_Roughly_ stated, 'day 0' (and there is one, which is an issue) is 31 December, 1967, because that allowed 'internal dates' to be relatively small during 'early implementation time', when memory was expensive. Quoth Dick.

It is also the case that 'day 1' is the first day of a four-year cycle. Each four year cycle has 1461 days, so the first thing which is done is to divide the internal data by 1461. That will leave a quotient which is the number of four-year periods, so add 4 times the quotient to 1968, to get the first approximation of the date. Then, if the date is greater than 366, subtract 1 and divide by 365 or so. Add that quotient to the year. Now, 'what's left' is the day of the year. Look in a look-up table for the month, being aware that 'things are different if this is a leap year'.