Is $\limsup_n\,(f(n)-2p_n)=\infty$ for $f(n)=\min_{0<i<n}(p_{n+i}+p_{n-i})$? (Erdős #454)

Let $p_k$ denote the $k$th prime and define $f(n)=\min_{0<i<n}(p_{n+i}+p_{n-i})$, the smallest sum of two primes 'symmetric' about $p_n$ in index. Since the $i=0$ term is $2p_n$, we always have $f(n)\le 2p_n$ is NOT forced — the question is how far below or above $2p_n$ the minimum can sit. Erdős asked whether $\limsup_n(f(n)-2p_n)=\infty$. For each $n$, $f(n)$ is a finite minimum over $i=1,\ldots,n-1$ using a precomputed prime table, so $g(n):=f(n)-2p_n$ is directly computable. GOAL (computational): compute $g(n)$ for all $n\le N$, track its running maximum $M(N)=\max_{n\le N}g(n)$, and report whether $M(N)$ grows (evidence the limsup is infinite) or stalls.

Acceptance. PARTIAL / PRIMARY (finite, machine-checkable): the sequence of records $M(N)=\max_{n\le N}(f(n)-2p_n)$ for $N$ as large as feasible (state $N$; e.g. $n$ up to $10^8$ or beyond with a segmented sieve), delivered with (i) a runnable script that sieves primes and computes $f(n)$, and (ii) the list of record-setting $n$ and $g(n)$ values. A steadily-growing record sequence is direct evidence that the limsup is infinite; a plateau is evidence against. FULLY RESOLVES: a proof that $\limsup_n(f(n)-2p_n)$ is (or is not) infinite. State $N$ and the sieving method.