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\documentclass[a4paper,11pt]{amsart}
\usepackage[margin=1in]{geometry}
\usepackage[T1]{fontenc}
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\usepackage{amsmath,amssymb,amsthm,mathtools}
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\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
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\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newcommand{\N}{\mathbb{N}}
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\newcommand{\HH}{\mathcal{H}}
\newcommand{\EE}{\mathcal{E}}
\newcommand{\VV}{\mathcal{V}}
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\title[Lower bounds for $H(n)$]{A constant-factor lower bound for $H(n)$}
\subjclass[2020]{05C65, 05D10}
\keywords{hypergraph, exact cover, extremal construction, computer-assisted proof}
\author{GPT-5.4 Pro}
\date{March 10, 2026}
\begin{document}
\begin{abstract}
We construct explicit hypergraphs showing that
\[
H(n)\ge \frac{26}{25}\,k_n
\qquad\text{for every }n\ge 15,
\]
where \(k_1=1\) and \(k_n=\floor{n/2}+k_{\floor{n/2}}+k_{\floor{(n+1)/2}}\).
The witnesses have exactly \(n\) distinct edges, no isolated vertices, and are generated by a pure-Python recursive constructor with no nonstandard dependencies.
The proof has two ingredients: a substitution theorem for support gadgets, and a finite bootstrap built from exact small frames, seven explicit higher-arity boosters, and a balanced four-way family.
All finite checks are reduced to exhaustive verification of explicit inequalities by an archived standalone verifier, and every gadget used by the constructor is written out in the paper.
We also record a separate Lubell-family argument giving the asymptotic lower bound \(\liminf_{n\to\infty} H(n)/k_n \ge 2\ln 2\), together with an explicit table of the harmonic multiplier \(\gamma_t\) for small \(t\) and a self-contained proof that
\[
k_n=\frac12 n\log_2 n+O(n).
\]
\end{abstract}
\maketitle
\section{Introduction}
Let \(G=(V,\HH)\) be a finite hypergraph.
We say that \(G\) \emph{contains a partition of size \(m\)} if there are \(D\subseteq V\) and \(\cP\subseteq \HH\) with \(\abs{D}=m\) such that every vertex of \(D\) belongs to exactly one edge of \(\cP\).
Write \(H(n)\) for the largest \(k\in\N\) such that there exists a hypergraph \((V,\HH)\) with \(\abs{V}=k\), no isolated vertices, and no partition of size greater than \(n\).
This extremal function arises in Brian and Larson's finitary reformulation of a problem about incompatible ideals \cite{BrianLarson}, and it appears verbatim as an open FrontierMath challenge \cite{Epoch}.
Brian and Larson proved the recursive lower bound
\[
H(n)\ge k_n,
\qquad
k_1=1,
\qquad
k_n=\floor{n/2}+k_{\floor{n/2}}+k_{\floor{(n+1)/2}}
\quad(n\ge 2).
\]
The problem posed on the FrontierMath page asks for a constant-factor improvement on this benchmark, already visible at \(n=15\), and asks for an explicit algorithm that outputs a witness hypergraph for each \(n\) \cite{Epoch}.
The purpose of this paper is to provide such a construction.
The paper combines the best features of two complementary approaches.
The first is a clean \emph{support-gadget substitution theorem}, which turns admissible set systems on a small meta-ground-set into recursive hypergraph constructions.
The second is a \emph{finite bootstrap}: below \(60\) edges we use exact small frames on at most five blocks, together with seven explicit higher-arity boosters; from \(60\) onward a balanced four-way family propagates a uniform multiplicative gain forever.
The official verifier-facing constructor is standard-library Python only: the finitely many exact small frames needed below \(60\) are precomputed once and for all, and then hard-coded into the final script.
All finite computational claims used in Section~\ref{sec:uniform} are verified by the ancillary script \texttt{verify\_hn\_26\_25.py}, included with the source and reproduced in Appendix~\ref{app:verify}.
Our principal theorem is the following.
\begin{theorem}\label{thm:main}
There exists an explicit sequence of hypergraphs \((G_n)_{n\ge 1}\) with the following properties.
For every \(n\ge 1\), the hypergraph \(G_n\) has exactly \(n\) distinct edges, has no isolated vertices, and contains no partition of size greater than \(n\).
Moreover, for every \(n\ge 15\),
\[
\abs{V(G_n)}\ge \frac{26}{25}\,k_n.
\]
Consequently,
\[
H(n)\ge \frac{26}{25}\,k_n
\qquad(n\ge 15).
\]
The uniform factor \(26/25\) is already in effect at \(n=15\), and the minimum ratio in the finite bootstrap window \(15\le n<60\) is attained at \(n=17\):
\[
\abs{V(G_{15})}=45>43=k_{15},
\qquad
\abs{V(G_{17})}=52=\frac{26}{25}k_{17}.
\]
\end{theorem}
The construction proving Theorem~\ref{thm:main} is the one implemented by the function \texttt{solution(n)} in Appendix~\ref{app:program}.
It uses only the explicit tables in Appendices~\ref{app:exactframes}, \ref{app:boosters}, and \ref{app:bootstrap}.
No optimization routine is invoked by the submitted program.
The support-gadget calculus also yields a separate asymptotic statement.
This second family is not the official verifier-facing one, but it gives a clean conceptual explanation for the asymptotic constant \(2\ln 2\).
\begin{theorem}\label{thm:asymptotic}
For every fixed \(t\ge 2\) there is an explicit recursive family of hypergraphs showing that
\[
H(n)\ge \frac{h_t-1}{\log_2 t}\,n\log_2 n - O_t(n),
\qquad
h_t\coloneq 1+\frac12+\cdots+\frac1t.
\]
Consequently,
\[
\liminf_{n\to\infty}\frac{H(n)}{k_n}\ge 2\ln 2.
\]
\end{theorem}
The remainder of the paper is organized as follows.
Section~\ref{sec:substitution} develops the general support-gadget substitution theorem.
Section~\ref{sec:uniform} builds the explicit \(26/25\) family and proves Theorem~\ref{thm:main}.
