(2\omega(n))]^2$, and
$I_1(p^k)\ll(k+1)p^k\log(2p^k)$ for prime powers.
We also construct sequences showing that $I_r(n)/n$ is unbounded
for every fixed $r$; for $r=1$, all terms may be generators.
\end{abstract}
\maketitle

\section{Introduction and main results}

Write $C_n$ additively. A sequence over $C_n$ is a finite multiset;
repetitions are allowed. Fix a generator $g$ and write
$S=(a_1g)\cdots(a_\ell g)$ with $a_i\in[1,n]\cap\Z$. Its index is
\[
 \ind(S)=\min_{u\in\U(n)}\frac1n\sum_{i=1}^{\ell}\res{ua_i}{n},
 \qquad \U(n)=(\Z/n\Z)^\times,
\]
where $\res{x}{n}$ is the least positive residue, so $\res0n=n$.
For $r\ge1$, a sequence is \emph{index-$r$-free} if it has no nonempty
subsequence of index in $\{1,\ldots,r\}$. Following \cite{Liu}, let
$I_r(n)$ be the least integer such that every sequence of length at least
$I_r(n)$ is not index-$r$-free. In particular, an index-one subsequence
is a nonempty subsequence whose positive representatives sum to exactly
$n$ in one unit direction. A zero sum modulo $n$ alone does not suffice.

The index-one problem originates in work of Lemke and Kleitman
\cite{LK}. The proposed equality $I_1(n)=n$ fails for composite orders:
Gao, Li, Peng, Plyley and Wang \cite{GLPPW} gave counterexamples, and Liu
\cite[Theorem 1.5]{Liu} proved, for fixed $d>1$ and $r\ge1$, that
\begin{equation}\label{eq:old-lower}
 I_r(n)\ge(1+1/d)n-O_{d,r}(\sqrt n)
 \qquad(d\mid n, n\longrightarrow\infty).
\end{equation}
Subsequence sums of index-one-free sequences are also studied in
\cite{PYL}. We consider the forcing length for arbitrary sequences,
including terms of proper additive order.

Our first result gives a near-linear forcing length once the allowed
index exceeds a bound depending only on the number of distinct prime
factors. Write $\omega(n)$ for this number. All constants below are
independent of the prime factors and their exponents unless a
dependence is displayed, and all logarithms are natural.

\begin{theorem}\label{thm:near-linear}
There is an absolute constant $C$ such that, for every
$\varepsilon>0$, every $n\ge2$ and every integer $r\ge1$,
\begin{equation}\label{eq:near-linear-main}
 r\ge C\bigl[\omega(n)\log(2\omega(n))\bigr]^2
 \quad\Longrightarrow\quad I_r(n)\ll_\varepsilon n^{1+\varepsilon}.
\end{equation}
Moreover, for each integer $k\ge1$, there is an integer $R_k$,
depending only on $k$, such that
\begin{equation}\label{eq:bounded-prime-main}
 I_r(n)\ll_k n\log^{k+1}(2n)
 \qquad\bigl(\omega(n)\le k,\ r\ge R_k\bigr).
\end{equation}
The implied constants depend only on $\varepsilon$ in
\eqref{eq:near-linear-main} and only on $k$ in
\eqref{eq:bounded-prime-main}, uniformly in $n$ and $r$.
\end{theorem}

In particular, one fixed index $R_2$ gives
\[
 I_{R_2}(p^a q^b)\ll p^a q^b\log^3(2p^a q^b)
\]
for all distinct primes $p,q$ and all positive integers $a,b$, with
no restriction on their relative sizes. This result is unconditional
and includes sequences with terms of proper additive order.
The proof combines an integer subset-sum interval with a choice of
unit controlled by the Jacobsthal function, then sums reciprocal
weights over all prime-power valuation classes.

For index one we obtain stronger bounds in several arithmetic
families, including a conditional refinement under Dirichlet GRH.
We also retain a uniform estimate for every allowed index $r$.

\begin{theorem}\label{thm:upper}
There is an absolute constant $C$ with the following properties.
\begin{enumerate}
\item For every prime $p$ and every $k\ge1$, with $n=p^k$,
\begin{equation}\label{eq:pp-main}
 I_1(n)\le C(k+1)n\log(2n).
\end{equation}
In particular $I_1(n)\ll n\log^2(2n)$ uniformly on prime powers,
and $I_1(p^k)\ll_k p^k\log(2p^k)$ for each fixed $k$.
\item For every $n\ge2$ and every integer $r\ge1$,
\begin{equation}\label{eq:general-r-main}
 n\le I_r(n)\le C\left(n+\frac{n^{3/2}}{\sqrt r}\right).
\end{equation}
In particular $I_r(n)\ll n^{3/2}/\sqrt r$ for $1\le r\le n$.
For index one, writing $P^{-}(n)$ for the least prime divisor,
\begin{equation}\label{eq:general-main}
 I_1(n)\le C\min\left\{n^{3/2},\,
 n\log(2n)\left(1+\frac{\sqrt n}{P^{-}(n)}\right)\right\}.
\end{equation}
\item If $p\ne q$ are primes, $a\ge1$, $P=p^a$ and $q\ge P$, then,
for $n=Pq$,
\begin{equation}\label{eq:mixed-main}
 I_1(n)\le C n\log(2n)
 \left(a+1+\frac1p\sqrt{\frac qP}\right).
\end{equation}
For every $\varepsilon>0$ one also has
\begin{equation}\label{eq:weyl-main}
 I_1(n)\ll_\varepsilon n^{4/3+\varepsilon}.
\end{equation}
Under the Dirichlet generalized Riemann hypothesis,
\begin{equation}\label{eq:grh-main}
 I_1(n)\ll n\log^3(2n)
\end{equation}
uniformly throughout this family.
\item For every prime $p$, with $n=2p^2$,
\begin{equation}\label{eq:twice-square-main}
 I_1(n)\le Cn\log(2n).
\end{equation}
This bound is unconditional and includes sequences with nonunit terms.
\end{enumerate}
\end{theorem}

For example, \eqref{eq:general-main} gives
$O_{B,k}(n^{3/2-1/k}\log(2n))$ when $n$ has $k\ge2$ prime factors,
counted with multiplicity, all in $[x,Bx]$. This gives exponent $7/6$
for balanced $pqr$ and $p^2q$, and exponent $5/4$ for balanced $p^2q^2$.
Equation \eqref{eq:mixed-main} gives
$O((a+1)n\log(2n))$ throughout $p^a\le q\le p^{a+2}$.
In particular the balanced semiprime bound is $O(n\log(2n))$.
The family $2p^2$ in \eqref{eq:twice-square-main} is outside the
larger-prime hypothesis for odd $p$. Its proof uses a parity-sensitive
integer budget and reciprocal weights that sum all valuation classes
with a convergent geometric series.

The conditional estimate \eqref{eq:grh-main} connects the forcing problem
with uniform character-sum estimates for Dirichlet $L$-functions. It
holds without an upper restriction on $q/p^a$. For example, for $n=4q$
with $q\ge5$ prime, our unconditional estimate is
$O_\varepsilon(n^{4/3+\varepsilon})$, while Dirichlet GRH gives
$O(n\log^3(2n))$. Its proof combines character orthogonality with
smoothed prime-sum estimates in Section \ref{sec:analytic}.

The estimate \eqref{eq:general-r-main} is uniform in the allowed index
$r$: it gives $I_r(n)=O(n)$ whenever $r$ is bounded below by a fixed
positive multiple of $n$. Exact large-$r$ ranges also follow from
known high-multiplicity theorems; see Corollary \ref{cor:large-index}.

The lower bounds exhibit a different behavior for general composite
orders. Put
\[
 P(d)=\prod_{p\mid d}(1+1/p),\qquad s=\omega(d),
\]
where the product and $\omega(d)$ count distinct prime factors.

\begin{theorem}\label{thm:divisor-lower}
Fix integers $d>1$ and $r\ge1$. For every sufficiently large multiple
$n$ of $d$, there is an index-$r$-free sequence over $C_n$ of length
\begin{equation}\label{eq:divisor-main}
 |S|\ge P(d)n-O_{d,r}\bigl(n^{s/(s+1)}\bigr).
\end{equation}
More explicitly, once $n>(rd^2)^{s+1}$ and $d\mid n$, one may take
\begin{equation}\label{eq:divisor-explicit}
 |S|\ge P(d)n-\bigl(sP(d)+rd^2P(d)+d\bigr)n^{s/(s+1)}.
\end{equation}
Consequently, for every $M>0$ there exists $d>1$, depending only on $M$,
such that for every fixed $r\ge1$,
$I_r(n)>Mn$ for all sufficiently large multiples $n$ of $d$.
\end{theorem}

When $d=p^a$ is a prime power, the leading coefficient $1+1/p$ and
the error order in \eqref{eq:divisor-main} agree with
\eqref{eq:old-lower} applied with divisor $p$. For several
prime divisors the coefficient multiplies: it is $2$ for $d=6$, $12/5$
for $d=30$, and $96/35$ for $d=210$. In particular the conclusion
$I_1(n)\le n+n/P^{-}(n)$ proposed in \cite[Conjecture 4.1]{Liu}
fails on all sufficiently large multiples of $6$. Section
\ref{sec:divisor-lower} gives a concrete index-one-free sequence of length
$18000$ over $C_{12000}$.

The next result shows that superlinear behavior persists even when every
term is a generator. It is independent of Theorem \ref{thm:divisor-lower}.

\begin{theorem}\label{thm:unit-lower}
There is a sequence $n_j\longrightarrow\infty$ and index-one-free
sequences $S_j$ over $C_{n_j}$, all of whose terms have order $n_j$, such
that
\begin{equation}\label{eq:unit-main}
 \log\frac{|S_j|}{n_j}
 =\left(\frac{\log6}{2}+o(1)\right)
 \frac{\log\log n_j}{\log\log\log n_j}.
\end{equation}
Thus $I_1(n)$ has no absolute linear upper bound even on sequences
consisting only of generators.
\end{theorem}

As a structural complement, we bound the length in terms of the maximum
multiplicity. Write $\sigma_{-3/4}(n)=\sum_{d\mid n}d^{-3/4}$.

\begin{theorem}\label{thm:support}
Let $S$ be a nonempty index-one-free sequence over $C_n$, of length
$N$ and maximum multiplicity $h$. Then
\begin{equation}\label{eq:multiplicity-main}
 N\ll(nh)^{3/4}\sigma_{-3/4}(n),
\end{equation}
with an absolute implied constant. For every fixed $k\ge1$, if
$\omega(n)\le k$, then $N\ll_k(nh)^{3/4}$; in particular,
$N\ge n$ forces $h\gg_k n^{1/3}$. Consequently every index-one-free
sequence satisfies $|\supp(S)|\le n^{3/4+o(1)}$, and at prime-power
orders $|\supp(S)|\ll n^{3/4}$ with an absolute constant.
\end{theorem}

This applies to sequences of arbitrary length and multiplicities.
The proof combines integer subset-sum theorems of Bringmann--Wellnitz
\cite{BW} and Conlon--Fox--Pham \cite{CFP} with an interval argument
for repeated denominations. Section \ref{sec:dense} proves the stronger
bound obtained by keeping the maximum multiplicity in each exact-order
class separately.

Section \ref{sec:preliminaries} develops the basic lemmas.
Section \ref{sec:near-linear} proves the near-linear small-index theorem.
Section \ref{sec:upper-proofs} proves the index-one upper bounds,
including the GRH refinement. Section \ref{sec:lower-proofs} gives
the two lower constructions, and Section \ref{sec:dense} proves the
support and multiplicity bounds. The appendix proves the uniform index-$r$ upper bound
by integer descent.

The author acknowledges assistance from OpenAI tools in preparing this paper.

\section{Preliminaries}\label{sec:preliminaries}
\subsection{Integer intervals and truncated multiplicities}\label{sec:intervals}

For a sequence $S$, let $c_x=v_x(S)$ be the multiplicity of $x$, and
let $\Sig(S)$ be its set of
subsequence sums including the empty sum. When the entries are positive
integers, $\Sigma_{0,\Z}(S)$ denotes their integer subsequence sums.
Products of sequences mean concatenation. A nonempty zero-sum sequence
is an \emph{atom} if all its proper nonempty subsequences have nonzero
sum. An index-one sequence is necessarily an atom: a proper zero sum
would have a positive integer weight strictly between $0$ and $n$ in
the same direction.

For $r\ge1$, let $L_{U,r}(n)$ be the maximum length of an
index-$r$-free sequence whose terms are units modulo $n$, and write
$L_U(n)=L_{U,1}(n)$. These maxima are finite, since every type has
multiplicity at most $n-1$.

\begin{lemma}\label{lem:orders}
If $d\mid n$ and $H$ is the subgroup of order $d$ in $C_n$, then
$\ind_{C_n}(T)=\ind_H(T)$ for every sequence $T$ over $H$. Consequently,
for every $r\ge1$, an index-$r$-free sequence over $C_n$ has length at most
\begin{equation}\label{eq:order-sum}
 \sum_{\substack{d\mid n\\d>1}}L_{U,r}(d).
\end{equation}
\end{lemma}
\begin{proof}
The reduction $\U(n)\to\U(d)$ is surjective. Indeed, lift a unit modulo
$d$ and impose residue $1$ at each prime divisor of $n$ not dividing
$d$; the Chinese remainder theorem supplies a lift coprime to $n$.
For subgroup coordinates $x=(n/d)b$,
\[
 \res{u(n/d)b}{n}=(n/d)\res{ub}{d}.
\]
The normalized weights and their minima therefore agree. Partition the
terms by their exact additive order. The class of order $d$ becomes a
unit sequence in $C_d$, and is still index-$r$-free. The zero element
cannot occur. This gives \eqref{eq:order-sum}.
\end{proof}

We use the classical short-zero-sum theorem: every sequence over $C_n$
of length at least $n$ has a nonempty zero-sum subsequence of length at
most its maximum multiplicity; see \cite[Lemma 3.1]{GLPPW}.

\begin{lemma}\label{lem:index-cap}
Every index-$r$-free sequence $S=\prod_a a^{c_a}$ over $C_n$ satisfies
\begin{equation}\label{eq:index-cap}
 \sum_a\min\{c_a,2r+1\}\le n-1.
\end{equation}
\end{lemma}
\begin{proof}
For a zero-sum sequence $T$ containing no zero term, in any unit
direction $u$ the two positive weights in directions $u$ and $-u$
sum to $|T|n$. Their normalized weights are positive integers, so
\[
 \ind(T)\le\lfloor |T|/2\rfloor.
\]
Retain at most $2r+1$ actual copies of each type. If their total
length were at least $n$, the short-zero-sum theorem would give a
zero sum of length at most $2r+1$, hence of index at most $r$.
A zero term is already an index-one singleton. This proves the bound.
\end{proof}

The next exact ranges are direct consequences of known
high-multiplicity results, stated here in terms of $I_r(n)$.

