With D, condensation to the ultra-dense form should be easier and release more energy, and spontaneous D–D fusion releasing energy locally should be occurring as well. Annihilation reactions from which muons are eventually formed however might not easily deposit their energy locally without thick shielding or special shielding.
Since the bonding is slightly stronger in D(0) than in protium p(0), it is likely that deuterons (which are bosons) condense to d(0) more easily than protons (fermions) do to p(0), and that d(0) is more resistant against excitation and fragmentation
There exists one clear difference in the cluster forms for protium p(0) and deuterium D(0). This concerns the distances observed in the CE experiments. The CE experiments on D(0) clusters give distances of approximately 2.3 pm (s = 2) and 0.56 pm (s = 1), while those for p(0) give distances of 2.3 pm (s = 2) and 5.0 pm (s = 3).
I'm not sure if the resistivity decrease is related to the superconductivity of the chain clusters of H(0), since above a certain temperature these are supposed to disappear due to reaching their critical temperature. I recall that the resistivity decrease observed in Celani wires persists regardless of temperature, until all or most of the hydrogen is removed from it with heat and vacuum. It could be that the properties of UDH are different when it is absorbed inside the material than on the surface, but this hasn't been investigated so far.