Section~\ref{sec:lubell} records the Lubell-family asymptotics and proves Theorem~\ref{thm:asymptotic}, including an explicit table of the harmonic multipliers and a self-contained proof that \(k_n=\frac12 n\log_2 n+O(n)\).
The appendices list the exact small frames, the boosters, the four-way residue gadgets, the finite bootstrap table, the official Python constructor, and the ancillary verifier.
\section{Support gadgets and substitution}\label{sec:substitution}
For a hypergraph \(G=(V,\HH)\) and a chosen edge family \(\cP\subseteq \HH\), define
\[
\uu_G(\cP)\coloneq \abs{\set{v\in V:\text{$v$ belongs to exactly one edge of }\cP}}.
\]
The partition problem is equivalent to a uniform bound on \(\uu_G\).
\begin{lemma}\label{lem:partition-u}
A hypergraph \(G\) contains no partition of size greater than \(n\) if and only if
\[
\uu_G(\cP)\le n
\qquad\text{for every }\cP\subseteq \HH.
\]
\end{lemma}
\begin{proof}
If \((D,\cP)\) is a partition of size \(m\), then every vertex of \(D\) is counted by \(\uu_G(\cP)\), so \(\uu_G(\cP)\ge m\).
Conversely, if \(\uu_G(\cP)>n\), let \(D\) be the set of vertices counted by \(\uu_G(\cP)\); then \((D,\cP)\) is a partition of size \(\uu_G(\cP)\).
\end{proof}
Fix \(t\ge 2\).
A \emph{support pattern} on \([t]\coloneq \set{1,\dots,t}\) is a subset of \([t]\) of size at least \(2\).
A finite multiset of support patterns will be denoted by \(\cF\).
Its members will encode new vertices that meet entire blocks of edges simultaneously.
\begin{definition}
Let \(\mathbf n=(n_1,\dots,n_t)\in \mathbb{Z}_{\ge 0}^t\).
For \(I\subseteq T\subseteq [t]\), define
\[
\omega_{\cF}(T,I)\coloneq
\abs{\set{S\in \cF:\ S\subseteq T\text{ and }\abs{S\cap I}=1}},
\]
counting multiplicity.
We say that \(\cF\) is an \emph{\(\mathbf n\)-frame} if
\[
\omega_{\cF}(T,I)\le \sum_{j\in T\setminus I} n_j
\qquad\text{for every }I\subseteq T\subseteq [t].
\]
\end{definition}
\begin{remark}
Capacity vectors are allowed to have zero coordinates.
This convention is needed later for the residue gadgets \(R_0\) and \(R_1\), whose stated capacity vectors contain zeros.
\end{remark}
\begin{definition}
Let \(\cF\) be a support multiset on \([t]\).
Let \(G_i=(V_i,\HH_i)\), \(1\le i\le t\), be hypergraphs with pairwise disjoint vertex sets and pairwise disjoint edge sets.
The \emph{substitution hypergraph}
\[
\cF[G_1,\dots,G_t]
\]
has edge set \(\HH_1\sqcup\cdots\sqcup\HH_t\), and its vertex set consists of \(V_1\sqcup\cdots\sqcup V_t\) together with one new vertex \(x_S\) for each occurrence of a support pattern \(S\in \cF\).
The new vertex \(x_S\) is incident to every edge of \(\HH_i\) for every \(i\in S\).
\end{definition}
\begin{theorem}[substitution theorem]\label{thm:substitution}
Let \(\mathbf n=(n_1,\dots,n_t)\in \mathbb{Z}_{\ge 0}^t\), and let \(\cF\) be an \(\mathbf n\)-frame on \([t]\).
Assume that each \(G_i\) contains no partition of size greater than \(n_i\).
Then the substituted hypergraph
\[
G\coloneq \cF[G_1,\dots,G_t]
\]
contains no partition of size greater than \(n_1+\cdots+n_t\).
Equivalently,
\[
\uu_G(\cP)\le n_1+\cdots+n_t
\qquad\text{for every }\cP\subseteq \EE(G).
\]
\end{theorem}
\begin{proof}
Fix \(\cP\subseteq \EE(G)\), and write \(\cP_i\coloneq \cP\cap \HH_i\).
Define
\[
T\coloneq \set{i:\abs{\cP_i}\le 1},
\qquad
I\coloneq \set{i:\abs{\cP_i}=1}.
\]
If \(i\notin T\), then \(\abs{\cP_i}\ge 2\), and the vertices of \(V_i\) counted by \(\uu_G(\cP)\) are counted by \(\uu_{G_i}(\cP_i)\), hence contribute at most \(n_i\).
If \(i\in T\setminus I\), then \(\abs{\cP_i}=0\), so the block \(V_i\) contributes nothing.
If \(i\in I\), then again block \(V_i\) contributes at most \(n_i\).
So the total contribution from the old block vertices is at most
\[
\sum_{i\notin T} n_i + \sum_{i\in I} n_i.
\]
Now consider a new support vertex \(x_S\).
It lies in exactly one selected edge if and only if \(S\subseteq T\) and \(\abs{S\cap I}=1\).
Hence the total contribution from support vertices is exactly
\[
\omega_{\cF}(T,I)\le \sum_{j\in T\setminus I} n_j,
\]
because \(\cF\) is an \(\mathbf n\)-frame.
Adding the two estimates gives
\[
\uu_G(\cP)\le
\sum_{i\notin T} n_i + \sum_{i\in I} n_i + \sum_{j\in T\setminus I} n_j
=
\sum_{i=1}^t n_i.
\]
Lemma~\ref{lem:partition-u} finishes the proof.
\end{proof}
\begin{corollary}\label{cor:frame-recurrence}
Suppose that for \(1\le i\le t\) there is a hypergraph \(G_i\) with exactly \(n_i\) edges, no isolated vertices, no partition of size greater than \(n_i\), and \(a_i\) vertices.
If \(\cF\) is an \((n_1,\dots,n_t)\)-frame, then \(\cF[G_1,\dots,G_t]\) has exactly \(n_1+\cdots+n_t\) edges, no isolated vertices, no partition of size greater than \(n_1+\cdots+n_t\), and
\[
a_1+\cdots+a_t+\abs{\cF}
\]
vertices.
