The end of daylight saving time here on the East Coast of the U.S. got me thinking more generally about the calendar year. Each solar year consists of approximately 365.24217 mean solar days. That’s pretty close to 365.25, which is why it makes sense to have an extra day every four years. However, the Gregorian calendar is a little more precise: There are 97 leap years every 400 years, averaging out to 365.2425 days per year.
Can you make a better approximation than the Gregorian calendar? Find numbers L and N (where N is less than 400) such that if every cycle of N years includes L leap years, the average number of days per year is as close as possible to 365.24217.
I used continued fractions.
$365 + \cfrac{1}{4} = 365.25$
$365 + \cfrac{1}{4 + \cfrac{1}{7}} = 365 + \dfrac{7}{29} \approx 365.24137$
$365 + \cfrac{1}{4 + \cfrac{1}{7 + \cfrac{1}{1}}} = 365 + \dfrac{8}{33} \approx 365.24242$
$365 + \cfrac{1}{4 + \cfrac{1}{7 + \cfrac{1}{1 + \cfrac{1}{2}}}} = 365 + \dfrac{23}{95} \approx 365.24211$
$365 + \cfrac{1}{4 + \cfrac{1}{7 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1}}}}} = 365 + \dfrac{31}{128} \approx 365.24219$
$365 + \cfrac{1}{4 + \cfrac{1}{7 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2}}}}}} = 365 + \dfrac{85}{351} \approx 365.24217$
Every $N=351$ years should have $L=85$ leap years.