well, no actually, not that simple just to put in a resistor.
voltage drop over a resistor is proportional to the current - TOM estimates (and probably actually has measured carefully) that to be ~450ma per cylinder. E=IR. You could do the math and figure out a R value that would give you a voltage drop of 1.2v, (something like 2.5 ohms) but the actual drop would be dependent on instantaneous current draw, and dependent on the specific ignition system. If the draw were substantively different, the resistance would be wrong. I understand the interest in using what we have (6v, or lipos, or whatever), but in this situation it does seem simplest, and best, just to use a 4 cell pack. It is what the ignition system was designed to use. The other issue is that voltage drop across the resistor will generate a bit of heat, albeit not in the ignition.