\end{corollary}
\begin{proof}
The edge set is the disjoint union of the edge sets of the blocks, so the edge count is additive.
Every old vertex remains non-isolated, and every new support vertex meets every edge in at least one block from its support.
The vertex count is immediate, and the partition bound is Theorem~\ref{thm:substitution}.
\end{proof}
\begin{lemma}\label{lem:frame-additivity}
If \(\cF\) is an \(\mathbf n\)-frame and \(\mathcal{G}\) is an \(\mathbf m\)-frame on the same ground set \([t]\), then \(\cF\uplus \mathcal{G}\) is an \((\mathbf n+\mathbf m)\)-frame.
\end{lemma}
\begin{proof}
For every \(I\subseteq T\subseteq [t]\),
\[
\omega_{\cF\uplus\mathcal{G}}(T,I)
=
\omega_{\cF}(T,I)+\omega_{\mathcal{G}}(T,I)
\le
\sum_{j\in T\setminus I}(n_j+m_j).
\]
\end{proof}
\begin{remark}
The entire problem is thus reduced to constructing useful support frames.
The strong \(26/25\) theorem below uses a finite bank of exact frames and boosters.
The asymptotic theorem later uses the Lubell family.
\end{remark}
\section{The uniform factor \texorpdfstring{$26/25$}{26/25}}\label{sec:uniform}
We now describe the explicit family proving Theorem~\ref{thm:main}.
There are three ingredients.
First, there is a finite bank of exact small frames on at most five blocks.
These were obtained by solving the corresponding integer programs once and for all, and then frozen as explicit support multisets.
They are listed in Appendix~\ref{app:exactframes}.
Second, there are seven higher-arity boosters:
\[
B_{15},\ B_{16},\ B_{17},\ B_{19},\ B_{31},\ B_{32},\ B_{33}.
\]
These are listed explicitly in Appendix~\ref{app:boosters}.
Third, there is the balanced four-way family, consisting of the classical \(13\)-vertex core
\[
Q_0\coloneq 12+13+14+23+24+34+123+124+134+234+3\cdot 1234
\]
on capacity vector \((3,3,3,3)\), together with the twelve residue gadgets \(R_r\), \(0\le r\le 11\), from Appendix~\ref{app:boosters}.
The core \(Q_0\) is exactly the Lubell frame \(\cL_4\), since \(M_4=3\).
Hence Proposition~\ref{prop:lubell} shows that \(Q_0\) is a \((3,3,3,3)\)-frame.
As in the appendices, we abbreviate a support pattern \(\{i_1,\dots,i_s\}\) by the concatenated string \(i_1\cdots i_s\), and coefficients denote multiplicity.
\begin{proposition}\label{prop:finite-bank}
Every exact small frame listed in Appendix~\ref{app:exactframes}, every booster listed in Appendix~\ref{app:boosters}, and every residue gadget \(R_r\) listed in Appendix~\ref{app:boosters} is a valid frame for its stated capacity vector.
\end{proposition}
\begin{proof}
This is a finite exhaustive check.
For a support multiset \(\cF\) on \(t\) blocks with capacity vector
\(\mathbf n=(n_1,\dots,n_t)\), frame-validity means checking
\[
\omega_{\cF}(T,I)\le \sum_{j\in T\setminus I} n_j
\qquad\text{for every }I\subseteq T\subseteq [t].
\]
There are exactly
\[
\sum_{T\subseteq [t]}2^{\abs{T}} = 3^t
\]
such pairs \((T,I)\).
The standalone script \texttt{verify\_hn\_26\_25.py} in Appendix~\ref{app:verify}
exhaustively evaluates these inequalities for every exact frame in
Appendix~\ref{app:exactframes}, every booster in Appendix~\ref{app:boosters},
and every residue gadget \(R_r\) in Appendix~\ref{app:boosters}.
The largest gadget used in the uniform construction is \(B_{33}\), which lives on \(9\) blocks, so the largest exhaustive check has \(3^9=19683\) cases.
Running the verifier confirms that every listed gadget is a valid frame for its stated capacity vector.
\end{proof}
\begin{definition}
Define integers \(A_n\) recursively as follows.
Set \(A_1\coloneq 1\).
For \(2\le n<60\), let \(A_n\) be the value listed in Appendix~\ref{app:bootstrap}.
For \(n\ge 60\), write
\[
n=4m+r,
\qquad
0\le r\le 3,
\]
and let \((a,b,c,d)\) be the nondecreasing balanced partition of \(n\) into four positive parts:
\[
(a,b,c,d)=
\begin{cases}
(m,m,m,m), & r=0,\\
(m,m,m,m+1), & r=1,\\
(m,m,m+1,m+1), & r=2,\\
(m,m+1,m+1,m+1), & r=3.
\end{cases}
\]
Then define
\[
A_n\coloneq A_a+A_b+A_c+A_d+e_r(m),
\]
where
\begin{align}
e_0(m)&\coloneq \floor{\frac{13m}{3}},\label{eq:e0}\\
e_1(m)&\coloneq \floor{\frac{13m+1}{3}},\label{eq:e1}\\
e_2(m)&\coloneq \floor{\frac{13m+5}{3}},\label{eq:e2}\\
e_3(m)&\coloneq \floor{\frac{13m+6}{3}}.\label{eq:e3}
\end{align}
\end{definition}
The formulas \eqref{eq:e0}--\eqref{eq:e3} are exactly the bonuses produced by \(\floor{n/12}\) copies of the four-way core \(Q_0\) together with the appropriate residue gadget \(R_{n\bmod 12}\).
\begin{theorem}\label{thm:constructive-An}
For every \(n\ge 1\) there exists a hypergraph \(G_n\) with exactly \(n\) distinct edges, no isolated vertices, no partition of size greater than \(n\), and
\[
\abs{V(G_n)}=A_n.
\]
The construction is explicit and recursive.
\end{theorem}
\begin{proof}
We argue by induction on \(n\).
For \(n=1\), take a single edge containing a single vertex.
Assume the statement proved for all smaller positive integers.
If \(2\le n<60\), the corresponding row of Appendix~\ref{app:bootstrap} specifies one of four possibilities.
