just guessing, but it seems that the propulsion generated by the propellor results (at least directly) from the normal force of the blade surface opposing the force of resistance caused by the density of the air molecules pushing back with their normal force right?
the propellor turns because of the force applied by the engine, which from the blade pov is, we can say, essentially verticle. the more this force is applied to the blade, the more the normal force of the blade has to push to find equallibrium against the air(since the blade is at an angle this would be pronounced), contingent on the density of the air. so the faster the turn, the more the push, the faster the plane goes? (minus the other effects like drag and gravity etc)
so wouldn't less rpm typically result in less air displaced and thus, less push and less rate?
the only measure i can think of where rpm reduction could mean increased fuel economy is at low altitude or in situations where not as much thrust was required to reach the desired speed.
with denser air, the blades wouldn't have to turn as much to produce the same force. so conceivably - if you picked a reasonable cruising speed you might be able to maintain it wih less rpm. of course going to max rpm would result in better performance at any altitude by that standard i would think.
i wish i knew more about airplanes systems and dynamics particular to airplanes to understand the difference better, but from a simple force diagram standpoint alone it doesn't seem like it would help except at low alt and only then if you were happy cruising at less than the max obtainable speed. maybe someone can set me straight