\begin{corollary}\label{cor:large-index}
For every $n\ge2$,
\[
 I_r(n)=n\qquad
 \left(r\ge\max\{1,\lfloor(n-1)/4\rfloor\}\right).
\]
For odd $n\ge5$, the same equality holds already when
$r\ge\max\{1,\lfloor(n-1)/6\rfloor\}$.
\end{corollary}
\begin{proof}
Suppose every length-$n$ sequence of maximum multiplicity at least
$H$ has an index-one subsequence. For smaller multiplicity, the
short-zero-sum theorem and unit reversal give a zero sum of index
at most $\lfloor(H-1)/2\rfloor$. Thus $I_r(n)\le n$ for
$r\ge\max\{1,\lfloor(H-1)/2\rfloor\}$.
Use $H=\lceil n/2\rceil$ from \cite[Theorem 1.3]{GLPPW}, and
$H=\lceil n/3\rceil$ for odd $n\ge5$ from
\cite[Theorem 2.1]{ZLY}. The lower bound is the ordinary
zero-sum-free sequence $g^{n-1}$ for a generator $g$.
\end{proof}

We use Kneser's addition theorem in its usual form: if $A,B$ are finite
nonempty subsets of an abelian group and $H$ is the stabilizer of
$A+B$, then $|A+B|\ge|A+H|+|B+H|-|H|$; see \cite{DeVos}.

\begin{lemma}\label{lem:diameter}
If $A\subseteq C_h$ contains $0$, generates $C_h$, and $|A|=k$, then
$\ell A=C_h$ for $\ell=\lceil2h/k\rceil$.
\end{lemma}
\begin{proof}
The case $h=1$ is immediate. Otherwise suppose $\ell A\ne C_h$, and let
$H$ be its stabilizer, of size $j$. Write $t=|A+H|/j$. Since $A$
generates the group, $t\ge2$, and $k\le tj$. Iterated Kneser in the
quotient by $H$ gives
\[
 \frac{|\ell A|}{j}\ge\ell(t-1)+1
 \ge\frac{2h(t-1)}{k}+1
 \ge\frac hj\frac{2(t-1)}t+1>\frac hj,
\]
a contradiction. The iteration is legitimate since a stabilizer of an
intermediate sumset stabilizes the final sumset as well.
\end{proof}

\begin{lemma}\label{lem:integer-interval}
Let $B$ be a set of $k$ distinct positive integers, with $\gcd(B)=1$ and
$h=\max B$. Put
\[
 \ell=\lceil2h/k\rceil,\qquad \rho=\ell+1,\qquad L=\ell h.
\]
If $c_b\ge\rho$ for $b\in B$, then
\begin{equation}\label{eq:integer-interval}
 [L,L+h+\sum_{b\in B}(c_b-\rho)b]\cap\Z
 \subseteq\Sigma_{0,\Z}\left(\prod_{b\in B}b^{c_b}\right).
\end{equation}
Moreover $\rho\le4h/k$ and $L\le2h^2/k+h$.
\end{lemma}
\begin{proof}
Reduction of $B$ modulo $h$ has $k$ distinct residues, contains $0$ and
generates $C_h$. By Lemma \ref{lem:diameter}, every residue has an
integer representation of weight at most $\ell h$, using at most
$\ell$ entries of $B\setminus\{h\}$. For any integer $z\in[L,L+h]$,
choose such a representation $w\equiv z\pmod h$, and add $(z-w)/h$
copies of $h$. This number lies between $0$ and $\ell+1$. Thus $\rho$
copies of each denomination suffice for $[L,L+h]$. Each further copy
of a denomination at most $h$ extends the right endpoint by its value,
since the interval and its translate overlap. The stated bounds follow
from $k\le h$.
\end{proof}

\begin{lemma}\label{lem:prime-window}
For all sufficiently large integers $n$, put
$H=\lfloor\sqrt n/16\rfloor$. Let $X$ be a multiset of primes in
$(H/2,H]$, each of multiplicity at most $4\sqrt n$. If
$|X|\ge128\sqrt n$, then $n\in\Sigma_{0,\Z}(X)$.
\end{lemma}
\begin{proof}
Let $q$ have largest multiplicity $M$ and put $\ell=\lfloor M/8\rfloor$.
If $M<16$, then $|X|\le15H<\sqrt n$, a contradiction. Thus
$M/16\le\ell\le M/8$ and $2\ell+3\le M/2$. Writing $c_a$ for the
other multiplicities, we have
\[
 \sum_{a\ne q}\min(c_a,\ell)
 \ge\frac\ell M(|X|-M)\ge\frac{|X|-M}{16}\ge q-1.
\]
Reserve $q-1$ non-$q$ copies, at most $\ell$ of each type, and
$2\ell+3$ copies of $q$. If $d_a$ is the number of reserved copies of
$a\ne q$, form
$A_j=\{0\}\cup\{a\bmod q:d_a\ge j\}$ for $1\le j\le\ell$.
The nonzero residues in each layer are distinct, because the original
window has width less than $q$. Cauchy--Davenport gives
$A_1+\cdots+A_\ell=C_q$, since
$\sum_j(|A_j|-1)=q-1$.

Each residue is therefore represented by reserved non-$q$ copies with
integer weight at most $L=\ell H$. Adding reserved $q$'s covers
$[L,L+2q]\cap\Z$; fewer than $2\ell+3$ copies suffice because $H<2q$.
The width $2q$ exceeds every remaining denomination. Also
$L\le MH/8\le n/32$, whereas fewer than $3\sqrt n$ copies were
reserved. For large $n$, $H\ge\sqrt n/32$, so the remaining weight is
greater than
\[
 \frac{\sqrt n}{64}(128\sqrt n-3\sqrt n)>n.
\]
Adding the remaining copies one at a time extends the integer interval
beyond $n$.
\end{proof}

\begin{proposition}\label{prop:truncation}
Every index-one-free unit sequence over $C_n$ satisfies
\begin{equation}\label{eq:unit-trunc}
 \sum_{x\in\U(n)}\min\{c_x,\lfloor4\sqrt n\rfloor\}
 \ll\varphi(n)\log(2n).
\end{equation}
Every index-one-free sequence over $C_n$ satisfies
\begin{equation}\label{eq:all-trunc}
 \sum_{x\in\supp(S)}\min\{c_x,\lfloor4\sqrt{\ord(x)}\rfloor\}
 \ll n\log(2n).
\end{equation}
\end{proposition}
\begin{proof}
Truncate each unit multiplicity at $\lfloor4\sqrt n\rfloor$, obtaining
$X$. Use the prime window of Lemma \ref{lem:prime-window}, removing
primes dividing $n$, and denote the remaining primes by $\mathcal P$.
The prime number theorem gives
$|\mathcal P|\gg\sqrt n/\log(2n)$ for large $n$: at most
$\omega(n)\le\log_2 n$ primes were removed. Quantitative versions of
the required prime estimates are given in \cite{Dusart}.
In every unit direction fewer than $128\sqrt n$ copies of $X$ lie in
$\mathcal P$, or the lemma gives an index-one subsequence. Averaging
over $\U(n)$, each unit type visits every allowed prime once, so
\[
 |\mathcal P||X|<128\varphi(n)\sqrt n.
\]
This proves \eqref{eq:unit-trunc}. The finitely many small $n$ are
absorbed by increasing the absolute constant, using $c_x<n$.
Apply this result to every exact-order class and use Lemma
\ref{lem:orders} and $\sum_{d\mid n}\varphi(d)=n$ to obtain
\eqref{eq:all-trunc}.
\end{proof}

\subsection{Two-generator zero sums}\label{sec:two-types}

The following elementary representation allows heights that need not
be units.

\begin{lemma}\label{lem:virtual}
Suppose $1^r b^s$ is a mixed atom over $C_n$, where $b\in\U(n)$ and
$b\ne1$. If $s=1$, it has index one. If $s>1$, there are positive
integers $u,v$ such that
\begin{equation}\label{eq:virtual}
 0<u<s,\qquad ru+sv=n,\qquad v\equiv bu\pmod n,
 \qquad\gcd(u,s)=1.
\end{equation}
\end{lemma}
\begin{proof}
Take $b\in[1,n-1]$. Write $r+bs=nt$. If a prime divides both $t$ and
$s$, it divides $r$, and division by that prime gives a proper zero sum.
Thus $(t,s)=1$. For $s>1$, choose $0<u<s$ with $tu\equiv1\pmod s$ and
set $v=(n-ru)/s$. This is an integer and
$bu-v=n(tu-1)/s$ is a multiple of $n$. If $v<0$, the counts
$(-v,u)$ give a proper zero sum, since $-v<r$ and $u<s$. If $v=0$,
then $n\mid bu$, impossible since $u<s<n$. Hence $v>0$.
For $s=1$, minimality gives $r+b=n$.
\end{proof}

\begin{proposition}\label{prop:capacity-threshold}
Let $\ell=P^{-}(n)$ and let $a,b$ be distinct generators of $C_n$.
If $a^A b^B$ is index-one-free but contains a nonempty zero sum, then
\begin{equation}\label{eq:capacity-threshold}
 A<n/\ell,\qquad B<n/\ell.
\end{equation}
\end{proposition}
\begin{proof}
Normalize $a$ to $1$. Any atom $(r,s)$ contained in the sequence has
index greater than one. In \eqref{eq:virtual}, $u$ is therefore not a
unit. A common prime factor of $u$ and $n$ divides $v$, so
$u,v\ge\ell$ and $s>\ell$.

Consider
\begin{equation}\label{eq:neighbor}
 (r',s')=(r+v,s-u).
\end{equation}
This is a positive zero sum and is itself an atom. Indeed, the vectors
$(r,s)$ and $(v,-u)$ form a basis of the lattice
$\{(R,S)\in\Z^2:R+bS\equiv0\pmod n\}$, since their determinant is
$-n$. If a nonzero lattice point has $0\le R\le r+v$ and
$0\le S\le s-u$, minimality of $(r,s)$ forces $R>r$. The integer
$\beta=(sR-rS)/n$ is positive. In the displayed basis write
\[
 R=kr+\beta v,\qquad S=ks-\beta u.
\]
Since $S\ge0$, $k\ge1$. Thus $R\ge r+v$, with equality only for
$k=\beta=1$. This proves minimality of \eqref{eq:neighbor}.
Moreover
\[
 \ell(r+v)\le ru+sv=n.
\]
If $A\ge n/\ell$, the neighboring atom remains in the original
inventory: $r'\le A$ and $1\le s'<s\le B$. If it still has index
greater than one, repeat. Its second count decreases strictly, so this
process must reach an index-one atom, a contradiction. This proves
$A<n/\ell$, and the other inequality follows by interchanging the
two types.

\end{proof}

\begin{lemma}\label{lem:square}
Suppose either $n=p^k$, or $n=p^k q$ with distinct primes $p,q$ and
$q\ge p^k$. Every atom over $C_n$ supported on at most two units has
an index-one subsequence in its square.
\end{lemma}
\begin{proof}
Single-type atoms and the case $s=1$ in Lemma \ref{lem:virtual} are
immediate. Use its positive heights $u,v$. In the mixed case $q\nmid u$:
otherwise $q\mid v$ and $r+s\le p^k\le q$, whereas $s>u\ge q$.
If $u$ is a unit there is nothing to prove. Otherwise $p\mid u,v$,
and $p\nmid r,s$ by $(u,s)=1$ and the original zero-sum relation.

If $v>r$, set $R=r$ and $S=s$. The positive heights
$(U_1,V_1)=(u+s,v-r)$ satisfy
$rU_1+sV_1=n$ and $V_1\equiv bU_1\pmod n$, and avoid $p$.
If $0<v<r$ and $ru\ge n/2$, put
\[
 j=\lfloor(s-1)/u\rfloor,\qquad R=r+jv,\quad S=s-ju.
\]
Then $1\le S<u$, $R<n/u\le2r$, $S\le s$, and
\[
 (U_1,V_1)=(u-S,v+R),\quad
 RU_1+SV_1=n,\quad V_1\equiv bU_1\pmod n.
\]
These heights again avoid $p$. If $sv>n/2$, use instead
\[
 j=\lfloor(r-1)/v\rfloor,\quad R=r-jv,\quad S=s+ju,
 \quad (U_1,V_1)=(u+S,v-R).
\]
Here $1\le R<v$, $S<n/v<2s$, and the same identities hold. Equality
$v=r$ is impossible because $p\mid v$ and $p\nmid r$. These cases
already complete the proof for a prime power.

For $n=p^kq$, a further repair is needed only if $q\mid U_1,V_1$.
In the translation case this forces $r+s\le p^k\le q$ and, from
$U_1=u+s<2s$, gives $s>q/2$ and $r<q/2$. Replace the heights by
$(u+hs,v-hr)$, choosing $h=2$ for odd $p$ and $h=3$ for $p=2$.
They are positive: $V_1\ge q$ and $(h-1)r<q$. Their residues modulo
$p$ are nonzero, and modulo $q$ are $(h-1)s,-(h-1)r$, also nonzero.

In the first shear case, $R+S\le p^k\le q$, and
$r=(j+1)R-jV_1>0$ implies $R>q/2$, hence $S<q/2$.
Use $(u-hS,v+hR)$ with the same choice of $h$. Its first coordinate
is positive because $U_1\ge q>(h-1)S$. Modulo $p$ the two coordinates
are $-hs,hr$, and modulo $q$ they are $-(h-1)S,(h-1)R$.
They are units. In the symmetric shear case,
$s=(j+1)S-jU_1>0$ gives $S>q/2$ and $R<q/2$; use
$(u+hS,v-hR)$. The verification is identical with the roles exchanged.
For $p=2$, $q$ is odd, so the factor $h-1=2$ is nonzero modulo $q$.

In every case the final positive heights $U,V$ satisfy
$RU+SV=n$, $V\equiv bU\pmod n$, $(U,n)=1$, with
$R\le2r$ and $S\le2s$. The integer equality ensures $U,V<n$;
they are the actual positive representatives, and the counts lie in
the square of the original atom.
\end{proof}


\subsection{Short lattice coordinates}

We first record the short-lattice argument in a form valid for arbitrary
moduli. A vector is called primitive in a lattice if it is not a positive
integer multiple of another vector of that lattice. Its integer
coordinates need not be coprime.

\begin{lemma}\label{lem:short-vector}
Let $b\in\U(n)$, let $1\le m<n$, and suppose $m^2>n$ and $1^m b^m$
has no nonempty zero sum. Set $Q=\lfloor n/m\rfloor$. The lattice
\[
 \Lambda_b=\{(x,y)\in\Z^2:y\equiv bx\pmod n\}
\]
has a unique primitive positive vector $(q,v)$ in $[1,Q]^2$. It can be
completed to a basis $(q,v),(s,-r)$ with
\begin{equation}\label{eq:lattice-basis}
 0\le s<q,\quad r>0,\quad qr+vs=n.
\end{equation}
If $D=(q,v)$, then
\begin{equation}\label{eq:lattice-gcd}
 D\mid n,\qquad (q,n)=(v,n)=D.
\end{equation}
\end{lemma}
\begin{proof}
Place the $m+1$ distinct representatives of $0,b,\ldots,mb$ in $m$
half-open intervals of length $n/m$. Two give a vector with
$1\le q\le m$ and $0<|v|<n/m$. Negative $v$ gives a forbidden zero
sum, so $v>0$. Shorten on the same lattice ray to a primitive vector;
then $qv<n$. Complete it to a lattice basis, choose the determinant
to be $-n$, and reduce the other first coordinate modulo $q$. This
gives \eqref{eq:lattice-basis}; $r>0$ follows from $vs<qv<n$.
If $q>n/m$, then $0<r\le n/q<m$ and $s<q\le m$. Here $s=0$ would
force $n\mid r$, impossible. Thus $1^r b^s$ is a forbidden zero sum.
Consequently $q\le n/m$, and both coordinates are at most $Q$.