\smallskip
\noindent\emph{Binary case.}
If the row is \(\mathrm{Bin}(a,b)\), then \(a+b=n\), the support multiset is \(\min(a,b)\) copies of the support \(12\), and Corollary~\ref{cor:frame-recurrence} applied to \(G_a\) and \(G_b\) gives a witness with
\[
A_a+A_b+\min(a,b)=A_n
\]
vertices.
\smallskip
\noindent\emph{Exact-frame case.}
If the row is \(F_{(n_1,\dots,n_t)}\), then the corresponding support multiset is the one listed for that capacity vector in Appendix~\ref{app:exactframes}.
By Proposition~\ref{prop:finite-bank} it is a valid frame, so Corollary~\ref{cor:frame-recurrence} applied to the smaller witnesses \(G_{n_1},\dots,G_{n_t}\) yields a witness with
\[
A_{n_1}+\cdots+A_{n_t}+\abs{F_{(n_1,\dots,n_t)}}=A_n
\]
vertices.
\smallskip
\noindent\emph{Booster case.}
If the row is \(B_{15}\), \(B_{16}\), \(B_{17}\), \(B_{19}\), \(B_{31}\), \(B_{32}\), or \(B_{33}\), the same argument applies using the explicitly listed support multiset from Appendix~\ref{app:boosters}.
\smallskip
\noindent\emph{Balanced four-way case.}
If the row is \(Q(a,b,c,d)\), let \(q\coloneq \floor{n/12}\) and \(s\coloneq n\bmod 12\).
The gadget \(Q_0\) is a \((3,3,3,3)\)-frame, and \(R_s\) is a valid frame for its residual capacity vector by Proposition~\ref{prop:finite-bank}.
By Lemma~\ref{lem:frame-additivity}, \(q\) copies of \(Q_0\) together with \(R_s\) form a frame on capacities \((a,b,c,d)\).
Corollary~\ref{cor:frame-recurrence} then yields a witness with
\[
A_a+A_b+A_c+A_d+\abs{q\cdot Q_0\uplus R_s}=A_n
\]
vertices.
\smallskip
Thus the induction closes for all \(n<60\).
Now assume \(n\ge 60\).
Write \(n=4m+r\) and let \((a,b,c,d)\) be the balanced four-way split.
By the inductive hypothesis we have witnesses \(G_a,G_b,G_c,G_d\) with the stated properties.
Let \(q\coloneq \floor{n/12}\) and \(s\coloneq n\bmod 12\).
As above, \(q\) copies of \(Q_0\) together with \(R_s\) form a valid frame on capacities \((a,b,c,d)\).
Corollary~\ref{cor:frame-recurrence} therefore yields a witness with
\[
A_a+A_b+A_c+A_d+\abs{q\cdot Q_0\uplus R_s}
=
A_a+A_b+A_c+A_d+e_r(m)
=
A_n
\]
vertices.
This completes the induction.
\end{proof}
To compare \(A_n\) with the benchmark \(k_n\), we need the four-way identities satisfied by \(k_n\).
\begin{lemma}\label{lem:k-four-way}
For every \(m\ge 1\),
\begin{align*}
k_{4m} &=4k_m+4m,\\
k_{4m+1} &=3k_m+k_{m+1}+4m,\\
k_{4m+2} &=2k_m+2k_{m+1}+4m+1,\\
k_{4m+3} &=k_m+3k_{m+1}+4m+2.
\end{align*}
\end{lemma}
\begin{proof}
The defining recurrence for \(k_n\) immediately gives
\[
k_{2m}=m+2k_m,
\qquad
k_{2m+1}=m+k_m+k_{m+1}.
\]
Applying these identities once more gives the displayed four-way formulas.
\end{proof}
\begin{lemma}\label{lem:floor-26-25}
For every integer \(m\ge 15\),
\begin{align*}
e_0(m)&\ge \frac{26}{25}\cdot 4m,\\
e_1(m)&\ge \frac{26}{25}\cdot 4m,\\
e_2(m)&\ge \frac{26}{25}(4m+1),\\
e_3(m)&\ge \frac{26}{25}(4m+2).
\end{align*}
\end{lemma}
\begin{proof}
Using \(\floor{x}\ge x-1\), we obtain
\begin{align*}
e_0(m)-\frac{26}{25}\cdot 4m
&\ge \frac{13m}{3}-1-\frac{104m}{25}
= \frac{13m-75}{75},\\
e_1(m)-\frac{26}{25}\cdot 4m
&\ge \frac{13m+1}{3}-1-\frac{104m}{25}
= \frac{13m-50}{75},\\
e_2(m)-\frac{26}{25}(4m+1)
&\ge \frac{13m+5}{3}-1-\frac{104m+26}{25}
= \frac{13m-28}{75},\\
e_3(m)-\frac{26}{25}(4m+2)
&\ge \frac{13m+6}{3}-1-\frac{104m+52}{25}
= \frac{13m-81}{75}.
\end{align*}
All four right-hand sides are affine functions of \(m\) with positive slope \(13/75\), and at \(m=15\) they equal
\[
\frac{120}{75},\qquad \frac{145}{75},\qquad \frac{167}{75},\qquad \frac{114}{75},
\]
respectively; hence they are nonnegative for every \(m\ge 15\).
\end{proof}
\begin{proposition}\label{prop:bootstrap-26-25}
For every integer \(n\) with \(15\le n<60\),
\[
A_n\ge \frac{26}{25}\,k_n.
\]
More precisely, Appendix~\ref{app:bootstrap} lists the exact slack
\[
\Delta_n\coloneq 25A_n-26k_n,
\]
and \(\Delta_n\ge 0\) for every \(15\le n<60\), with equality only at \(n=17\).
\end{proposition}
\begin{proof}
For each \(15\le n<60\), the ancillary verifier in Appendix~\ref{app:verify}
checks that the construction listed in Appendix~\ref{app:bootstrap} yields the stated value \(A_n\), and also checks the inequality \(25A_n\ge 26k_n\).
Since \(\Delta_n\) is defined by
\[
\Delta_n\coloneq 25A_n-26k_n,
\]
the printed slack values follow by arithmetic.
