gpt 6 astra ultra attempt on Math-1047

Minimal graphs with a prescribed number of spanning trees

Definitions and mathematical statement

For a finite simple undirected graph $G$, a spanning tree is a connected acyclic subgraph containing every vertex. Let $\tau(G)$ be the number of its spanning trees, with distinct edge sets counted separately. For each integer $n\ge3$, define \[ \alpha(n)=\min\{|V(G)|:G\text{ finite, simple and undirected},\ \tau(G)=n\}. \] The minimum exists: deleting any one edge of the cycle $C_n$ gives one of its exactly $n$ spanning trees, so $\alpha(n)\le n$. Any graph contributing to the minimum is connected because its tree count is positive. Loops, parallel edges and directed arcs are not allowed.

The conjecture asserts \[ \lim_{n\to\infty}\frac{\alpha(n)}{\log n}=0, \] where $n$ tends to infinity through all integers and $\log$ may be taken to be the natural logarithm. Equivalently, for every $\varepsilon>0$ there exists $N_\varepsilon$ such that every integer $n\ge N_\varepsilon$ is the exact spanning-tree count of a simple graph on at most $\varepsilon\log n$ vertices.

The number $n$ here is the prescribed count, not the number of graph vertices. The conclusion is uniform over all sufficiently large counts; constructions only for a sparse subsequence of highly composite integers do not suffice. Nor is it enough to find a graph with at least $n$ spanning trees. The exact equality is what makes the minimization an arithmetic realization problem.

Short English statement

Can every sufficiently large integer be realized as an exact spanning-tree count using a number of vertices negligible compared with its logarithm?

Research attempt

Scope and process.

GPT-6 Astra Ultra, 1 October 2026. This attempt keeps the catalogue statement above, checks primary literature, and develops a restricted argument with an adversarial gap audit. The proofs below are the mathematical output of this attempt, not a claim of novelty or external certification.

Literature and attack plan.

Stong's primary paper [R1] proves the uniform upper bound $O((\log n)^{3/2}/\log\log n)$ and explicitly distinguishes it from the conjectured $o(\log n)$. We do not claim that even an $O(\log n)$ bound follows from that result. The attempt here determines an exact infinite subsequence, proves the universal information-size obstruction, and examines why gluing graphs fails to supply the required uniform little-oh bound.

A universal counting obstruction.

A simple graph on $v\ge2$ labelled vertices has at most $v^{v-2}$ spanning trees, since each of its trees is also a tree of $K_v$, and Prüfer coding counts the latter. For completeness, the encoding successively removes the smallest labelled leaf and records its neighbour until two vertices remain. It produces a word of length $v-2$ over the $v$ labels. Decoding chooses at each step the smallest label absent from the remaining word, joins it to the first word entry, and removes that entry; finally it joins the last two labels. These procedures are inverse, so exactly $v^{v-2}$ labelled trees exist.

Thus every graph realizing $n\ge3$ satisfies \[ \log n\le(v-2)\log v. \tag{1} \] For sufficiently large $n$, write $L=\log n$. If $v\ge L$, then $v\ge L/\log L$. If $v<L$, $1$ gives $L\le v\log L$, giving the same inequality. Consequently \[ \alpha(n)\ge\frac{\log n}{\log\log n} \quad\text{for all sufficiently large }n. \tag{2} \] This lower bound is compatible with $o(\log n)$; its ratio to $\log n$ tends to zero.

An exact subsequence.

For $v\ge3$, put $n_v=v^{v-2}$. The complete graph gives $\alpha(n_v)\le v$. No graph of smaller order can attain that count: for integers $2\le u<v$, $u^{u-2}<v^{v-2}$, as follows by increasing both positive base and nonnegative exponent, with strictness once $v\ge3$. Graphs on one or two vertices have at most one spanning tree. Therefore \[ \alpha(v^{v-2})=v. \tag{3} \] It follows both that \[ \frac{\alpha(n_v)}{\log n_v} =\frac{v}{(v-2)\log v}\longrightarrow0 \] and that the lower-bound scale in $2$ is attained asymptotically: \[ \liminf_{n\to\infty} \frac{\alpha(n)\log\log n}{\log n}=1. \tag{4} \] For the upper direction of $4$, use $3$ and $\log\log(v^{v-2})=\log(v-2)+\log\log v$; division by $\log v$ tends to one. The lower direction is $2$.

Exact multiplication by vertex gluing.

Let connected simple graphs $G,H$ share exactly one vertex and otherwise have disjoint vertices and edges. A spanning tree of their union restricts to a spanning tree on each graph. Indeed, a path from a vertex of $G$ to the shared vertex cannot leave $G$ before first reaching that vertex; connectedness of the whole tree therefore forces connectedness of the restriction, while acyclicity is inherited. Conversely the union of two spanning trees is connected and acyclic, since a cycle crossing between the pieces would have to revisit the unique shared vertex. Thus \[ \tau(G\vee H)=\tau(G)\tau(H),\qquad \alpha(ab)\le\alpha(a)+\alpha(b)-1 \quad(a,b\ge3). \tag{5} \] Iterating gives $\alpha(a^t)\le1+t(\alpha(a)-1)$.

Why the gluing approach stops.

For a fixed factor $a$, division of the last inequality by $\log(a^t)=t\log a$ only bounds the ratio by a positive constant. Letting the efficient block size grow can help selected factorizations, but an arbitrary integer need not admit such factors; primes admit none at all. Neither the complete-graph subsequence nor $5$ turns into a construction for every sufficiently large integer. Changing "exactly $n$" to "at least $n$" would remove the arithmetic difficulty and change the problem.

Verification and final status.

The constructions remain simple graphs: identifying one vertex introduces neither loops nor parallel edges. For $v=3$, $3$ gives the triangle with three trees; for $v=4$, it gives $K_4$ with sixteen trees. The limiting calculation in $4$ concerns a liminf, not a limit or a uniform upper bound. The exact subsequence, product identity, and obstruction $2$ are proved. The all-integer realization bound in the original conjecture is UNRESOLVED by this attempt.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3121 (OPG-37420), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Minimal graphs with a prescribed number of spanning trees. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/minimal_graphs_with_a_prescribed_number_of_spanning_trees.
• R1 R. Stong, Minimal graphs with a prescribed number of spanning trees, Australasian Journal of Combinatorics 82$2$ $2022$, 182--196. Primary abstract and introduction checked. https://ajc.maths.uq.edu.au/pdf/82/ajc_v82_p182.pdf.