Any two vectors in this square have determinant of absolute value
less than $n$, because $Q^2<n$, and the determinant is a multiple of
$n$. They lie on the same ray, giving uniqueness of the primitive
vector. The basis determinant implies $D\mid n$. Since $b$ is a unit,
$(q,n)=(v,n)$; this common divisor divides $D$, and $D\mid n$ gives
the reverse divisibility. This proves \eqref{eq:lattice-gcd}.
\end{proof}

\section{Near-linear bounds for small indices}\label{sec:near-linear}

Let $J(n)$ be the least positive integer such that every block of
$J(n)$ consecutive integers contains an integer coprime to $n$, and set
$J(1)=1$. Equivalently, $J(n)-1$ is the maximal length of a block
containing no integer coprime to $n$; this additive-one convention
does not affect the bound quoted below. Also put
\[
 \mathcal A(n)=\prod_{p^a\parallel n}(a+6),\qquad
 \sigma(n)=\sum_{d\mid n}d.
\]
Here $\sigma(n)$ is the sum-of-divisors function. We prove the following
quantitative form of Theorem \ref{thm:near-linear}.

\begin{proposition}\label{prop:jacobsthal-upper}
There is an absolute constant $C$ such that, for every $n\ge2$ and
every integer $r\ge10^6J(n)$,
\begin{equation}\label{eq:jacobsthal-upper}
 I_r(n)\le C\sigma(n)\mathcal A(n)\log(2n).
\end{equation}
\end{proposition}

We use Iwaniec's bound for the Jacobsthal function
\cite[Corollary, p.~226]{Iwaniec}:
\begin{equation}\label{eq:jacobsthal-iwaniec}
 J(n)\ll\bigl[\omega(n)\log(2\omega(n))\bigr]^2.
\end{equation}
In particular, $J(n)\ll\log^4(2n)$. Also $J(d)\le J(n)$ whenever
$d\mid n$, since every block of $J(n)$ consecutive integers contains
an integer coprime to $n$, hence to $d$.

\subsection{Two-type atoms and short coordinates}

\begin{lemma}\label{lem:progression-unit}
Suppose $\gcd(z,B,n)=1$, with $B\ne0$, and put
$R=\prod_{p\mid n,\,p\nmid B}p$. Every block of $J(R)$ consecutive
integers $k$ contains one for which $z+Bk$ is coprime to $n$.
\end{lemma}
\begin{proof}
If $p\mid B$ and $p\mid n$, then $p\nmid z$. For every prime
$p\mid R$, the sole forbidden residue is $k\equiv-zB^{-1}\pmod p$.
The Chinese remainder theorem translates these forbidden residues
simultaneously into the condition that $k+\kappa$ be divisible by
one of the primes dividing $R$. The definition of $J(R)$ now applies.
The same assertion holds when $R=1$.
\end{proof}

\begin{lemma}\label{lem:small-index-atom}
Let $1^A b^B$ be a mixed atom over $C_n$, with $b\in\U(n)$ and
$b\ne1$. Its index is at most $r$ if either $\min(A,B)\le r$, or
\begin{equation}\label{eq:small-index-atom}
 r\ge4J(n),\qquad AB\le\frac{rn}{4(J(n)+2)}.
\end{equation}
\end{lemma}
\begin{proof}
Both counts are less than $n$. Normalize the type with the larger
count to $1$. The positive weight is a multiple of $n$ strictly less
than $(\min(A,B)+1)n$, proving the first assertion.

Thus assume $A,B>r$. Lemma \ref{lem:virtual} supplies positive
$u,v$ with
\[
 Au+Bv=n,\qquad v\equiv bu\pmod n,\qquad (u,B)=1.
\]
Put $P=\prod_{p\mid n,\,p\mid B}p$. Choose
$j\in[\lceil r/2\rceil,r]$ coprime to $P$; this interval contains
at least $J(P)$ consecutive integers. The conditions
\[
 x=ju+Bk>0,\qquad y=jv-Ak>0
\]
define an open interval of real $k$ of length
$jn/(AB)\ge2(J(n)+2)$. It contains a block of $J(n)$ consecutive
integers. Since $\gcd(ju,B,n)=1$, Lemma
\ref{lem:progression-unit} gives such a $k$ for which $x$ is a unit
modulo $n$. Then $y\equiv bx\pmod n$ is a unit as well, and
\[
 Ax+By=jn.
\]
The inequalities $A,B>r\ge j$ imply $0<x,y<n$. Hence these are the
actual positive representatives in direction $x$, and the atom has
index at most $j\le r$.
\end{proof}

For the remainder of this section, let $S=\prod_a a^{c_a}$ be an
index-$r$-free unit sequence, where $r\ge10^6J(n)$. Define
\begin{equation}\label{eq:near-linear-cutoff}
 T=64\lceil\sqrt n\rceil,\qquad
 \mathcal H=\{a:c_a\ge T\},\qquad Q=\lfloor n/T\rfloor.
\end{equation}
If $T\ge n$, then $\mathcal H$ is empty. Otherwise every sequence
$a^T b^T$ with distinct $a,b\in\mathcal H$ has no nonempty zero sum.
Indeed, a zero sum would contain a mixed atom with counts
$A,B\le T$, since $T<n$. But
\[
 AB\le T^2\le16384n<\frac{rn}{4(J(n)+2)},
\]
so Lemma \ref{lem:small-index-atom} gives a contradiction.

Lemma \ref{lem:short-vector}, applied with $m=T$, therefore gives
unique lattice-primitive coordinates
\begin{equation}\label{eq:near-linear-coordinates}
 q_{ab}a\equiv v_{ab}b\pmod n,\qquad
 1\le q_{ab},v_{ab}\le Q.
\end{equation}
Reversing $a,b$ interchanges the coordinates; take $(1,1)$ on the
diagonal. If $D=(q_{ab},v_{ab})$, then
\[
 D\mid n,\qquad (q_{ab},n)=(v_{ab},n)=D.
\]
In particular, writing $(q_{ab},v_{ab})=(Dx,Ds_a)$ and $M=n/D$,
we have $(x,s_a)=1$ and $x,s_a\in\U(M)$.

We may henceforth assume that $n$ exceeds an absolute threshold so that
\begin{equation}\label{eq:near-linear-geometry}
 2Q^2+Q<n,\qquad 8Q<T,\qquad J(n)Q<n/4.
\end{equation}
The first two inequalities follow from \eqref{eq:near-linear-cutoff},
and the last follows from \eqref{eq:jacobsthal-iwaniec}.
The implied constant in \eqref{eq:jacobsthal-iwaniec} is absolute,
so the finite exceptional set of moduli is independent of $r$.
These moduli are absorbed into an absolute constant using
$|S|\le(n-1)\varphi(n)$.

\subsection{An integer budget with an anchor}

Fix $b\in\mathcal H$, $D\mid n$ with $1<D\le Q$, and
$1\le x\le K=\lfloor Q/D\rfloor$. Consider those $a\in\mathcal H$
whose coordinates relative to $b$ are $(Dx,Ds_a)$, and put
\[
 F_D=\{p\mid n:p^{v_p(n)}\mid D\}.
\]
Partition these types by the valuations $v_p(s_a)$ for $p\in F_D$.
For a fixed valuation class, put
$L=\prod_{p\in F_D}p^{v_p(s_a)}$ and partition further according to
the unit
\begin{equation}\label{eq:near-linear-phase}
 \gamma=(s_a/L)a^{-1}\pmod n.
\end{equation}
Since $s_a\in\U(M)$, any prime $p\mid n$ dividing $s_a$ does not
divide $M=n/D$, so $p^{v_p(n)}\mid D$. Thus $p\in F_D$, and its
full contribution to $s_a$ is removed in $L$. Consequently $s_a/L$
and $\gamma$ are units modulo $n$.
Also $L$ is a unit modulo $M$, and all phases reduce to
$xL^{-1}b^{-1}\pmod M$. Since $M\mid n$, the Chinese remainder
theorem shows that the reduction map $\U(n)\to\U(M)$ is surjective,
so every fibre has $\varphi(n)/\varphi(M)\le D$ elements.
Thus there are at most $D$ phases in each valuation class.

\begin{lemma}\label{lem:shifted-anchor}
For every nonempty group $\mathcal C$ with fixed $b,D,x$, valuations
and phase as above,
\begin{equation}\label{eq:shifted-anchor-budget}
 \sum_{a\in\mathcal C}c_a s_a<2M.
\end{equation}
\end{lemma}
\begin{proof}
Let $g=\gcd_{a\in\mathcal C}s_a=Lg_0$ and $w_a=s_a/g$.
Since $g_0\mid s_a/L$ and $s_a/L$ is coprime to $n$, both $g_0$
and $w_a=(s_a/L)/g_0$ are units modulo $n$. The distinct positive
integers $w_a$ have gcd $1$, and $(x,g)=1$. The direction
$u_0=\gamma g_0^{-1}\pmod n$ gives the actual core heights $w_a$.
Write $Y_0=\res{u_0b}{n}$. The congruence $xa\equiv s_ab\pmod M$
implies
\begin{equation}\label{eq:anchor-j}
 gY_0=x+jM,\qquad 0\le j<Dg.
\end{equation}
Here $gY_0-x>-M$, since $x\le K<M$, and $gY_0<gn=DgM$.

We claim $(j,Dg)=1$. Set $h_a=(xa-s_ab)/M$, using integer
representatives for $a,b$. Lattice primitivity implies $(h_a,D)=1$:
if a prime $p$ divided both, then
\[
 n=DM\mid\frac Dp(xa-s_ab),
\]
so $(Dx/p,Ds_a/p)$ would be another lattice vector on the same
positive ray, contradicting primitivity. Multiplying
$Mh_a=xa-s_ab$ by $u_0$, reducing modulo $DM$, and cancelling the
integer factor $M$ gives
\[
 u_0h_a\equiv\frac{xw_a-s_aY_0}{M}=-w_a j\pmod D.
\]
Since $u_0,w_a$ and $h_a$ are units modulo $D$, this proves
$(j,D)=1$. Equation \eqref{eq:anchor-j} and $(x,g)=1$ also give
$(j,g)=1$. In particular $1\le j<Dg$.

Let $\lambda_0\in[1,Dg-1]$ be the inverse of $j$ modulo $Dg$, and
set $\sigma_0=Dg-\lambda_0$. By Lemma \ref{lem:progression-unit},
there is $t\in[0,J(n)-1]$ such that
\[
 \lambda=\lambda_0+tDg\in\U(n),\qquad 1\le k=t+1\le J(n)\le r.
\]
Define the integers
\begin{equation}\label{eq:shifted-target}
 \tau=\frac{M-x\sigma_0}{g},\qquad
 Y=\frac{M+\lambda x}{g}.
\end{equation}
Their integrality follows by reducing
$x+jM\equiv0\pmod g$ and $\lambda j\equiv1\pmod{Dg}$ modulo $g$.
Moreover
\begin{equation}\label{eq:anchor-geometry}
 D\bigl((D+2)K^2+K\bigr)\le2Q^2+Q<n,
\end{equation}
since $D\ge2$ and $DK\le Q$. Hence
$\tau>0$ and $1\le\sigma_0<Dg\le DK\le Q<T$.
The core heights in direction $\lambda u_0$ satisfy
\[
 0<\lambda w_a<J(n)Ds_a\le J(n)Q<n.
\]
The anchor height in that direction is exactly $Y$, because
\[
 g(\lambda Y_0-Y)=M(\lambda j-1)\in gn\Z,
 \qquad 0<Y<\frac Mg+J(n)Dx\le\frac n2+J(n)Q<n.
\]
In particular, no inversion of the possibly nonunit integer $g$ is
being used. Since $\lambda+\sigma_0=(t+1)Dg$, the target identity is
\begin{equation}\label{eq:anchor-target-identity}
 \lambda\tau+\sigma_0Y
 =\frac{(\lambda+\sigma_0)M}{g}=(t+1)DM=kn.
\end{equation}

Apply Lemma \ref{lem:integer-interval} to the denominations $w_a$.
Write $h=\max w_a\le K/g$ and let $L_0$ be its interval's left
endpoint. At most $4K\le4Q<T/2$ copies of each type are reserved,
and
\[
 gL_0\le2K^2+K<M-DK^2<M-x\sigma_0=g\tau
\]
by \eqref{eq:anchor-geometry}. If the sum in
\eqref{eq:shifted-anchor-budget} were at least $2M$, the unreserved
weight would be at least
\[
 \frac1{2g}\sum_{a\in\mathcal C}c_as_a\ge M/g>\tau.
\]
The resulting integer interval therefore contains $\tau$. Select
actual core copies with total $w$-weight $\tau$, and add
$\sigma_0$ copies of $b$. This anchor is outside $\mathcal C$,
since its diagonal coordinates have $D=1$, and its available
multiplicity is at least $T$. Equation
\eqref{eq:anchor-target-identity} gives a nonempty subsequence of
index at most $k\le r$, a contradiction.
\end{proof}

\subsection{Reciprocal weights and summation over orders}

\begin{proposition}\label{prop:near-linear-high}
Under \eqref{eq:near-linear-cutoff},
\begin{equation}\label{eq:near-linear-high}
 W=\sum_{a\in\mathcal H}c_a\ll n\mathcal A(n)\log(2n).
\end{equation}
\end{proposition}
\begin{proof}
Assume \eqref{eq:near-linear-geometry} and put $\alpha=3/2$.
For a pair with coordinates $(Dx,Ds)$ define
\[
 w_{ab}=\frac{s}{x}\,
 \alpha^{\sum_{p\in F_D}(v_p(x)-v_p(s))}.
\]
The reversed coordinates give $w_{ab}w_{ba}=1$, and $w_{aa}=1$.
Fix an anchor $b$. For fixed $D>1$, $x$, and valuations of $s$,
Lemma \ref{lem:shifted-anchor} and the bound of $D$ on the number
of phases give total $s$-weight less than $2MD=2n$.
Write $x=x_0\prod_{p\in F_D}p^{e_p}$, where $e_p=v_p(x)$ and
$x_0$ is coprime to every prime in $F_D$. If $e_p>0$, then
$(x,s)=1$ forces $v_p(s)=0$; together with the factor $p^{-e_p}$
from $1/x$, this contributes $(\alpha/p)^{e_p}$. If $e_p=0$,
summing over $v_p(s)=f\ge0$ contributes
$\sum_{f\ge0}\alpha^{-f}=3$. For each fixed exponent vector,
the remaining sum of $1/x_0$ over
$x_0\le K/\prod_{p\in F_D}p^{e_p}$ is at most $H_K$, where
$K=\lfloor Q/D\rfloor$. Extending these nonnegative sums to all
$e_p\ge0$ gives
\begin{align}
 \sum_{a:(q_{ab},v_{ab})=D}c_aw_{ab}
 &\le2nH_K\prod_{p\in F_D}
 \left(3+\sum_{e\ge1}(\alpha/p)^e\right)\notag\\
 &\le2n\,6^{|F_D|}H_K.                       \label{eq:all-prime-row}
\end{align}
where $H_K=\sum_{j=1}^K1/j$. Indeed,
$3+\alpha/(p-\alpha)\le6$ for every prime $p$, including $p=2$.

