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Math-5470-A1 · revision r1 · on Math-5470

GPT 6 Astra Ultra attempt on Math-5470

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 165: the asymptotic constant in $R(3,k)$
  2. Statement and outcome
  3. The scalar weights
  4. Shearer's graph induction
  5. An elementary lower construction
  6. Assessment and attribution
  7. COMPLETION ESTIMATE

Erdős Problem 165: the asymptotic constant in $R(3,k)$

Statement and outcome

The current problem asks for an asymptotic formula for $R(3,k)$, the least $n$ such that every graph on $n$ vertices contains a triangle or an independent $k$-set. This is not the sparse-ruler question appearing in the old local file with this number.

We reconstruct Shearer's upper bound \[ R(3,k)\le(1+o(1))\frac{k^2}{\log k}, \tag{1} \] including the scalar estimates needed in its induction. We also give an elementary lower bound of order $(k/\log k)^{3/2}$. The latter is substantially weaker than the published $(1/2+o(1))k^2/\log k$ lower bound of Hefty--Horn--King--Pfender. Their construction is not reproduced or used as a proved-here result.

The scalar weights

Define $f(0)=1$ and, for integers $d\ge1$, \[ f(d)=\frac{1+(d^2-d)f(d-1)}{d^2+1}, \qquad b_d=f(d-1)-f(d). \tag{2} \] We need that $f$ decreases and that $b_d$ is nonnegative and nonincreasing. Induction gives $f(d)\ge1/(d+1)$: the recurrence applied to $f(d-1)\ge1/d$ gives $f(d)\ge d/(d^2+1)\ge1/(d+1)$. Then $b_d=((d+1)f(d-1)-1)/(d^2+1)\ge0$.

For convexity, put $g_d=(3d+2)/(2d^2+6d+2)$. We have $f(0)=g_0$. Substituting $g_{d-1}$ into $2$ gives $(3d-2)/(2d^2+2d-2)\ge g_d$, the cross-multiplied difference being $4d(d-1)\ge0$. Hence $f(d)\ge g_d$. For $d\ge2$, direct subtraction gives \[ b_d-b_{d+1} =\frac{2(d^2+3d+1)f(d)-(3d+2)} {d(d-1)(d^2+2d+2)}\ge0. \] The remaining comparison is $b_1=1/2\ge b_2=1/10$.

There is also the exact identity \[ d f(d)-(d-1)f(d-1) =\frac1d-\frac{1+d(d-1)f(d-1)}{d(d^2+1)}. \tag{3} \] It first implies $d f(d)\le H_d$. For $d\ge2$, the subtracted term is at most $d^{-3}+H_{d-1}/d^2$, a summable sequence. Summing $3$ proves \[ d f(d)=\log d+O(1). \tag{4} \] This uses a summable $O(\log d/d^2)$ error; an error strictly below $1/d^2$ is neither needed nor generally valid.

Shearer's graph induction

For a triangle-free graph $G$, write $d_v=\deg(v)$ and $S(G)=\sum_v f(d_v)$. We prove $\alpha(G)\ge S(G)$ by induction on its order. Delete $i$ and its neighbours, leaving $H_i$. For a vertex $k$ at distance two from $i$, let $n_i(k)=|N(k)\cap N(i)|$. The gain available from choosing $i$ and then an independent set in $H_i$ is \[ L_i=1-f(d_i)-\sum_{j\in N(i)}f(d_j) +\sum_{\operatorname{dist}(i,k)=2} [f(d_k-n_i(k))-f(d_k)]. \] It suffices that some $L_i\ge0$. Convexity of the scalar sequence gives $f(d_k-n_i(k))-f(d_k)\ge n_i(k)b_{d_k}$. Triangle-freeness ensures that all relevant two-edge paths end at distance two, so changing the order of summation yields \[ \sum_i\sum_{\operatorname{dist}(i,k)=2} [f(d_k-n_i(k))-f(d_k)] \ge\sum_k b_{d_k}\sum_{j\in N(k)}(d_j-1) \ge\sum_k d_k(d_k-1)b_{d_k}. \] The last inequality follows edge by edge: the difference for an edge $jk$ is $(d_j-d_k)(b_{d_k}-b_{d_j})\ge0$. Consequently \[ \sum_i L_i\ge \sum_i[1-(d_i+1)f(d_i)+d_i(d_i-1)b_{d_i}]=0 \] by $2$; degree-zero terms equal zero as well. Some $L_i$ is nonnegative, completing the induction.

If $\alpha(G)<k$, every neighbourhood is independent, hence $\Delta(G)\le k-1$. Since $f$ decreases, $k>\alpha(G)\ge n f(k-1)$. Applying $4$ gives $1$.

An elementary lower construction

For large $k$, let $N=\lfloor\frac18(k/\log k)^{3/2}\rfloor$ and $p=2\log k/k$. In $G(N,p)$ the expected number of triangles is at most $N^3p^3/6\le N/48$. Markov's inequality makes the probability of more than $N/2$ triangles at most $1/24$. Meanwhile the probability of an independent $k$-set is at most \[ \binom Nk(1-p)^{\binom k2} \le\exp\left(k\log(eN/k)-p k(k-1)/2\right)=o(1). \] Thus some graph has no independent $k$-set and at most $N/2$ triangles. Delete one vertex from each triangle, removing at most $N/2$ vertices. The surviving induced graph is triangle-free, has at least $N/2$ vertices, and still has no independent $k$-set. This proves the stated weaker lower order.

Assessment and attribution

The sharp upper argument is a reconstruction of Shearer's work, using the exposition in Section 5 of Harris--Hefty--Horn--King--Pfender, https://arxiv.org/abs/2609.01782, with the scalar estimates supplied explicitly. The current lower record is documented in https://arxiv.org/abs/2510.19718. Problem: https://www.erdosproblems.com/165. The unrelated contents of the old gpt_pro_5.2/165.tex have not been reused. The asymptotic leading constant, and even a reconstruction here of a matching-order lower bound, remain unresolved in this attempt. No novelty claim is made.

COMPLETION ESTIMATE

COMPLETION: 45%