\end{proof}
\begin{theorem}\label{thm:uniform-26-25}
For every integer \(n\ge 15\),
\[
A_n\ge \frac{26}{25}\,k_n.
\]
Hence Theorem~\ref{thm:main} holds.
\end{theorem}
\begin{proof}
The range \(15\le n<60\) is Proposition~\ref{prop:bootstrap-26-25}.
Now let \(n\ge 60\), write \(n=4m+r\) with \(0\le r\le 3\), and note that \(m\ge 15\).
If \(r=0\), then by the definition of \(A_n\), Lemma~\ref{lem:floor-26-25}, and Lemma~\ref{lem:k-four-way},
\[
A_{4m}=4A_m+e_0(m)\ge \frac{26}{25}(4k_m+4m)=\frac{26}{25}k_{4m}.
\]
If \(r=1\), then
\[
A_{4m+1}=3A_m+A_{m+1}+e_1(m)\ge \frac{26}{25}(3k_m+k_{m+1}+4m)=\frac{26}{25}k_{4m+1}.
\]
If \(r=2\), then
\[
A_{4m+2}=2A_m+2A_{m+1}+e_2(m)\ge \frac{26}{25}(2k_m+2k_{m+1}+4m+1)=\frac{26}{25}k_{4m+2}.
\]
If \(r=3\), then
\[
A_{4m+3}=A_m+3A_{m+1}+e_3(m)\ge \frac{26}{25}(k_m+3k_{m+1}+4m+2)=\frac{26}{25}k_{4m+3}.
\]
This completes the induction.
\end{proof}
\section{Lubell frames and asymptotic context}\label{sec:lubell}
We now record the general Lubell family.
This family is conceptually cleaner than the finite bootstrap and explains the asymptotic constant \(2\ln 2\), although it does not by itself produce the stronger uniform factor \(26/25\) already at \(n=15\).
For \(t\ge 2\), define
\[
M_t\coloneq \operatorname{lcm}\set{\binom{t-1}{r}:1\le r\le t-1}.
\]
Let \(\cL_t\) be the support multiset on \([t]\) that contains every \(j\)-subset of \([t]\) with multiplicity
\[
\frac{M_t}{\binom{t-1}{j-1}}
\qquad(2\le j\le t).
\]
Equivalently,
\[
\cL_t\coloneq \biguplus_{j=2}^{t} \frac{M_t}{\binom{t-1}{j-1}}\cdot \binom{[t]}{j}.
\]
\begin{proposition}\label{prop:lubell}
For every \(t\ge 2\), the support multiset \(\cL_t\) is an \((M_t,\dots,M_t)\)-frame.
Moreover,
\[
\abs{\cL_t}=M_t\,t\,(h_t-1),
\qquad
h_t\coloneq 1+\frac12+\cdots+\frac1t.
\]
\end{proposition}
\begin{proof}
Fix \(I\subseteq T\subseteq [t]\), and write
\[
i\coloneq \abs{I},
\qquad
z\coloneq \abs{T\setminus I}.
\]
If \(i=0\) or \(z=0\), then \(\omega_{\cL_t}(T,I)=0\), so there is nothing to prove.
Assume \(i\ge 1\) and \(z\ge 1\).
A support pattern of size \(r+1\) contributes to \(\omega_{\cL_t}(T,I)\) exactly when it chooses one element of \(I\) and \(r\) elements of \(T\setminus I\).
Therefore
\[
\omega_{\cL_t}(T,I)
=
iM_t\sum_{r=1}^{z}\frac{\binom{z}{r}}{\binom{t-1}{r}}.
\]
Use the beta-integral identity
\[
\frac{1}{\binom{t-1}{r}}
=
t\int_0^1 x^r(1-x)^{t-1-r}\,dx.
\]
Then
\begin{align*}
\sum_{r=0}^{z}\frac{\binom{z}{r}}{\binom{t-1}{r}}
&=
t\int_0^1 \sum_{r=0}^{z}\binom{z}{r}x^r(1-x)^{t-1-r}\,dx\\
&=
t\int_0^1 (1-x)^{t-1-z}\,dx\\
&=
\frac{t}{t-z}.
\end{align*}
Subtracting the \(r=0\) term gives
\[
\sum_{r=1}^{z}\frac{\binom{z}{r}}{\binom{t-1}{r}}
=
\frac{z}{t-z}.
\]
Hence
\[
\omega_{\cL_t}(T,I)=iM_t\frac{z}{t-z}.
\]
Since \(i+z=\abs{T}\le t\), we have \(i\le t-z\), so
\[
\omega_{\cL_t}(T,I)\le zM_t=\sum_{j\in T\setminus I} M_t.
\]
Thus \(\cL_t\) is an \((M_t,\dots,M_t)\)-frame.
For the size,
\[
\abs{\cL_t}
=
\sum_{j=2}^{t}\binom{t}{j}\frac{M_t}{\binom{t-1}{j-1}}
=
M_t\sum_{j=2}^{t}\frac{t}{j}
=
M_t\,t\,(h_t-1).
\]
\end{proof}
The case \(t=3\) is the exact frame \(F_{(2,2,2)}\) from Appendix~\ref{app:exactframes}; the case \(t=4\) is the classical four-way core \(Q_0\) used in the uniform construction.
Thus the Lubell family unifies the clean asymptotic side with the explicit uniform one.
Table~\ref{tab:gamma} records the values of \(\gamma_t\) for small \(t\).
\begin{table}[h]
\centering
\small
\begin{tabular}{ccccc}
\toprule
\(t\) & \(M_t\) & \(|\mathcal{L}_t|\) & \(\gamma_t=2(h_t-1)/\log_2 t\) & approx.\\
\midrule
2 & 1 & 1 & \(1\) & \(1.000000\)\\
3 & 2 & 5 & \(\frac{5}{3\log_2 3}\) & \(1.051550\)\\
4 & 3 & 13 & \(\frac{13}{12}\) & \(1.083333\)\\
5 & 12 & 77 & \(\frac{77}{30\log_2 5}\) & \(1.105403\)\\
6 & 10 & 87 & \(\frac{29}{10\log_2 6}\) & \(1.121873\)\\
7 & 60 & 669 & \(\frac{223}{70\log_2 7}\) & \(1.134774\)\\
\(\infty\) & --- & --- & \(2\ln 2\) & \(1.386294\)\\
\bottomrule
\end{tabular}
\caption{Harmonic gadget parameters and asymptotic multipliers \(\gamma_t\).}
\label{tab:gamma}
\end{table}
\begin{remark}
Already at \(t=6\) the Lubell gadget achieves \(\gamma_6\approx 1.1219\), which strictly exceeds \(\gamma_4=13/12\approx 1.0833\).