For $D=1$ the same row bound holds with the factor $6^{|F_D|}=1$.
To see this directly, fix $q_{ab}=x$. In direction $xb^{-1}$ the
corresponding heights are $v_{ab}$ and the anchor height is $x$.
Their union has gcd $1$, since $(x,v_{ab})=1$, and all heights are
at most $Q$. Include the anchor only once if it already occurs.
Lemma \ref{lem:integer-interval} reserves fewer than $T/2$ copies
per type and starts below $2Q^2+Q<n$. Thus
$\sum_{a:q_{ab}=x,\,(q_{ab},v_{ab})=1}c_av_{ab}<2n$;
otherwise its interval reaches $n$. Sum after division by $x$.

Consequently every row is at most
\[
 2nH_Q\sum_{D\mid n}6^{|F_D|}
 =2nH_Q\prod_{p^a\parallel n}(a+6)
 =2nH_Q\mathcal A(n).
\]
For each prime power, the exponents $0,\ldots,a-1$ in $D$ have
weight $1$, while exponent $a$ has weight $6$, proving the product
identity. Multiply each row by $c_b$ and sum over $b$. The reciprocal
relation and $z+z^{-1}\ge2$ yield
\[
 W^2\le\sum_{a,b\in\mathcal H}c_ac_bw_{ab}
 \le2nH_Q\mathcal A(n)W.
\]
Divide by $W$ when it is positive. The bounded exceptional moduli
are absorbed into the absolute constant as above.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:jacobsthal-upper}]
By Proposition \ref{prop:truncation}, the types of multiplicity below
$T$ in a unit sequence contribute $O(\varphi(n)\log(2n))$, since
\[
 T\le32\lfloor4\sqrt n\rfloor,\qquad
 c_a\le32\min\{c_a,\lfloor4\sqrt n\rfloor\}\quad(c_a<T).
\]
Together with Proposition \ref{prop:near-linear-high} this gives
\[
 L_{U,r}(n)\ll
 \bigl(\varphi(n)+n\mathcal A(n)\bigr)\log(2n).
\]
For a general sequence, partition by exact additive order $d\mid n$.
Lemma \ref{lem:orders} applies, and $r\ge10^6J(d)$ because
$J(d)\le J(n)$. Since $\mathcal A(d)\le\mathcal A(n)$,
\begin{align*}
 I_r(n)-1
 &\ll\sum_{d\mid n}\bigl(\varphi(d)+d\mathcal A(d)\bigr)\log(2d)\\
 &\le\bigl(n+\sigma(n)\mathcal A(n)\bigr)\log(2n).
\end{align*}
This proves \eqref{eq:jacobsthal-upper}, after increasing the constant.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:near-linear}]
Equation \eqref{eq:jacobsthal-iwaniec} makes the index hypothesis of
Proposition \ref{prop:jacobsthal-upper} valid with an absolute
constant in \eqref{eq:near-linear-main}. For every $\varepsilon>0$,
$\mathcal A(n)\ll_\varepsilon n^\varepsilon$. Indeed, $a+6\le7^a$
for $a\ge1$, so all sufficiently large primes satisfy
$a+6\le p^{\varepsilon a}$. For each of the finitely many smaller
primes, $(a+6)p^{-\varepsilon a}$ is bounded. Also
$\sigma(n)\le n(1+\log n)$. Applying these estimates with a smaller
$\varepsilon$ proves \eqref{eq:near-linear-main}.

If $\omega(n)\le k$, then $\sigma(n)/n\ll_k1$ and
$\mathcal A(n)\ll_k\log^k(2n)$. The first follows by bounding
$\prod_{p\mid n}(1-1/p)^{-1}$ by the product over the first $k$
primes; the second follows from $a\le\log n/\log2$ for every
$p^a\mid n$. Choose a fixed integer
$R_k\ge C[k\log(2k)]^2$ and apply
\eqref{eq:jacobsthal-upper}. This proves
\eqref{eq:bounded-prime-main} for all $r\ge R_k$.
\end{proof}


\section{Index-one upper bounds}\label{sec:upper-proofs}
\subsection{A bound in terms of the least prime divisor}\label{sec:rough}

\begin{lemma}\label{lem:row}
Let $S$ be an index-one-free unit sequence. Suppose $A$ is a set of types
of multiplicity at least $T\ge8\lceil\sqrt n\rceil$, and for each
$a,b\in A$ there are coordinates satisfying
\[
 q_{ab}a\equiv v_{ab}b\pmod n,\qquad
 1\le q_{ab},v_{ab}\le Q\le\sqrt n/4,
\]
with $(q_{ab},v_{ab})=1$, both coordinates units, and
$(q_{ba},v_{ba})=(v_{ab},q_{ab})$. On the diagonal take $(1,1)$.
Then $W=\sum_{a\in A}c_a<2nH_Q$, where $H_Q=\sum_{j=1}^Q1/j$.
\end{lemma}
\begin{proof}
Fix $b$ and a nonempty group of types with $q_{ab}=j$. In the unit
direction $jb^{-1}$ their distinct heights are $v_{ab}$, and the
height of $b$ is $j$. The union of these denominations has gcd $1$.
If $b$ already belongs to the group, include it once. The interval
of Lemma \ref{lem:integer-interval} requires at most $4Q\le T/2$
copies of every type and starts below $2Q^2+Q<n$.
If $\sum_{a:q_{ab}=j}c_av_{ab}\ge2n$, the unreserved copies in the
group have total weight at least $n$, so the interval reaches $n$.
Therefore
\[
 \sum_{a:q_{ab}=j}c_av_{ab}<2n,
 \qquad
 \sum_{a\in A}c_a\frac{v_{ab}}{q_{ab}}<2nH_Q.
\]
Multiply by $c_b$ and sum over $b$. Reciprocal slopes on the reversed
pair give
\begin{equation}\label{eq:reciprocal}
 W^2\le\sum_{a,b\in A}c_ac_b\frac{v_{ab}}{q_{ab}}<2nH_QW.
\end{equation}
Indeed, $x+x^{-1}\ge2$ on each unordered pair, and the diagonal is
unchanged. Divide by $W$ if it is nonzero.
\end{proof}

\begin{proposition}\label{prop:rough-unit}
For every $n\ge2$, with $\ell=P^{-}(n)$,
\begin{equation}\label{eq:rough-unit}
 L_U(n)\ll\log(2n)
 \left[n+\varphi(n)\left(1+\frac{\sqrt n}{\ell}\right)\right].
\end{equation}
Consequently the second term in \eqref{eq:general-main} bounds $I_1(n)$.
\end{proposition}
\begin{proof}
For an index-one-free unit sequence choose the integer
\[
 T=\max\{8\lceil\sqrt n\rceil,4n/\ell\},
 \qquad A=\{a:c_a\ge T\}.
\]
If $T\ge n$, then $A$ is empty. Otherwise Proposition
\ref{prop:capacity-threshold} implies that every two-type sequence
on $A$ has no ordinary zero sum. Apply Lemma \ref{lem:short-vector}
with $m=T$. Its coordinates are at most $Q=\lfloor n/T\rfloor\le
\ell/4$, so neither is divisible by a prime factor of $n$.
Equation \eqref{eq:lattice-gcd} then makes their integer gcd $1$.
Reversing the pair exchanges the coordinates by uniqueness. Lemma
\ref{lem:row} pays the entire high part by $O(n\log(2n))$.

Put $B_0=\lfloor4\sqrt n\rfloor$. For $c_a<T$,
\[
 c_a\le\max\{1,T/B_0\}\min(c_a,B_0),\qquad
 T/B_0\ll1+\sqrt n/\ell.
\]
Proposition \ref{prop:truncation} gives the remaining term in
\eqref{eq:rough-unit}. This argument also covers $A=\varnothing$.

For a general sequence use \eqref{eq:order-sum}, keeping the
$\varphi(d)$ factor in the low part. Since $P^{-}(d)\ge\ell$ for
$d\mid n$, its length is at most a constant times
\[
 \log(2n)\left[\sigma(n)+n+\frac{n\sqrt n}{\ell}\right].
\]
Finally
\[
 \frac{\sigma(n)}n=\sum_{d\mid n}\frac1d
 \le1+\frac{\tau(n)-1}{\ell}
 \le1+\frac{2\sqrt n}{\ell}.
\]
This proves the assertion, including all terms of proper order.
The extra $1$ in passing from free length to forcing threshold is
absorbed in the absolute constant.
\end{proof}

\subsection{Prime powers and a prime power times a larger prime}\label{sec:pp}

We now keep the short coordinates even when their gcd is nontrivial.
Throughout this section let $n=p^k$, or $n=p^kq$ with distinct primes
$p,q$ and $q\ge p^k$. In the first case choose
$T=8\lceil\sqrt n\rceil$; in the second choose
\begin{equation}\label{eq:mixed-cutoff}
 T=\max\{8\lceil\sqrt n\rceil,\lceil2q/p\rceil\}.
\end{equation}
If $T\ge n$ there are no high types; all arguments concerning them
are then omitted. Otherwise put
\[
 A=\{a:c_a\ge T\},\quad m=\lfloor T/2\rfloor,\quad
 Q=\lfloor n/m\rfloor,\quad B=\lfloor(n-1)/T\rfloor.
\]
Here $m^2>n$ and $Q\le\sqrt n/4$. Lemma \ref{lem:square} shows that
the sequence with multiplicities $\lfloor c_a/2\rfloor$ contains no
ordinary zero sum on at most two types. Thus Lemma
\ref{lem:short-vector} gives, for all distinct $a,b\in A$, unique
primitive positive coordinates
\begin{equation}\label{eq:short-pp}
 q_{ab}a\equiv v_{ab}b\pmod n,\qquad
 1\le q_{ab},v_{ab}\le Q.
\end{equation}
Use $(1,1)$ on the diagonal. The reversed coordinates are exchanged.

\begin{lemma}\label{lem:rectangle}
For the coordinates in \eqref{eq:short-pp},
\begin{equation}\label{eq:rectangle}
 q_{ab}c_b<n,\qquad v_{ab}c_a<n.
\end{equation}
Consequently $q_{ab},v_{ab}\le B$. Their gcd $D$ is a power of $p$.
Writing $(q_{ab},v_{ab})=(Dr,Ds)$, both $r,s$ are units modulo $n$.
\end{lemma}
\begin{proof}
Normalize $b$ to $1$ and $a$ to $a/b$, and write
$(\alpha,\beta)=(q_{ab},v_{ab})$. Complete this vector to the basis
$(\alpha,\beta),(s_0,-r_0)$ of Lemma \ref{lem:short-vector}. If
$\alpha\ge n/c_b$, then $\alpha>1$, $1\le r_0\le n/\alpha\le c_b$,
and $0\le s_0<\alpha\le Q<c_a$. The possibility $s_0=0$ would force
$n\mid r_0$, impossible. If $D=1$, $\alpha$ is a unit by
\eqref{eq:lattice-gcd}, and $r_0\alpha+s_0\beta=n$ gives index one
with available copies. If $D>1$, then $p\mid\alpha,\beta$. In the
mixed case $Q<q$, so $D$ has no prime factor other than $p$.
Since the first-coordinate projection of the lattice is $\Z$, the
basis gives $(\alpha,s_0)=1$, so $p\nmid s_0$.
The positive direction $\alpha-s_0$ avoids $p$ and, in the mixed case,
is less than $q$. It is a unit. In this direction the heights are
$\alpha-s_0$ and $\beta+r_0$, and
\[
 r_0(\alpha-s_0)+s_0(\beta+r_0)=n.
\]
This is another forbidden certificate within the original counts.
Thus $\alpha<n/c_b$; reversing the pair proves the other inequality.
The diagonal follows from $c_a<n$.

The gcd assertion follows from \eqref{eq:lattice-gcd} and $Q<q$ in
the mixed case. If $D=p^e$ with $e<k$, the divided coordinates avoid
$p$ by \eqref{eq:lattice-gcd}. When $e=k$, which is possible only in
the mixed case, \eqref{eq:mixed-cutoff} gives
\[
 r,s\le B/p^k<p/2.
\]
They again avoid $p$. All divided coordinates are less than the
larger prime, so they avoid that prime as well.
\end{proof}

The following local lemma is the step that removes the gcd loss.
Its formulation concerns actual positive representatives in a single
unit direction.

\begin{lemma}\label{lem:anchor}
Let $D=p^e>1$ with $D\mid n$, $D\le B$, and put $M=n/D$ and
$K=\lfloor B/D\rfloor$. In a unit direction suppose distinct core
types have heights $v_i$, and a separate type has height $r+jM$, where
\begin{equation}\label{eq:anchor-data}
 1\le r,v_i\le K,\quad (r,n)=(v_i,n)=(r,v_i)=1,
 \quad 1\le j<D,\quad(j,D)=1.
\end{equation}
Suppose all these multiplicities are at least $T$. If the sequence is
index-one-free, then
\begin{equation}\label{eq:anchor-budget}
 \sum_i c_i v_i<2n/D.
\end{equation}
\end{lemma}
\begin{proof}
Let $g=\gcd_i v_i$, $x_i=v_i/g$, and $h=\max_i x_i$. Thus
$(g,n)=(r,g)=1$ and $\gcd_i x_i=1$. The Chinese remainder theorem
gives a unique $0\le\sigma<Dg$ satisfying
\begin{equation}\label{eq:sigma}
 r\sigma\equiv M\pmod g,\qquad j\sigma\equiv-1\pmod D.
\end{equation}
In particular $1\le\sigma<Dg\le DK\le n/T<T$ and $p\nmid\sigma$.
Put
\begin{equation}\label{eq:tau-lambda}
 \tau=(M-r\sigma)/g,\qquad \lambda=Dg-\sigma.
\end{equation}
The integer $\lambda>0$ avoids $p$. In the mixed case it is at most
$B<q$, so it avoids the other prime too. Thus $\lambda g^{-1}$ is a
legal change of direction modulo $n$.

