This question turns out to be quite a lot more involved that I thought at first. But in the end, a lot of calculation just confirms (assuming I made no mistake) that no, shorter cranks do not allow you to get away with fewer gears: they allow you to pedal at higher cadence while retaining somewhat reasonable efficiency, but in the end they're never more efficient.
As a most idealised approximation, as per Adam Rice's answer, crank length simply forms part of the effective gear ratio, i.e. the speed and output power goes like
vbike = ω·ρgear = vped⁄lcrank·ρgear
Pdrive = τped·ω
- ω is the cadence
- Pdrive is the power you put through the drivetrain (ignoring friction losses in the gears/chain, which are low in a well-maintained bike)
- lcrank is the crank length
- vbike is the speed you're riding at
- vped is the speed the pedals/feet move at (relative to the BB)
- Fped is the force on the pedals, assuming perfect coordination
- ρgear is the gear ratio (including wheel size – you could factor this out, but might as well consider it just a constant factor)
- τped is the pedal torque
For the sake of comparison, we should consider speed and power as constants. But the formulas above don't say anything about the loss mechanisms, which is what choice of cadence is actually about.
In terms of pedalling, there are mostly two factors that incur energy loss L:
Too hard gears require burning energy in muscle-overloading, to get out of the saddle and do body movements that don't feed the power very well into the drivetrain. This term is governed mostly by pedal force, i.e. for a given output power you have
Lgrind ∝ Fped
= Pdrive⁄(ω · lcrank)
Too light gears require the leg mass to be accelerated back and forth at rapid pace. Essentially, you need to provide an extra centripetal force that doesn't contribute to propulsion at all, just to keeping the legs moving. Unlike with the grinding loss, this is a kinematic force, i.e. the loss is the actual power that's put into it (force times movement-speed)
Lspin ∝ FCP·vped
= (vped2⁄lcrank) · vped
[I'm not sure about these exact formulas – specifically whether linear proportionalities are really appropriate. But I tried a couple of variations, and they don't change the rest of the answer much.]
Let's condense the unknown power and proportionality factors – which depend strongly on the rider – into a single constant α, then we have the total loss
L ∝ α⁄(ω · lcrank) + ω3·lcrank2
∝ ⁴√α⁄(ω · lcrank) + (ω⁄⁴√α)3·lcrank2
or, with β = ⁴√α,
L ∝ (ω⁄β)-1 · 1⁄lcrank + (ω⁄β)3·lcrank2.
Plotting this for different crank lengths looks thus:
This tells us, somewhat unsurprisingly, that for a shorter crank length the optimum cadence will be faster. More interestingly, it tells us that you get a deeper “valley” if the crank length is long – i.e., longer cranks are more efficient! –That, of course, only holds within reason: you can't make the cranks arbitrary long, at some point the legs just wouldn't be able to follow, and even before you get ærodynamic issues. (To the other side, the muscles and joints probably aren't happy with a super-short action either.)
Anyway the mechanical optimum efficiency isn't that different – note that I vary the crank length by a ludicrous factor of 4 in the above plot, but minimum loss only varies by 25%. That explains why some disciplines do opt for shorter-than-ideal cranks, because the advantages (æro, ground clearance etc.) may well outweight the slight loss in pedalling efficiency.
It's mostly the optimum cadence that is different, namely
ωopt ∝ 1⁄⁴√lcrank3
This still doesn't answer the title question yet: does crank length affect the gear range a bike needs? – Well, looking at the plot above, I first thought that it demonstrates length does affect the needed range: with short cranks, you get a “shallower” curve, i.e. short cranks allow higher difference between the highest cadence that's usable and the lowest that is. IOW, you can get away with having a bit fewer gears, and then just use those “outskirts” of the cadency range where the efficiency is still ok... right?
However, on closer inspection this doesn't really hold up: in practice, you'll anyway have to choose the gears so you can operate in the optimum cadence region, i.e. you rescale the cadences overall by another constant. Given some constant target speed,
ρopt ∝ 1⁄ωopt
∝ ⁴√lcrank3 = lcrank¾
(Side note: therefore, crank length should not be understood as just a linear factor in the gear ratio – it is a factor, but less than linear.)
If we do that, centering the plot always around the optimum cadence, we see that for any discrepancy from the optimum, longer cranks result in less losses: