days-between-dates Solver equation for 17b

[Albert Chan]

(10-07-2018 07:14 AM)Dieter Wrote:  Check:
g(1582,10,15) = 578041
j (1582,10, 4 ) = 578040

That's the 11 days for a calendar switch in 1752, or 13 days today.

Since Julian calendar "leap" faster, above two calendar systems must match in the past.

Above 10 days gap implied 15th century, two systems are 9 days apart.
Every 400 years to the past, two calendars are 3 days closer ...

Thus, two calendar match in the 15 - 9/3 * 4 = 3rd century A.D.

--> Trivia: From 3/1/0200 to 2/28/0300, g(date) == j(date)
--> Year 201 to 299 (99 years), Julian calendar match Gregorian.

Of course, this is just arithmetic. Back then, Gregorian calendar did not exist
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We can use above trivia to quickly find out days apart for X century:
X = 3,4,5,6, 7,8,9,10, ... => days apart = 0,1,1,2, 3,4,4,5, ...

Y = X - 3 (note: treat March 1, 100*(X-1) as start of century X)
Days apart = 3 * floor(Y/4) + floor((Y % 4) * 2/3 + 0.5)

For year 2018, Y = X - 3 = 21 - 3 = 18
Days apart = 3 * floor(18/4) + floor(11/6) = 3*4 + 1 = 13 (match above quote)

For Julian Day Number 0 (Julian date: Jan 1, 4713 B.C.)
year = 1 - 4713 = -4712
Y = X - 3 = -47 - 3 = -50
Days apart = 3 * floor(-50/4) + floor(11/6) = -39 + 1 = -38

So, JDN of 0 correspond to Gregorian date of Nov 24, 4714 B.C (shorted 7+31 days)