2 cars start racing at t=0. When will the faster travelling at a constant speed of s1 circuits/hour say lap the slower travelling at speed s2? At time t the faster has travelled s1*t circuits and the slower has gone s2*t circuits. We require the faster to have travelled exactly 1 circuit more than the slower, at which point it is lapping. So s1*t=s2*t + 1 or t=1/(s1-s2).
In astronomy we have orbits instead of circuits and usually orbital periods (T1=1/s1 and T2=1/s2) instead of speeds and we use 1/t=s1-s2=1/T1-1/T2. This makes t look like a "harmonic mean difference" so is vaguely Pythagorean given that the harmonic mean hm given by 1/hm=1/T1+1/T2 is one of the Pythagorean means.
Consider the big hand of the clock with s1=1 circuit per hour, and the little hand going at 1/12 circuit per hour. The synodic period is 1/(1-1/12)=12/11 hours. Or with T1=1 and T2=12 , 1/t=1-1/12=11/12 so t=12/11 again.
Consider the sun and moon as seen from earth - the moon orbits the earth in 27.322 days (T1) and the sun appears to do so in 365.242 days(T2). The synodic period of the moon is therefore the 1/(1/27.322-1/365.242) = 29.531 days.
With 1/S=1/T1-1/T2 we have S=T1*T2/(T2-T1) = T1*(1 + T1/(T2-T1)), showing algebraically that for T2 much larger than T1, S is slightly more than T1. The easiest way to consider the formula is really just 1=S(1/T1-1/T2) as the RHS is the distance between the two after time S - i.e. 1 orbit, so body 1 is "lapping" body 2.
For earth there is only a 4 minute difference between the sidereal (23.93 hours) and synodic (24 hours) day. 1/(1/23.93 - 1/(365.25*24))=24.00. The tidal or lunar day is 24 hours 50 minutes or 24.83 hours =24/(1-1/29.53)=24.84, near enough. Or maybe 1/(1/23.93 - 1/(27.322*24))=24.836, i.e. 50.2 minutes, probably more logical.
Chereau's explanation of the "philosophic cross" ("La croix philosophique", 1806) gives a solar year of 365 days 48 min 48 sec, missing out the 5 hours, which would make it an accurate 365.24222 decimal days. Chereau reckons there are 13 lunar months in the year. 365.24222/13=28.09555 days = 28 days 2 hours 17 mintes and 36 seconds - remarkably close to his figure which has 56 seconds, possibly a typo. Not that any lunar month is 28.095 days long, but interesting.
Note the Metonic cycle of 19 years or 228 months means 235 lunations occur. 19*365.2422=6939.602 and 235*29.531=6939.785, only 0.183 days or 4.4 hours longer. We need 19.0005 years then to match the 235 lunar cycles, or 19 years +4-5 hours. 163 years and 2016 lunations is a better approximation - 59534.479 v 59534.496 but even then we need another 0.017 days or about 20 minutes extra. 334 years /4131 lunations is better, the best I could find - based on 12 +123/334 lunations per year - that's 12.3682668 compared with 365.2421988/29.53058868=12.3682635. 11 years is pretty good too 11*365.2422=4017.6642 , 136 * 29.531=4016.216, so 11 years is long by 1.448 days. 8*365.2422=2921.9376, 99*29.531=2923.569, so we need 8 years and 1.63 days. 5 years/ 62 lunations , 3 years/37 lunations, and even 2 years/25 lunations also not bad, but that 19 years is special! It's a famous patteren for eclipses and also a cycle for tides to repeat at a given location.