We check the endpoints before using it. Since $D\ge2$,
\begin{equation}\label{eq:anchor-size}
 D\bigl((D+2)K^2+K\bigr)
 \le2(DK)^2+DK\le2n^2/T^2+n/T<n.
\end{equation}
Consequently $(D+2)K^2+K<M$, and $r\sigma<DK^2$ implies $\tau>0$.
In the new direction the core heights are exactly $\lambda x_i$:
they lie in $(0,DK)\subset(0,n)$. The anchor height is exactly
$Dr+\tau$. Indeed, \eqref{eq:sigma} implies
\[
 \sigma(r+jM)+g\tau\equiv0\pmod n,
\]
and hence its residue after multiplying by $\lambda g^{-1}$ is
$Dr+\tau$. This positive integer is less than $n$, as is also
verified by the identity
\begin{equation}\label{eq:exact-certificate}
 \lambda\tau+\sigma(Dr+\tau)=Dg\tau+Dr\sigma=DM=n.
\end{equation}

It remains to represent the integer $\tau$ with available core copies.
If there are $s$ core types, Lemma \ref{lem:integer-interval} supplies
an interval starting at $L=\lceil2h/s\rceil h$, using reserves
$\rho=\lceil2h/s\rceil+1\le4K<T/2$ per type. Moreover
\[
 gL\le2K^2+K<M-DK^2<M-r\sigma=g\tau.
\]
If \eqref{eq:anchor-budget} fails, the unreserved normalized weight is
at least $\frac12\sum_i c_ix_i\ge M/g>\tau$. The interval therefore
contains $\tau$, represented by counts $d_i\le c_i$.
Take these core copies and $\sigma<T$ anchor copies in the new
direction. Equation \eqref{eq:exact-certificate} gives total weight
exactly $n$, contradicting freeness. No entry of height $g$ or $1$ has
been added: $g$ is used only to normalize integer denominations and
to change a unit direction.
\end{proof}

\begin{proposition}\label{prop:high-budget}
For the high types in this section,
\begin{equation}\label{eq:high-budget}
 \sum_{a\in A}c_a\le2n(k+1)H_Q.
\end{equation}
\end{proposition}
\begin{proof}
Fix $b\in A$. Pairs with $D=(q_{ab},v_{ab})=1$ satisfy exactly the
denomination-group estimate in the proof of Lemma \ref{lem:row}, and
contribute less than $2nH_Q$ to
$\sum_a c_av_{ab}/q_{ab}$.

For $D=p^e>1$, write $(q_{ab},v_{ab})=(Dr,Ds_a)$ with $(r,s_a)=1$,
and put $M=n/D$. Lemma \ref{lem:rectangle} ensures all divided
coordinates are units and at most $K=\lfloor B/D\rfloor$. Define
\[
 z_a=rab^{-1}s_a^{-1}\equiv1+t_aM\pmod n.
\]
Primitivity in the original lattice implies $(t_a,D)=1$. Indeed, if
$p\mid t_a$, then
\[
 ra-s_ab\equiv t_aMs_ab\pmod n
\]
implies $pM\mid ra-s_ab$, since $pM\mid n$. Hence
$n\mid(D/p)(ra-s_ab)$, so $(Dr/p,Ds_a/p)$ would still belong to
the original lattice, contrary to the primitivity of $(Dr,Ds_a)$.
Fix $r$ and a phase $t=t_a$. The unit direction
\[
 u_0=(1+tM)^{-1}rb^{-1}
\]
makes all core heights exactly the distinct integers $s_a$. The anchor
height is $r+jM$, with
$j\equiv-r(1+tM)^{-1}t\pmod D$, so $(j,D)=1$.
Lemma \ref{lem:anchor} applies to this group and yields
\[
 \sum_{a\text{ in the group}}c_as_a<2n/D.
\]
There are at most $\varphi(D)$ phases. Dividing by $r$, summing over
$r\le\lfloor B/D\rfloor$, and then over $D=p^e>1$, gives
\[
 \sum_{a:D>1}c_a\frac{v_{ab}}{q_{ab}}
 \le2n\sum_{\substack{1\le e\le k\\p^e\le B}}
 \frac{\varphi(p^e)}{p^e}H_{\lfloor B/p^e\rfloor}
 \le2nkH_Q.
\]
Adding the $D=1$ estimate, multiplying by $c_b$ and using reciprocal
pairing as in \eqref{eq:reciprocal} proves \eqref{eq:high-budget}.
The anchor is used separately to prove each necessary inequality;
certificates from different directions are never concatenated.
\end{proof}

\begin{proof}[Proof of \eqref{eq:pp-main} and \eqref{eq:mixed-main}]
For $n=p^k$, the low part has $c_a<T=8\lceil\sqrt n\rceil$, so
\eqref{eq:unit-trunc} bounds it by $O(\varphi(n)\log(2n))$.
Proposition \ref{prop:high-budget} gives
\[
 L_U(p^k)\ll(k+1)p^k\log(2p^k).
\]
The finite cases with $T\ge n$ are absorbed in the constant. Sum over
the exact-order classes $p^j$, $1\le j\le k$. Since
$\sum_{j=1}^kp^j\le2p^k$, Lemma \ref{lem:orders} proves
\eqref{eq:pp-main}. Also $k\le\log_2 n$, giving its uniform consequence.

For $n=p^kq$, the cutoff \eqref{eq:mixed-cutoff} and
\eqref{eq:unit-trunc} give the low-part estimate
\[
 O\left(n\log(2n)\left(1+\frac1p\sqrt{q/p^k}\right)\right).
\]
Together with \eqref{eq:high-budget} this proves the corresponding
unit bound. Apply it to the classes $p^jq$, $1\le j\le a$, when
the ambient order is $p^aq$. The necessary sums satisfy
\[
 \sum_{j=1}^a(j+1)p^jq\le2(a+1)p^aq,
 \qquad
 \sum_{j=1}^a p^{j/2-1}q^{3/2}
 \ll\frac{p^aq}{p}\sqrt{q/p^a}.
\]
The prime class $q$ and the pure $p$-power classes are paid by
\eqref{eq:pp-main}. Lemma \ref{lem:orders} now gives
\eqref{eq:mixed-main} for arbitrary sequences.
\end{proof}

\subsection{Twice a prime square}\label{sec:twice-square}

We prove \eqref{eq:twice-square-main}. Throughout this section $p$ is
odd and $n=2p^2$; the single order arising from $p=2$ is absorbed into
the absolute constant.

\begin{lemma}\label{lem:sixfold}
Every atom supported on at most two units of $C_{2p^2}$ has an
index-one subsequence in its sixth power.
\end{lemma}
\begin{proof}
Only a mixed atom $1^r b^s$, with $s>1$, requires proof. Lemma
\ref{lem:virtual} gives positive heights $u,v$ with
\begin{equation}\label{eq:six-heights}
 ru+sv=n,\qquad v\equiv bu\pmod n,\qquad
 (u,s)=(v,r)=1.
\end{equation}
For the second coprimality, a prime dividing $v,r$ would divide $n$,
then $s$ from $r+bs\equiv0$, and then $u$ from $v\equiv bu$, a
contradiction. The counts $r,s$ have the same parity, as do $u,v$.

First suppose $p\mid u,v$, allowing any positive zero-sum counts and
heights satisfying \eqref{eq:six-heights}. Write $u=pA,v=pB$. Then
\[
 rA+sB=2p,\qquad p\nmid rs.
\]
If $u,v$ are even, then $A,B\ge2$, $r+s\le p$, and $r,s$ are odd.
The heights $(u+s,v-r)$ are positive units with the same counts.
If $u,v$ are odd and $r,s$ are even, use these heights when $v>r$.
Otherwise $r\ge p$ and $s<p\le u$, so use $(u-s,v+r)$.
These also are positive units with the same counts.

It remains that all four integers are odd. If $v>2r$, use
$(u+2s,v-2r)$; if $u>2s$, use $(u-2s,v+2r)$. Otherwise
$r\ge pB/2$ and $s\ge pA/2$ imply $AB\le2$. Thus $A=B=1$,
$r+s=2p$, and $r,s\ge p/2$, with neither equal to $p$.
For $r<p<s$, take
\[
 (R,S,U,V)=(r+p,s-p,r,r+2p);
\]
for $s<p<r$, take
\[
 (R,S,U,V)=(r-p,s+p,s+2p,s).
\]
In both cases
$RU+SV=n$, $V\equiv bU\pmod n$, and $U,V$ are positive units,
while $R\le3r,S\le3s$. For example, the first choice follows from
the neighboring counts $(r+v,s-u)$ and the height shift
$(u-S,v+R)$. This proves a threefold repair whenever $p\mid u,v$.

If the original $u$ is already a unit, finish. The only remaining case
has $u,v$ even and $p\nmid u$. Then $r,s$ are odd. If $v>r$, use
the unchanged counts and heights $(u+s,v-r)$. If $v<r$ and
$ru\ge n/2$, put
\[
 j=\lfloor(s-1)/u\rfloor,\quad R=r+jv,\quad S=s-ju,
 \qquad (U,V)=(u-S,v+R).
\]
Here $1\le S<u$, $R<n/u\le2r$ and $S\le s$.
The coprimality and zero-sum relations are preserved, since
\begin{gather*}
 (U,S)=(u,S)=(u,s)=1,\qquad
 (V,R)=(v,R)=(v,r)=1,\\
 R+bS=(r+bs)+j(v-bu)\equiv0\pmod n.
\end{gather*}
If $ru<n/2$, put instead
\[
 j=\lfloor(r-1)/v\rfloor,\quad R=r-jv,\quad S=s+ju,
 \qquad (U,V)=(u+S,v-R).
\]
Then $1\le R<v$, $S<n/v<2s$ and $R\le r$.
The same gcd equalities hold, and now
$R+bS=(r+bs)-j(v-bu)\equiv0\pmod n$.
Equality $v=r$ is excluded by parity. In every case the new heights
are positive odd integers, satisfy the same determinant and congruence
identities, and obey $(U,S)=(V,R)=1$. If they avoid $p$, finish;
otherwise apply the preceding threefold repair. The resulting counts
are at most $6r,6s$. A positive identity of weight $n$ makes both
unit heights less than $n$, so they are the actual representatives.
\end{proof}

The next lemma handles an even ordinary gcd without inverting it
modulo $2p^2$.

\begin{lemma}\label{lem:parity-budget}
Put $M=p^2$. Let $r,s_i\in[1,K]$ with the $s_i$ distinct,
$(r,s_i)=1$, and $r,s_i$ of opposite parity. Suppose all $s_i$ have
the same $2$-adic valuation and
\[
 2K<p,\qquad 4K^2+K<M.
\]
Let $b,a_i$ be actual unit types with $ra_i\equiv s_i b\pmod M$,
each of multiplicity at least $T\ge8K+8$. In an index-one-free
sequence their original multiplicities satisfy
\begin{equation}\label{eq:parity-budget}
 \sum_i c_{a_i}s_i<2M=n.
\end{equation}
\end{lemma}
\begin{proof}
Let $g=(s_i:i)$ and $x_i=s_i/g$. The $x_i$ are odd, have gcd $1$,
and have maximum $h\le K/g$; also $(g,p)=(g,r)=1$.
Let $u_0$ be the odd lift modulo $2M$ of
$rg^{-1}b^{-1}\pmod M$. It is a unit and
$\res{u_0a_i}{n}=x_i$. Writing $y=\res{u_0b}{n}$ gives
\[
 gy=r+jM,\qquad 1\le j<2g,\qquad (j,2g)=1.
\]
Indeed $g,r$ have opposite parity, so $j$ is odd; any common divisor
of $j,g$ divides $r$. The anchor is separate from the core since
$y>M/g>h$.

Take $1\le\lambda<2g$ with $\lambda j\equiv1\pmod{2g}$, and put
\[
 \sigma=2g-\lambda,\qquad
 \tau=(M-r\sigma)/g,\qquad Y=(M+\lambda r)/g.
\]
These are integers. The number $\lambda$ is odd and less than $p$,
so $\lambda u_0$ is a legal unit. The actual core heights and anchor
height in this direction are respectively $\lambda x_i$ and $Y$,
because
\begin{equation}\label{eq:parity-certificate}
 \lambda\tau+\sigma Y=2M.
\end{equation}
Here $r\sigma<2K^2<M$ makes $\tau>0$, the identity gives $0<Y<n$,
and the anchor congruence follows from $\lambda j\equiv1\pmod{2g}$.
Also $\sigma<2g\le2K<T$.

Apply Lemma \ref{lem:integer-interval} to the denominations $x_i$.
With $k$ types, take $\ell=\lceil2h/k\rceil$, $\rho=\ell+1$ and
$L=\ell h$. Its reserve satisfies $\rho\le4h<T/2$, while
\[
 gL\le2K^2+K<M-2K^2<M-r\sigma=g\tau.
\]
If \eqref{eq:parity-budget} fails, the unreserved core has weight
at least $M/g>\tau$. The interval therefore represents the integer
$\tau$ using available copies. Combine it with $\sigma$ anchor copies
in direction $\lambda u_0$; \eqref{eq:parity-certificate} gives
weight exactly $n$, a contradiction.
\end{proof}

\begin{proof}[Proof of \eqref{eq:twice-square-main}]
First consider a free unit sequence and set
\[
 T=64\lceil\sqrt n\rceil,\quad m=\lfloor T/6\rfloor,\quad
 Q=\lfloor n/m\rfloor,\quad K=\lfloor Q/2\rfloor,\quad
 A=\{a:c_a\ge T\}.
\]
If $T\ge n$, the high part is empty. Otherwise $m\ge T/7$,
$m^2>n$, and $Q\le7\sqrt n/64<\sqrt n/8<p/2$.
Lemma \ref{lem:sixfold} implies that every pair of types in $A$
is ordinary-zero-sum-free in its $m$-by-$m$ capacity box. Lemma
\ref{lem:short-vector} supplies unique primitive coordinates
$(q_{ab},v_{ab})\in[1,Q]^2$; reverse pairs exchange coordinates.
Their ordinary gcd $D$ is $1$ or $2$ by \eqref{eq:lattice-gcd}.
On the diagonal take $(1,1)$.