So while the uniform \(26/25\) construction in Section~\ref{sec:uniform} uses the four-way core (\(t=4\)) because it needs explicit control at every \(n\ge 15\), the asymptotic picture is strictly better with \(t\ge 6\), and the limiting constant is \(2\ln 2\).
\end{remark}
The \(t=6\) Lubell gadget is worth recording explicitly.
Here \(M_6=\operatorname{lcm}(1,5,10,10,5,1)=10\), so \(\lambda_r=10/\binom{5}{r}\) for each \(r=1,\dots,5\):
\[
\lambda_1=\tfrac{10}{5}=2,\quad
\lambda_2=\tfrac{10}{10}=1,\quad
\lambda_3=\tfrac{10}{10}=1,\quad
\lambda_4=\tfrac{10}{5}=2,\quad
\lambda_5=\tfrac{10}{1}=10.
\]
Thus \(\mathcal{L}_6\) consists of:
\begin{itemize}[leftmargin=2.2em]
\item \(2\) copies of every \(2\)-element subset of \([6]\) (\(15\) patterns, \(30\) total),
\item \(1\) copy of every \(3\)-element subset (\(20\) patterns),
\item \(1\) copy of every \(4\)-element subset (\(15\) patterns),
\item \(2\) copies of every \(5\)-element subset (\(6\) patterns, \(12\) total),
\item \(10\) copies of the full \(6\)-set (\(10\) total).
\end{itemize}
The total is \(30+20+15+12+10=87=6\cdot 10\cdot(h_6-1)\).
The gadget is \((10,10,10,10,10,10)\)-admissible by Proposition~\ref{prop:lubell}.
\begin{proposition}\label{prop:digit-sum}
Let \(s_2(m)\) denote the sum of the binary digits of \(m\), with \(s_2(0)=0\).
Then for every \(n\ge 1\),
\[
k_n=n+\sum_{j=0}^{n-1}s_2(j).
\]
Moreover,
\[
k_n=\frac12 n\log_2 n+O(n),
\qquad\text{and therefore}\qquad
\lim_{n\to\infty}\frac{k_n}{n\log_2 n}=\frac12.
\]
\end{proposition}
\begin{proof}
Define
\[
f(n)\coloneq \sum_{j=0}^{n-1}s_2(j),
\qquad
a_n\coloneq n+f(n).
\]
Using
\[
s_2(2j)=s_2(j),
\qquad
s_2(2j+1)=s_2(j)+1,
\]
we obtain
\[
f(2m)=\sum_{j=0}^{m-1}\bigl(s_2(2j)+s_2(2j+1)\bigr)=2f(m)+m.
\]
Hence
\[
a_{2m}=2m+f(2m)=m+2a_m.
\]
Also
\[
a_{m+1}=a_m+1+s_2(m),
\]
so
\[
a_{2m+1}=a_{2m}+1+s_2(2m)
=m+2a_m+1+s_2(m)
=m+a_m+a_{m+1}.
\]
Thus \((a_n)\) satisfies the same recurrence and initial condition as \((k_n)\):
\[
a_1=1,
\qquad
a_{2m}=m+2a_m,
\qquad
a_{2m+1}=m+a_m+a_{m+1}.
\]
By induction, \(a_n=k_n\) for all \(n\), proving the exact formula.
We now prove the asymptotic.
Let
\[
f(n)\coloneq \sum_{j=0}^{n-1}s_2(j),
\qquad\text{so }k_n=n+f(n).
\]
For each bit position \(i\ge 0\), let \(w_i(n)\) be the number of integers \(0\le j<n\) whose \(i\)-th binary digit equals \(1\).
Then
\[
f(n)=\sum_{i\ge 0} w_i(n),
\]
and only the indices \(0\le i\le \floor{\log_2 n}\) can contribute.
Fix \(i\ge 0\) and write \(n=q\cdot 2^{i+1}+r\) with \(0\le r<2^{i+1}\).
In each full block of length \(2^{i+1}\) the \(i\)-th digit is \(1\) for exactly \(2^i\) values, and in the final partial block it contributes \(\max(0,r-2^i)\) ones.
Hence
\[
w_i(n)=q\cdot 2^i+\max(0,r-2^i).
\]
Using \(q=\floor{n/2^{i+1}}\), we have
\[
\frac{n}{2}-2^i \le q\cdot 2^i \le \frac{n}{2},
\qquad
0\le \max(0,r-2^i)\le 2^i,
\]
so for every \(0\le i\le L\coloneq \floor{\log_2 n}\),
\[
\bigl|w_i(n)-\tfrac{n}{2}\bigr|\le 2^i.
\]
Summing over \(i=0,1,\dots,L\) gives
\[
\left|f(n)-\frac{(L+1)n}{2}\right|
\le
\sum_{i=0}^{L}2^i
=
2^{L+1}-1
<2n,
\]
because \(2^L\le n<2^{L+1}\).
Therefore
\[
f(n)=\frac{(L+1)n}{2}+O(n).
\]
Since \(L+1=\log_2 n+O(1)\), we conclude
\[
f(n)=\frac12\,n\log_2 n+O(n),
\qquad
k_n=n+f(n)=\frac12\,n\log_2 n+O(n),
\]
and hence
\[
\lim_{n\to\infty}\frac{k_n}{n\log_2 n}=\frac12.
\]
\end{proof}
\begin{theorem}\label{thm:fixed-t}
Fix \(t\ge 2\).
