To check the above conjuncture empirically, we need only to determine the persistence of all numbers of the form 2n. It is conjunctured that 2n for n>15 has at most 1 zero in base 3 notation. This were check in 1994 for all n<=500. If 2n contains a zero in its base 3 notation, then the persistence of the corresponding n twos is 2. Which would imply pmax(3)=3.
The numerical bound of n<=500 was, by own computations, pushed to n<=1011. This fact was reached by the use of a computer. To gain this, not the total digit-sequenz of 2n is needed. Saving the whole digitsequenz of for example 210^10 would need more thana 1 Gigabyte of available space. To save resources, and make the computation possible on a normal PC, only the last 100 digits in base 3 notation were checked. In the checked intervall the first zero was always before the 80 digit. If we compare this with the 6*109 digits of 210^10, the above conjuncture seems very likely. But a proof isn't in sight so far.