If $D=1$, both coordinates are odd units, and the denominator-group
argument in Lemma \ref{lem:row} gives, for each fixed anchor $b$,
\begin{equation}\label{eq:twice-good}
 \sum_{a:D=1}c_a v_{ab}/q_{ab}<2nH_Q.
\end{equation}
If $D=2$, write $(q_{ab},v_{ab})=(2r_{ab},2s_{ab})$.
The reduced coordinates are coprime and have opposite parity: if
both were odd, $r_{ab}a-s_{ab}b$ would be even and divisible by
the odd number $M$, contradicting lattice primitivity.
For fixed $r=r_{ab}$ and $e=v_2(s_{ab})$, the numerators $s_{ab}$
are distinct. Indeed a reduced relation determines $a$ modulo $M$,
and a unit modulo $M$ has a unique odd lift. Lemma
\ref{lem:parity-budget} applies and gives
\[
 \sum_{a:r_{ab}=r,\ v_2(s_{ab})=e}c_a s_{ab}<n.
\]

Choose $\alpha=3/2$ and assign the positive rational pair weight
\[
 w_{ab}=\frac{v_{ab}}{q_{ab}}
 \alpha^{v_2(q_{ab})-v_2(v_{ab})}.
\]
It has $w_{ab}w_{ba}=1$ and $w_{aa}=1$; for $D=1$ it is the
unchanged slope. For $D=2$ and odd $r$, the numerators have
$e\ge1$, so their total row cost is at most
$nH_K\sum_{e\ge1}\alpha^{-e}=2nH_K$.
For even $r=2^f r_0$, the numerator is odd; summing over odd $r_0$
and $f\ge1$ costs at most
$nH_K\sum_{f\ge1}(\alpha/2)^f=3nH_K$.
Thus every row has
\[
 \sum_{a\in A}c_a w_{ab}<2nH_Q+5nH_K\le7nH_Q.
\]
Writing $W=\sum_{a\in A}c_a$, reciprocity gives
\[
 W^2\le\sum_{a,b\in A}c_ac_b w_{ab}<7nH_QW.
\]
This pays all original high multiplicities with one harmonic sum.
Proposition \ref{prop:truncation} pays the low part by
$O(\varphi(n)\log(2n))$, since $T$ is an absolute constant times
$\sqrt n$. Hence $L_U(2p^2)\ll n\log(2n)$.

Finally the proper nonzero exact orders are $2,p,2p,p^2$.
The order-$2$ class has length at most one. The prime and prime-square
classes have $O(d\log(2d))$ bounds from Section \ref{sec:pp}.
For the order-$2p$ class, the elementary mixed estimate from that
section gives $O(p^{3/2}\log(2p))$, already smaller than the required
$O(p^2\log(2p))$ ambient bound. Thus no sharper twice-prime estimate
is needed. Lemma \ref{lem:orders} includes every nonunit term with
the same $O(n\log(2n))$ total cost.
\end{proof}


\subsection{Character estimates and the GRH bound}\label{sec:analytic}

Let $M_{2,U}(d)$ be the maximum length of a unit sequence modulo $d$
having no ordinary nonempty zero sum on at most two distinct types.
At moduli where Lemma \ref{lem:square} holds, halving each multiplicity
in an index-one-free sequence shows
\begin{equation}\label{eq:half-transfer}
 L_U(d)\le2M_{2,U}(d)+\varphi(d).
\end{equation}
A two-type atom in the halved sequence would have its square inside
the original inventory, contradicting that lemma.

\begin{lemma}\label{lem:character-row}
Suppose $2\le m\le\lfloor(d-1)/2\rfloor$, and let $A_m$ be the set
of types having multiplicity at least $m$ in such a two-type-zero-sum-free
sequence. Write $K=|A_m|$ and $\Phi=\varphi(d)$. For a nonnegative
weight $w$ supported on $1\le h\le m$, set
\[
 W=\sum_{(h,d)=1}w(h),\qquad
 B_\chi=\sum_{h\le m}w(h)\chi(h),\qquad
 \lambda=\max_{\chi(-1)=-1}|B_\chi|^2.
\]
Then $K\le\Phi\lambda/(W^2+\lambda)$ whenever $W>0$.
\end{lemma}
\begin{proof}
There are no equations $ha+jb=0$ with $a,b\in A_m$ and positive unit
counts $h,j\le m$. This includes $a=b$, because $h+j<d$.
Character orthogonality on $\U(d)$ gives
\[
 0=K^2W^2+\sum_{\chi\ne\chi_0}
 \chi(-1)|B_\chi|^2\left|\sum_{a\in A_m}\chi(a)\right|^2.
\]
Discard the even nonprincipal terms, which are nonnegative, and use
Parseval to bound the remaining sum of squared coefficients by
$\Phi K-K^2$. Thus $K^2W^2\le\lambda(\Phi K-K^2)$.
\end{proof}

\begin{proposition}\label{prop:analytic}
Uniformly for $d$ having at most two distinct prime divisors,
\[
 M_{2,U}(d)\ll_\varepsilon d^{4/3+\varepsilon}.
\]
Under the Dirichlet generalized Riemann hypothesis,
\[
 M_{2,U}(d)\ll d+\varphi(d)\log^3(2d).
\]
\end{proposition}
\begin{proof}
Every multiplicity is less than $d$. Any two distinct multiplicities
have sum at most $d-1$: otherwise the progressions
$\{0,a,\ldots,c_aa\}$ and $\{0,-b,\ldots,-c_bb\}$ intersect at a
nonzero residue, giving a forbidden zero sum. Thus at most one type
has multiplicity exceeding $\lfloor(d-1)/2\rfloor$, and its tail
costs less than $d$. It suffices to sum the layer bounds of Lemma
\ref{lem:character-row} up to this cutoff; finitely many small layers
cost $O(d)$.

For the unconditional bound use $w(h)=(1-h/m)^2$ for $h\le m$.
For every nonprincipal character the Mellin identity is
\[
 B_\chi=\frac1{2\pi i}\int_{(2)}
 L(s,\chi)m^s\frac2{s(s+1)(s+2)}\,ds.
\]
Petrow and Young \cite[Theorem 1.1]{PY} prove, for every primitive
character of conductor $f$,
\begin{equation}\label{eq:weyl-L}
 L(1/2+it,\chi)\ll_\delta(f(1+|t|))^{1/6+\delta}.
\end{equation}
An induced character adds at most two bounded Euler factors on
$\Re s=1/2$. Shifting the integral there, its cubic denominator gives
\[
 |B_\chi|\ll_\delta\sqrt m\,d^{1/6+\delta},
 \qquad \lambda\ll_\delta m d^{1/3+2\delta}.
\]
Inclusion--exclusion over at most two primes gives
$W=m\varphi(d)/(3d)+O(1)$, uniformly. Since $\varphi(d)/d\ge1/3$,
$W\gg m$ above an absolute constant. Thus
$K\ll_\delta d^{4/3+2\delta}/m$. Summing the layers and absorbing
the harmonic logarithm into $d^\varepsilon$ proves the first assertion.

For the conditional assertion use the von Mangoldt weight
\[
 w(h)=\Lambda(h)\frac hm\left(1-\frac hm\right)
 \qquad(1\le h\le m).
\]
We give the uniform estimates, including the passage to induced
characters. Only odd characters occur in the maximum in Lemma
\ref{lem:character-row}; their primitive inducing characters are
nonprincipal and odd as well.

First let $\psi$ be primitive and odd, of conductor $f$. Mellin
inversion with kernel $1/((s+1)(s+2))$ gives
\begin{equation}\label{eq:grh-mellin}
 B_\psi=\frac1{2\pi i}\int_{(2)}
 -\frac{L'}L(s,\psi)\frac{m^s}{(s+1)(s+2)}\,ds.
\end{equation}
Shift to $\Re s=-1/2$, taking horizontal edges at a sequence of heights
avoiding zeros. The crossed poles are precisely the nontrivial zeros,
counted with multiplicity: $L(s,\psi)$ has no pole, its first trivial
zero is at $-1$, and the kernel poles $-1,-2$ lie to the left of the
new line. On that line the functional equation and the absolutely
convergent logarithmic derivative at $3/2-it$ imply
\[
 \left|\frac{L'}L(-1/2+it,\psi)\right|
 \ll\log\bigl(f(2+|t|)\bigr).
\]
The kernel has quadratic decay there. The remaining integral is
therefore $O(m^{-1/2}\log(2f))$, with an absolute constant, and
\begin{equation}\label{eq:grh-explicit}
 B_\psi=-\sum_\rho\frac{m^\rho}{(\rho+1)(\rho+2)}
        +O\bigl(m^{-1/2}\log(2f)\bigr).
\end{equation}
This is the usual explicit-formula contour argument; see
\cite[Section 2]{LLS} for the completed-function framework.

Here is an explicit uniform bound for the zero sum. Set
\[
 \mathcal L(s,\psi)=
 (f/\pi)^{(s+1)/2}\Gamma((s+1)/2)L(s,\psi).
\]
Let $b_\psi$ be the coefficient of $s$ in the exponential factor of
the Hadamard product for $\mathcal L(s,\psi)$. The functional equation
gives, also for complex $\psi$,
\[
 \Re b_\psi+\sum_\rho\Re\frac1\rho=0;
\]
see \cite[(2.3)--(2.4)]{LLS}. Taking real parts of the logarithmic
derivative and using GRH therefore gives at $s=2$
\begin{align*}
 \sum_{\rho=1/2+i\gamma}
 \frac{3/2}{(3/2)^2+\gamma^2}
 &=\Re\frac{\mathcal L'}{\mathcal L}(2,\psi)\\
 &=\frac12\log(f/\pi)
   +\frac12\frac{\Gamma'}\Gamma(3/2)
   +\Re\frac{L'}L(2,\psi)
 \ll\log(2f).
\end{align*}
The last logarithmic derivative is absolutely bounded, since
$|L'/L(2,\psi)|\le\sum_{h\ge2}\Lambda(h)/h^2<\infty$.
Moreover $|\rho+1||\rho+2|\gg1+\gamma^2$. Consequently
\begin{equation}\label{eq:grh-zero-sum}
 \sum_\rho\frac1{|(\rho+1)(\rho+2)|}\ll\log(2f),
 \qquad |B_\psi|\ll\sqrt m\log(2f).
\end{equation}

Now let $\chi$ modulo $d$ be induced by $\psi$. Their weighted sums
differ only at powers of primes dividing $d$ but not $f$. Because
$0\le t(1-t)\le1/4$ on $[0,1]$ and $\omega(d)\le2$,
\begin{align}
 |B_\chi-B_\psi|
 &\le\frac14\sum_{\substack{\ell\mid d,\ \ell\text{ prime}\\\ell\nmid f}}
       \log\ell\left\lfloor\frac{\log m}{\log\ell}\right\rfloor
 \le\frac12\log m.\label{eq:grh-induced}
\end{align}
Since $f\mid d$ and $m<d$, this proves
\begin{equation}\label{eq:grh-odd-bound}
 |B_\chi|\ll\sqrt m\log(2d),\qquad
 \lambda\ll m\log^2(2d)
\end{equation}
uniformly over all the odd characters in question.

Finally, the principal mass is uniformly large. The prime number
theorem and partial summation give
\[
 \sum_{h\le m}\Lambda(h)\frac hm\left(1-\frac hm\right)
 =\frac m6+o(m).
\]
Removing prime powers whose primes divide $d$ costs at most
$\omega(d)\log m/4\le\frac12\log m$, by the same estimate as
\eqref{eq:grh-induced}. Thus $W\ge m/12$ for all $m\ge m_0$, where
$m_0$ is absolute and independent of $d$. Lemma
\ref{lem:character-row} and \eqref{eq:grh-odd-bound} now yield
\[
 |A_m|\ll\frac{\varphi(d)\log^2(2d)}m\qquad(m\ge m_0).
\]
Summing these layers contributes $O(\varphi(d)\log^3(2d))$.
The layers $m<m_0$ cost $O(d)$, as does the single-type tail already
isolated above. This proves the conditional assertion with an
absolute implied constant.
\end{proof}

\begin{proof}[Proof of \eqref{eq:weyl-main} and \eqref{eq:grh-main}]
Every divisor of $p^aq$ is either $p^j$ or $p^jq$, and $q\ge p^j$.
Thus Lemma \ref{lem:square} and \eqref{eq:half-transfer} apply to
each exact-order class. With $\alpha=4/3+\varepsilon$,
$\sum_{d\mid p^aq}d^\alpha\ll_\varepsilon(p^aq)^\alpha$.
For the conditional bound use
$\sum_{d\mid p^aq}d\le3p^aq$ and
$\sum_{d\mid p^aq}\varphi(d)=p^aq$. Lemma \ref{lem:orders} and
Proposition \ref{prop:analytic} complete both proofs.
\end{proof}

\section{Lower bounds}\label{sec:lower-proofs}
\subsection{A construction on all large multiples of a fixed integer}
\label{sec:divisor-lower}

\begin{proposition}\label{prop:finite-lower}
Let $d\ge2$, $r\ge1$ and $Q\ge1$ be integers with $\rad(Q)\mid d$.
Let $m>rdQ$ and set $n=dm$, $H=m-rdQ$. Then
\begin{equation}\label{eq:finite-lower-sequence}
 S=\prod_{a\mid Q}\ \prod_{\substack{0\le i<d\\(i,a)=1}}
 (im+a)^{\lfloor H/a\rfloor}
\end{equation}
is index-$r$-free over $C_n$, and
\begin{equation}\label{eq:finite-lower-length}
 |S|=\sum_{a\mid Q}\frac{d\varphi(a)}a\left\lfloor\frac Ha\right\rfloor.
\end{equation}
Here $(0,1)=1$, so all $d$ lifts with $a=1$ are included. There is no
requirement that $Q$ divide $m$ or $n$.
\end{proposition}
\begin{proof}
All displayed types lie in $[1,n-1]$ and are distinct: $a\le Q<m$,
so reduction modulo $m$ determines $a$, then determines $i$.
Fix any $u\in\U(n)$ and write their actual heights as
\[
 w_{a,i}=\res{u(im+a)}n=ua+m(iu-dk_{a,i}),
 \qquad k_{a,i}\in\Z.
\]
Every prime dividing $a$ divides $d$ but divides neither $i$ nor $u$.
Thus $(iu-dk_{a,i},a)=1$. The rational numbers
\[
 \frac{w_{a,i}/a-u}{m}=\frac{iu-dk_{a,i}}a
\]
are reduced. Equality of two normalized heights $w_{a,i}/a$ first
forces equality of their positive reduced denominators, and then
$d\mid u(i-j)$, so the types are equal. Since $a\mid Q$, the
numbers
\[
 y_{a,i}=Qw_{a,i}/a
\]
are distinct positive integers, all congruent to $Qu$ modulo $m$.
At most one lies in $(0,m)$. For every other type,
\begin{equation}\label{eq:high-lower}
 w_{a,i}\ge am/Q.
\end{equation}

Suppose actual counts $e_{a,i}\le\lfloor H/a\rfloor$ give an integer
sum $jn$, $1\le j\le r$. Reduction modulo $m$, and $(u,m)=1$, show
that $B=\sum e_{a,i}a$ is a positive multiple of $m$.
If no copy of the possible exceptional type with $y<m$ is selected,
\eqref{eq:high-lower} gives $B\le jdQ<m$, a contradiction.
Otherwise its strictly positive height makes the contribution of all
other types satisfy $B_{\mathrm{high}}<jdQ\le rdQ$.
The one exceptional type contributes at most
$a\lfloor H/a\rfloor\le H=m-rdQ$. Hence again $B<m$, a contradiction.
This rules out every sum $n,2n,\ldots,rn$ in every unit direction,
and is equivalent to index-$r$-freeness.