There exists a constant \(C_t>0\) such that
\[
H(n)\ge \frac{h_t-1}{\log_2 t}\,n\log_2 n - C_t n
\qquad\text{for all }n\ge 1.
\]
\end{theorem}
\begin{proof}
Fix \(t\ge 2\), and write
\[
M\coloneq M_t,
\qquad
B\coloneq \abs{\cL_t}=Mt(h_t-1),
\qquad
a_t\coloneq \frac{h_t-1}{\log_2 t}.
\]
Define \(L_n^{(t)}\) recursively as follows.
For \(1\le n<tM\), set
\[
L_n^{(t)}\coloneq A_n.
\]
For \(n\ge tM\), let \(p_1,\dots,p_t\) be the balanced nondecreasing partition of \(n\) into \(t\) positive parts; this exists because \(n\ge tM\ge t\).
Set
\[
q\coloneq \floor{\frac{\floor{n/t}}{M}},
\]
and define
\[
L_n^{(t)}\coloneq L_{p_1}^{(t)}+\cdots+L_{p_t}^{(t)}+qB.
\]
We claim that \(H(n)\ge L_n^{(t)}\) for all \(n\ge 1\).
For \(1\le n<tM\), this is Theorem~\ref{thm:constructive-An}.
Now let \(n\ge tM\).
Then \(qM\le p_i\) for every \(i\), so \(q\) copies of \(\cL_t\) form a legal frame on capacities \(p_1,\dots,p_t\).
By Corollary~\ref{cor:frame-recurrence}, applied recursively to witnesses for the smaller values \(p_1,\dots,p_t\), we obtain a witness with \(L_n^{(t)}\) vertices.
Hence
\[
H(n)\ge L_n^{(t)}
\qquad\text{for all }n\ge 1.
\]
We now prove by induction that there exists \(C_t>0\) such that
\[
L_n^{(t)}\ge a_t\,n\log_2 n - C_t n
\qquad\text{for all }n\ge 1.
\]
Choose \(C_t\) large enough that this holds for all \(1\le n<tM\).
Assume now that \(n\ge tM\) and that the claim is already true for smaller values.
Then
\[
L_n^{(t)}
=
L_{p_1}^{(t)}+\cdots+L_{p_t}^{(t)}+qB
\ge
a_t\sum_{i=1}^t p_i\log_2 p_i - C_t n + qB.
\]
Since \(x\mapsto x\log_2 x\) is convex and \(\sum_i p_i=n\), the balanced partition minimizes the sum, so
\[
\sum_{i=1}^{t} p_i\log_2 p_i
\ge
n\log_2(n/t)
=
n\log_2 n - n\log_2 t.
\]
Also,
\[
q=\floor{\frac{\floor{n/t}}{M}}
\ge
\frac{n}{tM}-2,
\]
and therefore
\[
qB\ge (h_t-1)n - 2Mt(h_t-1).
\]
Substituting these bounds,
\begin{align*}
L_n^{(t)}
&\ge
a_t\bigl(n\log_2 n - n\log_2 t\bigr)-C_t n+(h_t-1)n-2Mt(h_t-1)\\
&=
a_t\,n\log_2 n - C_t n - 2Mt(h_t-1),
\end{align*}
because \(a_t\log_2 t=h_t-1\).
After increasing \(C_t\) once more, the induction closes.
Thus
\[
H(n)\ge L_n^{(t)}\ge a_t\,n\log_2 n - C_t n
\qquad\text{for all }n\ge 1.
\]
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:asymptotic}]
By Proposition~\ref{prop:digit-sum},
\[
k_n=\frac12 n\log_2 n + O(n).
\]
Also \(h_t=\ln t + O(1)\), so
\[
\frac{2(h_t-1)}{\log_2 t}\longrightarrow 2\ln 2
\qquad (t\to\infty).
\]
Fix \(\varepsilon>0\).
Choose \(t\) so large that
\[
\frac{2(h_t-1)}{\log_2 t}>2\ln 2 - \varepsilon.
\]
Then Theorem~\ref{thm:fixed-t} gives
\[
H(n)\ge \left(\frac{h_t-1}{\log_2 t}\right)n\log_2 n - O_t(n).
\]
Dividing by \(k_n\) and letting \(n\to\infty\) yields
\[
\liminf_{n\to\infty}\frac{H(n)}{k_n}
\ge
\frac{2(h_t-1)}{\log_2 t}
>
2\ln 2 - \varepsilon.
\]
Since \(\varepsilon>0\) was arbitrary, the theorem follows.
\end{proof}
\appendix
\section{Exact small frames used in the finite bootstrap}\label{app:exactframes}
In this appendix every support pattern \(\{i_1,\dots,i_s\}\) is abbreviated by the concatenated string \(i_1\cdots i_s\), and coefficients denote multiplicity.
These are exactly the support frames used in the rows marked \(F_{(\cdots)}\) in Appendix~\ref{app:bootstrap}.
They have all been precomputed and hard-coded into the final script, so the official constructor never solves any optimization problem at runtime.
