Does someone have an exact answer to this question about power amp output?

[Steve81]

What about the amplifier output power? Is it then also increased by a factor of four ?

No. To verify this, let's look at a set of measurements for the Dayton UM18 driver at data-bass:
http://www.data-bass.com/data?page=system&id=116&mset=128

Under the "Multi-Series Chart" section, we can see the system's voltage sensitivity when the UM18 driver is loaded into a 4.2CF sealed box. At 50Hz, the system puts out ~90dB with 2V @ 1m. At 100Hz, the system puts out ~88dB at 2V @ 1m. Going back to the "Comparable Charts" section, we can cross-reference these numbers with system impedance. At 50Hz, the system presents a ~7.5 ohm load, and at 100Hz, the system presents a ~6.2 ohm load. Given those details, it actually takes less power to deliver a given SPL with this system at 50Hz vs 100Hz.

 

[Verdinut]

In your example, you don't take the cone excursion factor for the two frequencies into consideration. Since the speaker is loaded by the sealed box, everything depends also on the T/S parameters of the driver.

I now believe that what Badmaieff and Davis advanced might have been respecting the laws of physics in theory.

However, in practice, the quadrupling of the cone excursion at half of a certain frequency does happen but in rare speaker box configurations, and in a particular design, it won't happen over all the low frequency range but only in a small isolated section of it. Actually, it can happen, for instance with a bass reflex box, in part of the frequency section just above the higher frequency impedance peak.

 

[Steve81]

In your example, you don't take the cone excursion factor for the two frequencies into consideration.

It does. Cone excursion in this case (a sealed box) is related to frequency and output as you originally asserted. 6dB more output equates to double the excursion; dropping down an octave (halving the frequency) while output remains the same equates to four times the excursion from the system. This is true (with caveats if you want to go deeper) for sealed boxes and infinite baffle installations.

An equation from Leo Beranek's Acoustics gives a pretty good idea of how these factors (and a couple other pertinent details) go together for these alignments:

pRMS = sqrt(2) * pi * f^2 * rho * Sd * xp / r

where:
pRMS = RMS acoustic pressure in Pascals, half space
f = frequency in Hz
rho = density of air (= 1.2041 kg / m^3 at 20 deg C, sea level)
Sd = surface area of driver in m^2
xp = peak displacement in meters (driver excursion)
r = distance in meters

To compute SPL, you calculate:
SPL = 20 * log10 (pRMS/ pREF ) where: pREF = 2 * 10^-5 Pascals

If you're lazy, this site dumbs things down a bit, but gets the job done for those who want to play around with the basics:
http://www.baudline.com/erik/bass/xmaxer.html

In the example I provided, the UM18 driver is delivering 2dB more output at 50Hz vs 100Hz for the same 2 volt input. IOW, the cone is delivering more than 4x the excursion (more output + 1 octave deeper) when being fed 2V at 50Hz vs being fed the same 2V at 100Hz. The impedance profile also means that the UM18 driver is soaking up more current (and thus more power) at 100Hz.