The number of allowed $i$ for a fixed $a$ is $d\varphi(a)/a$:
coprimality depends only on $\rad(a)$, which divides $d$.
This proves \eqref{eq:finite-lower-length}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:divisor-lower}]
Define
\[
 F(Q)=\sum_{a\mid Q}\frac{\varphi(a)}{a^2},
 \qquad G(Q)=\sum_{a\mid Q}\frac{\varphi(a)}a.
\]
Proposition \ref{prop:finite-lower} and $\lfloor x\rfloor\ge x-1$ give
\begin{equation}\label{eq:finite-count}
 |S|\ge(n-rd^2Q)F(Q)-dG(Q),\qquad G(Q)\le Q.
\end{equation}
If $Q=\prod_{p\mid d}p^{k_p}$, multiplicativity yields
\[
 F(Q)=\prod_{p\mid d}\left(1+\frac1p-\frac1{p^{k_p+1}}\right).
\]
Set $X=n^{1/(s+1)}$, $k_p=\lfloor\log_p X\rfloor$. Then
$Q\le X^s$, and $n>(rd^2)^{s+1}$ ensures $n>rd^2Q$, as required.
Since $p^{k_p+1}>X$,
\[
 0\le P(d)-F(Q)\le P(d)\sum_{p\mid d}p^{-k_p-1}
 \le sP(d)/X.
\]
Substitution in \eqref{eq:finite-count} gives
\eqref{eq:divisor-explicit} and hence \eqref{eq:divisor-main}.

Finally the numbers $P(d)$ are unbounded. For $d$ the product of the
primes at most $z$,
\[
 P(d)=\prod_{p\le z}(1-p^{-2})
       \prod_{p\le z}(1-p^{-1})^{-1}
 \ge\frac12\sum_{j\le\lfloor z\rfloor}\frac1j\longrightarrow\infty.
\]
Here the first product is at least
$\prod_{j=2}^{\infty}(1-j^{-2})=1/2$, and the finite Euler product
contains all reciprocals $1/j$ with $j\le z$. Given $M$, choose $d$
with $P(d)>M$, independently of $r$, and then let its multiples tend
to infinity. Since $I_r(n)>|S|$, the final assertion follows.
\end{proof}

\begin{example}
Take $d=6$, $Q=6$, $r=1$, and $m=2000$. Formula
\eqref{eq:finite-lower-sequence} gives the following index-one-free
sequence over $C_{12000}$:
\[
 \begin{aligned}
 &\prod_{i=0}^{5}(2000i+1)^{1964}\;
 \prod_{i\in\{1,3,5\}}(2000i+2)^{982}\;\cdot{}\\[-2pt]
 &\hspace{25mm}
 \prod_{i\in\{1,2,4,5\}}(2000i+3)^{654}\;
 \prod_{i\in\{1,5\}}(2000i+6)^{327}.
 \end{aligned}
\]
Its length is
$6\cdot1964+3\cdot982+4\cdot654+2\cdot327=18000$.
Thus $I_1(12000)\ge18001$, contradicting the proposed bound
$n+n/P^{-}(n)=18000$ at this order. This example is a direct instance
of the proof, with no exhaustive computation needed.
\end{example}

\subsection{Long free sequences consisting only of generators}
\label{sec:unit-lower}

For a real $K\ge2$ let $\mathcal P(K)=\prod_{\ell\le K,\ \ell\text{ prime}}\ell$.
The first observation uses a gap in the small positive units and a
multiplicative subgroup of the unit group.

\begin{lemma}\label{lem:unit-gap}
Let $n=AK$ with integers $A,K\ge2$. Let $T$ be a nontrivial
multiplicative subgroup of $\U(n)$ such that
\begin{enumerate}
\item every $a\in T$ is congruent to $1$ modulo $A$;
\item $\U(n)\cap[1,K]=\{1\}$ in positive representatives;
\item every nonidentity representative of $T$ is greater than $n/2$.
\end{enumerate}
Put $\delta=\min_{a\in T\setminus\{1\}}(n-\res an)$.
For every $1\le c<\delta$, $\prod_{a\in T}a^c$ is index-one-free.
At $c=\delta$ it has an index-one subsequence.
\end{lemma}
\begin{proof}
An index-one certificate in direction $u$ has a number of terms
divisible by $A$, since every term is congruent to $u$ modulo $A$.
It therefore uses at least $A$ copies. If $uT$ does not contain the
representative $1$, every height is a unit greater than $K$, giving
weight at least $A(K+1)>n$.
If $1\in uT$, then $u\in T$ and multiplication by $u$ permutes the
uniform sequence. A sum $n$ now uses either only $1$'s, requiring $n$
copies, or one nonidentity height $a>n/2$, requiring
$n-a\ge\delta$ copies of $1$. Neither is available when $c<\delta$.
At $c=\delta$, an element attaining the minimum together with
$\delta$ copies of $1$ gives weight $n$.
\end{proof}

To construct many subgroup elements at heights greater than $n/2$,
we first represent the sixth roots of unity by small positive costs.

\begin{lemma}\label{lem:six-roots}
For every prime $p\equiv1\pmod3$, $p>64$, there are positive integers
$x,y$ such that
\[
 p=x^2-xy+y^2,\qquad x/2<y<2x.
\]
Set $\lambda=(x+y)^{-1}\pmod p$ and
\[
 E_p=\{0,3x,3y,2(x+y),2y-x,2x-y\}.
\]
These six distinct integers lie in $[0,p)$, and
$\{1-\lambda e:e\in E_p\}$ is exactly the sixth-root subgroup of
$\F_p^\times$. Its only zero cost is $e=0$, and
$\max E_p=2(x+y)\le4\sqrt p$.
\end{lemma}
\begin{proof}
Let $\omega=e^{2\pi i/3}$. The norm on $\Z[\omega]$ is
$N(a+b\omega)=a^2-ab+b^2$. Rounding both real coordinates in the
basis $1,\omega$ gives error norm at most $3/4$, proving the Euclidean
algorithm and principality of ideals. A nontrivial cubic root in
$\F_p$ defines a surjection $\Z[\omega]\to\F_p$. Its kernel has
index $p$, so a generator has norm $p$. Multiply it by a complex unit
to put its argument in $[\pi/6,\pi/2]$. Its coordinates are then
$x,y>0$ with $x/2\le y\le2x$. Equality would give a norm three times
a square, impossible for $p>3$.

Modulo $p$, $(x+y)^2=3xy$. Thus
$r_1=1-3\lambda x$ and $r_2=1-3\lambda y$ have sum $-1$ and product
$1$, and are the two nontrivial cubic roots. They are distinct, since
$x=y$ would give $p=x^2$. The six roots
$1,r_1,r_2,-1,-r_1,-r_2$ have the listed costs. Their positive costs
follow from $x/2<y<2x$. Finally
$p=(x+y)^2-3xy\ge(x+y)^2/4$, so the maximum cost is at most
$4\sqrt p<p$. Distinct roots then imply distinct integer costs.
\end{proof}

\begin{proposition}\label{prop:six-construction}
Choose distinct primes $p_i\equiv1\pmod3$, $p_i>64$, with cost sets
$E_i$ and coefficients $\lambda_i$ as above, such that
\begin{equation}\label{eq:cost-condition}
 \sum_{i=1}^s\frac{\max E_i}{p_i}<\frac12.
\end{equation}
Put $K=\prod_i p_i$, $d_i=K/p_i$, $A_0=\mathcal P(K)/K$.
Choose a positive integer $c$ satisfying
\begin{equation}\label{eq:CRT-lower}
 A_0c\,d_i\equiv\lambda_i\pmod{p_i}\quad(1\le i\le s),
\end{equation}
and $A=A_0c\ge2K$. Set $n=AK$. For
$e=(e_i)\in\prod_i E_i$, put $B(e)=\sum_i d_ie_i$ and define
\[
 a_0=1,\qquad a_e=n-AB(e)+1\quad(e\ne0).
\]
These $6^s$ types form a multiplicative subgroup of $\U(n)$. Writing
$e_i^+=\min(E_i\setminus\{0\})$, the sequence with each type repeated
\begin{equation}\label{eq:six-capacity}
 c_*=n\min_i(e_i^+/p_i)-2
\end{equation}
times is index-one-free, and has length at least
$6^s(n/\max_i p_i-2)$.
\end{proposition}
\begin{proof}
Since $(A_0,K)=1$, the Chinese remainder theorem gives
\eqref{eq:CRT-lower}; all its residues for $c$ are units, so $(A,K)=1$.
Adding a multiple of $K$ to $c$ ensures $A\ge2K$. By
\eqref{eq:cost-condition}, $0<B(e)<K/2$ for $e\ne0$.
Thus $a_e\in(n/2,n)$, and every $a_e$ is $1$ modulo $A$.
Modulo $p_i$, it is $1-\lambda_i e_i$. Hence the types are distinct,
are units, and form under CRT the subgroup
$\{1\}\times\prod_i\mu_6(\F_{p_i})$ in
$\U(A)\times\prod_i\F_{p_i}^\times$.
Every prime at most $K$ divides $n$, so the small-unit condition of
Lemma \ref{lem:unit-gap} holds. Its minimum complementary height is
\[
 \delta=A\min_i(d_ie_i^+)-1
 =n\min_i(e_i^+/p_i)-1.
\]
Use that lemma with $c_*=\delta-1$. This is a positive integer, and
$e_i^+\ge1$ gives the claimed length estimate.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:unit-lower}]
For sufficiently large $s$, choose $s$ distinct primes in
$[256s^2,512s^2]$ satisfying $p_i\equiv1\pmod3$. The prime number theorem
in this fixed progression supplies more than $s$ choices; for explicit
estimates sufficient here see \cite{BMOR}. Lemma \ref{lem:six-roots}
gives
\[
 \sum_i\frac{\max E_i}{p_i}\le
 \sum_i\frac4{\sqrt{p_i}}\le\frac14,
\]
so Proposition \ref{prop:six-construction} applies. Take the least
positive CRT solution $1\le c<K$. For large $s$, $A_0c\ge2K$
automatically, by the prime number theorem.

Let $n_s=\mathcal P(K)c$. Then
\[
 \log K=(2+o(1))s\log s,\qquad
 \log n_s=\vartheta(K)+\log c=(1+o(1))K.
\]
The exact capacity \eqref{eq:six-capacity} is at least
$n_s/(512s^2)-2$ and less than $n_s$, so
$\log(|S_s|/n_s)=s\log6+O(\log s)$. Also
\[
 \frac{\log\log n_s}{\log\log\log n_s}=(2+o(1))s.
\]
This proves \eqref{eq:unit-main}.
\end{proof}


\section{Support and multiplicity bounds}\label{sec:dense}

We first state precisely the external integer subset-sum input. For a
nonempty positive-integer multiset $X$, write $m=|X|$, $M=\max X$,
$\mu=\max_t v_t(X)$ and $W=\sum X$. It is $\delta$-dense if
$m^2\ge\delta\mu M$. An integer $d>1$ is an
$\alpha$-almost divisor if at most $\alpha\mu W/m^2$ actual entries
are not divisible by $d$.

\begin{lemma}[Bringmann--Wellnitz \cite{BW}, Theorems 4.1--4.2]
\label{lem:BW}
If $\delta\ge1$, $\delta\ge16\alpha$ and $X$ is
$\delta$-dense, some $d\ge1$ makes
$Y=\{x/d:x\in X,\ d\mid x\}$ $\delta$-dense, without an
$\alpha$-almost divisor, and with $|Y|\ge3m/4$.
Furthermore, put
\[
 D(m,\mu)=1699200\log_2(2m)\log_2^2(2\mu),\quad
 A(\mu)=42480\log_2(2\mu),\quad B(\mu)=169920\log_2(2\mu).
\]
If $X$ is $D(m,\mu)$-dense and has no $A(\mu)$-almost divisor,
then its integer subsequence sums contain every integer in
\[
 [\lambda,W-\lambda],\qquad
 \lambda=B(\mu)\mu MW/m^2.
\]
\end{lemma}

Using base-two logarithms only enlarges the numerical sufficient
thresholds from the source. We use no algorithmic running-time assertion.

\begin{lemma}\label{lem:target-window}
Let a generator sequence over $C_q$ have maximum multiplicity at most
$h$. Suppose one unit direction has $m$ actual copies with heights at
most $M<q$. Set $L=\log_2(2h)$, $D=1699200\log_2(2m)L^2$ and
$B=169920L$. An index-one subsequence exists if
\begin{equation}\label{eq:target-window}
 m\ge4\sqrt{hq},\qquad m^2\ge DhM,\qquad
 \frac43 BhM^2/m\le q.
\end{equation}
\end{lemma}
\begin{proof}
Select exactly $m$ such copies and apply the first part of Lemma
\ref{lem:BW} to their heights $X$, with $\delta=D$ and
$\alpha=42480L$. The numerical condition $D\ge16\alpha$ holds.
For the resulting divided multiset $Y$, put $t=|Y|\ge3m/4$ and
$V=\sum Y$. Sorting positive integers of multiplicity at most $h$
gives
\begin{equation}\label{eq:weight-retention}
 V\ge t^2/(2h)\ge9m^2/(32h)\ge(9/2)q.
\end{equation}
In particular the division has not reduced the total below the target.
The second part of Lemma \ref{lem:BW} applies to $Y$, since its
parameters and logarithmic thresholds are no larger. Also
\[
 \lambda_Y\le BhM^2/(d^2t)\le\frac43 BhM^2/m\le q.
\]
Thus $q\in[\lambda_Y,V-\lambda_Y]$ is an actual integer sum in $Y$.
Every retained height was a unit modulo $q$, so $(d,q)=1$.
If $u$ was the original direction, $ud^{-1}$ is a legal unit and
has actual representatives $x/d$ for precisely those retained copies.
Their selected integer sum $q$ is the required certificate.
\end{proof}

\begin{proposition}\label{prop:dense-unit}
Every index-one-free set of generators of $C_q$ has cardinality
$O(q^{3/4})$, with an absolute implied constant.
\end{proposition}
\begin{proof}
Let $K$ be its cardinality. For all sufficiently large $q$ we have
$4^{\omega(q)}\le q$. Put $m=\lceil4\sqrt q\rceil\le5\sqrt q$ and
suppose $K>Cq^{3/4}$, with $C$ a sufficiently large absolute constant.
Then $K>4m$. Set $M=\lceil2qm/K\rceil$, so
$m\le M<q$ and $M\le4qm/K$. Inclusion--exclusion and $K\le\varphi(q)$
give
\[
 R_q(M):=|\U(q)\cap[1,M]|\ge M\varphi(q)/q-2^{\omega(q)},
 \qquad \frac{KR_q(M)}{\varphi(q)}\ge2m-2^{\omega(q)}\ge m.
\]
Thus some unit direction has at least $m$ distinct heights at most $M$.
With $B_0=169920$, we have
\[
 \frac{m^2}{M}\ge\frac K{\sqrt q},\qquad
 \frac43 B_0M^2/m\le\frac{320}{3}B_0q^{5/2}/K^2.
\]
For sufficiently large $C$ and then sufficiently large $q$, these
inequalities imply all the conditions of Lemma \ref{lem:target-window}
with $h=1$. This contradicts freedom. The finitely many remaining
orders are absorbed into the constant using $K\le q$.
\end{proof}

