At this USGS very good site about rms noise in sensors, a phrase regarding third octave bands (the original topic) appeared.

The figure displays the minimum vertical seismic noise directly as rms amplitudes in a bandwidth of one-sixth decade. By coincidence, these amplitudes may also be interpreted as average peak amplitudes in a bandwidth of one-third octave (a standard bandwidth in acoustics)

One sixth decade factor is $10^{1/6}=1.467799$; the 10th factor $10/6=1.666666$, and one third octave factor $2^{1/3}=1.259921$, none of them being comparable.

It happens actually to be comparable with one tenth decade factor $10^{1/10}=1.258925$.

Note the rms and peak distictions... Why the author state as coincident both log factors? In addition, the same paragraph continues with:

An example: the minimum vertical ground noise between the periods of 10 and 20 sec is at -180 dB relative to ${\rm 1 m/s^2}$, thus ${\rm 10^{-180/20}\;m/s^2 = 1\;nm/s^2}$ average peak in one-third octave.

Which is ok, but then:

Since the bandwidth considered is a full octave, the total average peak amplitude in this band is ${\rm\sqrt{3}\;nm/s^2}$.

In here i honestly dont know why a $\sqrt 3$ factor appeared...

I think they made a mistake. $2^{1/3} \approx 1.259921$ is not close enough to $10^{1/6} \approx 1.467799$ to say that there is any coincidence. Even the half-octave factor $2^{1/2} \approx 1.414214$ is closer.
• I think the OP is trying to compare $2^{1/3}=1.259921$ and $10^{1/10}=1.258925$ which are more nearly comparable: $$\left(2^{1/3}\right)^{10} = \left(2^{10}\right)^{1/3} \approx \left(10^3\right)^{1/3} = 10\implies 2^{1/3} \approx 10^{1/10}$$ – Dilip Sarwate Apr 10 '18 at 15:18
• @DilipSarwate the source does not mention $10^{1/10}.$ – Olli Niemitalo Apr 10 '18 at 16:25
• That factor comes from that if each third of the octave of interest have 1 nm/s² "average peak", which is some sort of an amplitude quantity (similar to RMS), then they sum as $\sqrt{1 + 1 + 1}$ nm/s² average peak if the noises in the three bands are independent. – Olli Niemitalo Apr 10 '18 at 21:47
• @robertbristow-johnson That's right. The underlying connection can be described by that each $2^{10} = 10^a$ and $2^{1/a} = 10^{1/10}$ solve as $a = \frac{10\ln(2)}{\ln(10)} \approx 3.010299956 \approx 3.$ – Olli Niemitalo Apr 11 '18 at 6:49