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\setlength{\tabcolsep}{5pt}
\begin{longtable}{>{\raggedright\arraybackslash}p{0.15\textwidth} c >{\raggedright\arraybackslash}p{0.66\textwidth}}
\caption{Exact small frames used below \(60\).}\label{tab:exactframes}\\
\toprule
capacity vector & size & support multiset \\
\midrule
\endfirsthead
\multicolumn{3}{l}{\textit{Table \ref{tab:exactframes} continued.}}\\
\toprule
capacity vector & size & support multiset \\
\midrule
\endhead
\midrule
\multicolumn{3}{r}{\textit{Continued on next page.}}\\
\endfoot
\bottomrule
\endlastfoot
$(2,2,2)$ & 5 & 12 + 13 + 23 + 2$\cdot$123 \\
$(4,4,6)$ & 11 & 3$\cdot$12 + 13 + 23 + 6$\cdot$123 \\
$(6,6,6)$ & 15 & 3$\cdot$12 + 3$\cdot$13 + 3$\cdot$23 + 6$\cdot$123 \\
$(2,2,2,3)$ & 9 & 12 + 13 + 23 + 123 + 124 + 134 + 234 + 2$\cdot$1234 \\
$(2,2,2,4)$ & 10 & 12 + 13 + 23 + 123 + 124 + 134 + 234 + 3$\cdot$1234 \\
$(4,4,6,9)$ & 22 & 3$\cdot$12 + 13 + 23 + 3$\cdot$123 + 2$\cdot$124 + 3$\cdot$134 + 2$\cdot$234 + 7$\cdot$1234 \\
$(6,6,6,6)$ & 26 & 2$\cdot$12 + 2$\cdot$13 + 2$\cdot$14 + 2$\cdot$23 + 2$\cdot$24 + 2$\cdot$34 + 2$\cdot$123 + 2$\cdot$124 + 2$\cdot$134 + 2$\cdot$234 + 6$\cdot$1234 \\
$(6,6,6,7)$ & 26 & 3$\cdot$12 + 2$\cdot$13 + 14 + 2$\cdot$23 + 24 + 2$\cdot$34 + 123 + 2$\cdot$124 + 2$\cdot$134 + 3$\cdot$234 + 7$\cdot$1234 \\
$(6,6,6,8)$ & 27 & 3$\cdot$12 + 3$\cdot$13 + 3$\cdot$23 + 2$\cdot$123 + 4$\cdot$124 + 4$\cdot$134 + 4$\cdot$234 + 4$\cdot$1234 \\
$(6,6,6,9)$ & 28 & 3$\cdot$12 + 3$\cdot$13 + 3$\cdot$23 + 2$\cdot$123 + 4$\cdot$124 + 4$\cdot$134 + 4$\cdot$234 + 5$\cdot$1234 \\
$(6,6,6,10)$ & 29 & 3$\cdot$12 + 3$\cdot$13 + 3$\cdot$23 + 2$\cdot$123 + 4$\cdot$124 + 4$\cdot$134 + 4$\cdot$234 + 6$\cdot$1234 \\
$(2,2,2,3,4)$ & 15 & 12 + 13 + 23 + 124 + 125 + 134 + 135 + 145 + 2$\cdot$234 + 2$\cdot$2345 + 3$\cdot$12345 \\
$(3,4,4,4,6)$ & 25 & 12 + 13 + 14 + 23 + 2$\cdot$34 + 45 + 123 + 2$\cdot$124 + 125 + 135 + 3$\cdot$345 + 1234 + 2$\cdot$1235 + 2$\cdot$1245 + 2345 + 4$\cdot$12345 \\
$(4,4,4,4,4)$ & 25 & 12 + 13 + 14 + 15 + 23 + 24 + 25 + 34 + 35 + 45 + 234 + 235 + 245 + 345 + 2$\cdot$1234 + 2$\cdot$1235 + 2$\cdot$1245 + 2$\cdot$1345 + 2345 + 2$\cdot$12345 \\
$(4,4,4,4,6)$ & 27 & 12 + 13 + 2$\cdot$14 + 23 + 24 + 34 + 2$\cdot$123 + 125 + 135 + 145 + 234 + 235 + 245 + 345 + 1234 + 1235 + 2$\cdot$1245 + 2$\cdot$1345 + 2$\cdot$2345 + 3$\cdot$12345 \\
$(5,6,6,6,6)$ & 35 & 2$\cdot$12 + 13 + 14 + 15 + 2$\cdot$23 + 24 + 2$\cdot$34 + 35 + 45 + 2$\cdot$124 + 125 + 135 + 145 + 2$\cdot$245 + 3$\cdot$345 + 2$\cdot$1234 + 3$\cdot$1235 + 8$\cdot$12345 \\
$(6,6,6,6,6)$ & 38 & 12 + 13 + 14 + 3$\cdot$15 + 23 + 24 + 25 + 34 + 35 + 45 + 2$\cdot$123 + 2$\cdot$124 + 2$\cdot$134 + 2$\cdot$234 + 2$\cdot$235 + 2$\cdot$245 + 2$\cdot$345 + 2$\cdot$1235 + 2$\cdot$1245 + 2$\cdot$1345 + 6$\cdot$12345 \\
$(6,6,6,6,10)$ & 42 & 2$\cdot$12 + 2$\cdot$13 + 2$\cdot$14 + 2$\cdot$23 + 2$\cdot$24 + 2$\cdot$34 + 123 + 2$\cdot$124 + 125 + 134 + 135 + 145 + 234 + 2$\cdot$235 + 245 + 2$\cdot$345 + 1234 + 3$\cdot$1235 + 3$\cdot$1245 + 3$\cdot$1345 + 2$\cdot$2345 + 5$\cdot$12345 \\
$(6,6,6,6,11)$ & 42 & 2$\cdot$12 + 2$\cdot$13 + 2$\cdot$14 + 2$\cdot$23 + 2$\cdot$24 + 2$\cdot$34 + 2$\cdot$123 + 2$\cdot$124 + 134 + 2$\cdot$234 + 5$\cdot$1235 + 5$\cdot$1245 + 6$\cdot$1345 + 5$\cdot$2345 + 2$\cdot$12345 \\
$(6,6,6,6,12)$ & 44 & 2$\cdot$12 + 2$\cdot$13 + 2$\cdot$14 + 2$\cdot$23 + 2$\cdot$24 + 2$\cdot$34 + 2$\cdot$123 + 2$\cdot$124 + 2$\cdot$134 + 2$\cdot$234 + 6$\cdot$1235 + 6$\cdot$1245 + 6$\cdot$1345 + 6$\cdot$2345 \\
$(6,6,6,6,13)$ & 44 & 2$\cdot$12 + 2$\cdot$13 + 2$\cdot$14 + 2$\cdot$23 + 2$\cdot$24 + 2$\cdot$34 + 2$\cdot$123 + 2$\cdot$124 + 2$\cdot$134 + 2$\cdot$234 + 6$\cdot$1235 + 5$\cdot$1245 + 6$\cdot$1345 + 5$\cdot$2345 + 2$\cdot$12345 \\
\end{longtable}
\endgroup
\section{Explicit boosters and four-way residue gadgets}\label{app:boosters}
The next table lists the seven higher-arity boosters used in the finite bootstrap.
They are the only gadgets in the paper living on more than five blocks.
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