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3 results for “gregorian”

Formulae (3)

\text{leap} \iff (y \bmod 4 = 0 \;\wedge\; y \bmod 100 \neq 0) \;\vee\; (y \bmod 400 = 0)

Tests whether a Gregorian year has 366 days.

h = \left(q + \left\lfloor\tfrac{13(m+1)}{5}\right\rfloor + K + \left\lfloor\tfrac{K}{4}\right\rfloor + \left\lfloor\tfrac{J}{4}\right\rfloor + 5J\right) \bmod 7

Finds the day of the week for any Gregorian date.

d = (19a + b) \bmod 30,\quad a = y \bmod 19

Core step of the Gregorian algorithm locating the Paschal full moon.