We also use a set version of the integer subset-sum theorem.

\begin{lemma}[Conlon--Fox--Pham {\cite[Theorem 6.1]{CFP}}]
\label{lem:cfp-interval}
There are absolute constants $C_0,C_1,C_2>0$ such that the following
holds. If $X\subseteq[1,M]\cap\Z$ has $m\ge C_0\sqrt M$ elements,
some integer $g\ge1$ makes
\[
 Y=\{x/g:x\in X,\ g\mid x\},\qquad
 |Y|\ge m-C_1\bigl(\log^3(2M)+M/m\bigr).
\]
The integer subset sums of $Y$ contain at least $M$ consecutive
integers, with first point at most $C_2M^2/(gm)$.
\end{lemma}

The cited theorem represents every point of the interval using at most
$C_2M/m$ distinct elements of $Y$. Since each is at most $M/g$, this
also gives the asserted bound on its first point. Reserving one copy
of every element of $Y$ therefore suffices for all these representations.

\begin{proposition}\label{prop:capacity-layer}
Let $S=\prod_a a^{c_a}$ be an index-one-free generator sequence over
$C_q$, and let $K_H=|\{a:c_a\ge H\}|$ for each integer $H\ge1$.
Then
\begin{equation}\label{eq:capacity-layer}
 K_H\ll q^{3/4}H^{-1/4}.
\end{equation}
Consequently, if $h$ is its maximum multiplicity, then
\begin{equation}\label{eq:generator-inverse}
 |S|\ll(qh)^{3/4}.
\end{equation}
Both implied constants are absolute.
\end{proposition}
\begin{proof}
There is nothing to prove for $H\ge q$, since $q$ copies of a
generator have index one. The case $H=1$ follows from Proposition
\ref{prop:dense-unit}, applied to one copy of each type. That same
proposition, together with $2^{\omega(q)}=q^{o(1)}$ and
$q/\varphi(q)=q^{o(1)}$, gives, uniformly for every $H$,
\begin{equation}\label{eq:capacity-window-error}
 \frac{K_H2^{\omega(q)}}{\varphi(q)}
 \ll q^{-1/4+o(1)}\le1
\end{equation}
for all sufficiently large $q$. This uses only the already proved
set bound.

Fix $A=16$ and a sufficiently large absolute $B$, depending on the
constants of Lemma \ref{lem:cfp-interval}. Suppose $2\le H<q$ and
\[
 K:=K_H>BK_0,\qquad K_0=q^{3/4}H^{-1/4}.
\]
Set $t=\sqrt{q/H}$, $m=\lceil At\rceil$ and $M=\lceil2qm/K\rceil$.
Since $t\ge1$ and $K_0/t=(qH)^{1/4}$, sufficiently large $q$ gives
$At\le m\le2At$ and $K>4m$. Hence $m\le M<q$ and $M\le4qm/K$.
Averaging the $K$ distinct types over unit directions and using
\eqref{eq:capacity-window-error} gives
\[
 \frac{KR_q(M)}{\varphi(q)}\ge KM/q-1\ge2m-1\ge m.
\]
Choose exactly $m$ types in such a direction $u$, with heights
$X\subseteq[1,M]$, and retain $H$ actual copies of each. The parameters
satisfy
\begin{equation}\label{eq:capacity-scales}
 \frac{M^2}{m}\le\frac{32Aq}{B^2},\qquad
 \frac{m^2}{M}\ge\frac{mK}{4q}.
\end{equation}
Whenever we retain heights divisible by an integer $g$, that integer
is coprime to $q$. In direction $ug^{-1}$, the retained heights are
exactly $x/g$, still positive and less than $q$.

\emph{Low multiplicity: $2\le H\le q^{1/3}$.}
Here $m\ge Aq^{1/3}$ and
\[
 \frac{m^2}{M}\ge\frac{AB}{4}q^{1/4}H^{-3/4}\ge\frac{AB}{4}.
\]
Choose $B$ so that $m\ge C_0\sqrt M$ and $C_1M/m\le m/4$.
For sufficiently large $q$, also $C_1\log^3(2M)\le m/4$.
Lemma \ref{lem:cfp-interval} therefore retains $s\ge m/2$ divided
types $Y$, each with $H$ actual copies. Reserve one copy of every
retained type. Their subset sums contain at least $M$ consecutive
integers with first point at most
\[
 C_2M^2/(gm)\le32AC_2q/B^2\le q.
\]
Each remaining copy has value at most $M/g\le M$, so adjoining
these copies extends the interval without an integer gap. Their
total weight is at least
\[
 (H-1)\frac{s(s+1)}2\ge\frac{(H-1)m^2}{8}
 \ge\frac{A^2q}{16}\ge q.
\]
Thus the extended interval contains $q$, giving an index-one
subsequence in the single direction $ug^{-1}$.

\emph{High multiplicity: $q^{1/3}\le H<q$.}
Put $g=\gcd(X)$, $Y=X/g$ and $v=\max Y\le M$. Apply Lemma
\ref{lem:integer-interval} to these $m$ distinct denominations.
Since $HK>Bq$, its reserved multiplicity satisfies
\[
 \rho\le4v/m\le16q/K\le16H/B\le H/2.
\]
Its interval starts at
\[
 L\le2v^2/m+v\le3M^2/m\le96Aq/B^2\le q.
\]
The unreserved copies have total weight at least
\[
 (H-\rho)\sum_{y\in Y}y\ge\frac H2\frac{m(m+1)}2
 \ge\frac{A^2q}{4}\ge q.
\]
The resulting integer interval again contains $q$ in direction
$ug^{-1}$. Both cases contradict freedom.

All choices of constants precede the lower bound on $q$. The finitely
many omitted orders are absorbed uniformly in $H$, since $H<q$ and
$K_H\le q$. This proves \eqref{eq:capacity-layer}. Finally,
\[
 |S|=\sum_{H=1}^h K_H
 \ll q^{3/4}\sum_{H=1}^h H^{-1/4}\ll(qh)^{3/4},
\]
proving \eqref{eq:generator-inverse}.
\end{proof}

\begin{proposition}\label{prop:multiplicity-bound}
Let $S$ be a nonempty index-one-free sequence over $C_n$. For each
$d\mid n$, $d>1$, let $h_d$ be the largest multiplicity of an element
of exact order $d$, taking $h_d=0$ if that class is empty. Then
\begin{equation}\label{eq:inverse-general}
 |S|\ll\sum_{\substack{d\mid n\\d>1}}(dh_d)^{3/4}.
\end{equation}
In particular, if $h$ is the maximum multiplicity, then
\begin{equation}\label{eq:inverse-pp}
 |S|\ll(nh)^{3/4}\qquad\text{when $n$ is a prime power}.
\end{equation}
All implied constants are absolute. For each fixed $\varepsilon>0$,
all copies of types whose original multiplicity is at most
$n^{1/3-\varepsilon}$ contribute only $o(n)$ terms.
\end{proposition}
\begin{proof}
By Lemma \ref{lem:orders}, the exact-order-$d$ class is an
index-one-free generator sequence over $C_d$. Applying
\eqref{eq:generator-inverse} to each nonempty class gives
\eqref{eq:inverse-general}. Since $h_d\le h$,
\[
 |S|\ll h^{3/4}\sum_{d\mid n}d^{3/4}
 =(nh)^{3/4}\sigma_{-3/4}(n).
\]
If $\omega(n)\le k$, then
\[
 \sigma_{-3/4}(n)\le\prod_{p\mid n}(1-p^{-3/4})^{-1}
 \le(1-2^{-3/4})^{-k}.
\]
This proves \eqref{eq:inverse-pp} and the length and inverse-multiplicity
assertions of Theorem \ref{thm:support}. Retaining one copy of every
type and using $\sigma_{-3/4}(n)\le\tau(n)=n^{o(1)}$ proves its
support assertions, where $\tau(n)$ counts the positive divisors of $n$.
Finally, apply the same estimate to the actual subsequence consisting
of all copies of types with original multiplicity at most
$n^{1/3-\varepsilon}$. Its length is at most
$n^{1-3\varepsilon/4+o(1)}=o(n)$.
\end{proof}

\appendix
\section{An integer descent and the general bound}
\label{sec:appendix}

We prove the uniform index-$r$ bound \eqref{eq:general-r-main}.
The descent retains one denomination of maximum multiplicity; this
ensures that the final divisor is a unit in the group application.

\begin{lemma}\label{lem:descent}
Let $X=\prod_{a\in B}a^{c_a}$ be an integer multiset with distinct
denominations $B\subset(H/2,H]\cap\Z$, length $N$, and maximum
multiplicity $M$. If $N\ge2048H$, there are a submultiset $X'$ and an
integer $D$ dividing all its entries and one original denomination
of multiplicity $M$, such that $|X'|\ge15N/16$ and
$\Sigma_{0,\Z}(X'/D)$ contains an integer interval $[L,V]$ satisfying
\begin{equation}\label{eq:descent-endpoints}
 L\le65MH^2/N+1,\qquad V-L>N^2/(32M).
\end{equation}
\end{lemma}
\begin{proof}
Fix $q_0$ of multiplicity $M$ and
$\ell=\lceil64MH/N\rceil$. Since $N\le MH$ and $N\ge2048H$,
\[
 M\ge2048,\quad64MH/N\le\ell\le65MH/N,\quad
 \ell\le M/16,\quad2\ell+3\le M/4.
\]
At a descent step, let the already divided common factor be $D_i$,
$H_i=H/D_i$, and $q_i=q_0/D_i$. All current denominations lie in
$(H_i/2,H_i]$, and all $M$ anchor copies are retained. For $a\ne q_i$
put $d_a=\min(\lfloor c_a/4\rfloor,\ell)$ and
\[
 A_j=\{0\}\cup\{a\bmod q_i:d_a\ge j\},\qquad1\le j\le\ell.
\]
The nonzero residues within each layer are distinct. If their sumset
is proper in $C_{q_i}$, let its stabilizer be $J=d_i\Z/q_i\Z$ with
$d_i\mid q_i$ and $d_i\ge2$. Iterated Kneser gives
\[
 \sum_{a:d_i\nmid a}d_a
 =\sum_j|A_j\setminus J|
 \le\sum_j(|A_j+J|-|J|)<q_i.
\]
For each type $d_a\ge(\ell/M)c_a-1$. There are at most $H_i$
denominations, so fewer than
$(M/\ell)(q_i+H_i)\le2MH_i/\ell$ copies have denomination not
divisible by $d_i$. Delete these and divide all retained entries by
$d_i$. The anchor is retained. Each $H_i$ decreases at least by half,
so the total loss over the whole descent is at most
\[
 \sum_i2MH_i/\ell\le4MH/\ell\le N/16.
\]
The descent terminates, since $q_i$ is a decreasing positive integer
and the residue cover holds when $q_i=1$.

At termination reserve the $d_a$ copies of each nonanchor type and
$2\ell+3$ anchor copies. This reserves at most
$|X'|/4+M/4\le|X'|/2$ copies. The layer cover gives each residue an
integer representation of weight at most $\ell H_i$. As in the prime
window argument, the anchor extends these to
$[\lceil\ell H_i\rceil,\lceil\ell H_i\rceil+2q_i]\cap\Z$.
The width exceeds $H_i$, so every remaining copy extends this interval.
At least $R\ge|X'|/2\ge15N/32>N/4$ copies remain. Distinct positive
integer denominations with at most $M$ copies each have total weight
at least $R^2/(2M)$. Adding these copies increases the interval
width by their full integer weight. Thus $V-L>N^2/(32M)$,
while the left is at most $\ell H_i+1\le65MH^2/N+1$.
All divided factors divide the original anchor, as required.
\end{proof}

\begin{proof}[Proof of \eqref{eq:general-r-main}]
First let $S$ be an index-$r$-free unit sequence over $C_d$, where
$1\le r\le d/2$, and write $W=|S|$. Set
\[
 H=\lfloor\sqrt{rd}\rfloor,\qquad
 E(d)=2^{\omega(d)}d/\varphi(d).
\]
Inclusion--exclusion gives
\[
 |\U(d)\cap(H/2,H]|\ge H\varphi(d)/(2d)-2^{\omega(d)+1}.
\]
There is an absolute bound $E(d)\ll d^{1/4}$: for primes
$p\ge23$, $2p/(p-1)\le p^{1/4}$, and the finitely many smaller
primes contribute an absolute constant. Thus for all sufficiently
large $d$, uniformly in $r$, one has $H\ge8E(d)$.
The window contains at least $H\varphi(d)/(4d)$ units.
If $N_u$ actual copies lie in the window in direction $u$, averaging
gives
\[
 \frac1{\varphi(d)}\sum_{u\in\U(d)}N_u\ge WH/(4d).
\]

Suppose some $N_u\ge2048d$. Since $H<d$, Lemma
\ref{lem:descent} applies, with maximum multiplicity $M<d$.
Its divisor $D$ divides an actual unit denomination, so $(D,d)=1$.
In the single legal direction $uD^{-1}$ the divided integers are
exactly the positive representatives of the retained original copies.
The covered integer interval satisfies
\[
 L\le65rd/2048+1<rd,\qquad
 V-L>\frac{(2048d)^2}{32d}>d.
\]
Consequently it contains a positive multiple $jd$ with $1\le j\le r$.
The corresponding actual submultiset is a zero sum of index at most
$j$, contrary to freedom. Therefore every $N_u<2048d$, and
\begin{equation}\label{eq:general-r-unit}
 L_{U,r}(d)\ll d^2/H\ll d^{3/2}/\sqrt r
 \qquad(1\le r\le d/2).
\end{equation}
The finitely many smaller $d$ are absorbed into the absolute constant:
their unit sequences have length at most $(d-1)\varphi(d)$, and
there are only finitely many admissible pairs $(d,r)$.

Now take an arbitrary index-$r$-free sequence over $C_n$.
By Lemma \ref{lem:index-cap}, all copies of types with
$c_a\le2r+1$ have total length at most $n-1$.
Every remaining type has additive order $d>2r+1$, since a type
of order $d$ has multiplicity less than $d$.
Partition these remaining copies by exact order. Lemma
\ref{lem:orders} identifies each class with an index-$r$-free unit
sequence over $C_d$, and \eqref{eq:general-r-unit} applies because
$r<d/2$. Hence
\[
 |S|\le n-1+\frac C{\sqrt r}
 \sum_{\substack{d\mid n\\d>2r+1}}d^{3/2}
 \le n-1+\frac{3C n^{3/2}}{\sqrt r}.
\]
Here $\sum_{d\mid n}(d/n)^{3/2}\le
\sum_{k\ge1}k^{-3/2}\le3$.
This proves the upper bound in \eqref{eq:general-r-main}; the
lower bound is $g^{n-1}$. At $r=1$ it supplies the $O(n^{3/2})$
term of \eqref{eq:general-main}, completing Theorem \ref{thm:upper}.
\end{proof}

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