A350539 Chronological Julian day number of the first day (Muharram 1) of Tabular Islamic year n.
1948440, 1948794, 1949149, 1949503, 1949857, 1950212, 1950566, 1950921, 1951275, 1951629, 1951984, 1952338, 1952692, 1953047, 1953401, 1953755, 1954110, 1954464, 1954819, 1955173, 1955527, 1955882, 1956236, 1956590, 1956945, 1957299, 1957654, 1958008, 1958362, 1958717, 1959071, 1959425
Offset: 1
Examples
a(1) = floor((1*10631+3)/30) + 1948086 = 1948440 (JDN). Year 1 has a(2) - a(1) = 354 days (a regular year). Year 1 began on weekday (a(1) mod 7) = 4 (Friday). Year 2 has a(3) - a(2) = 355 days (a leap year).
References
- Jean Meeus, Astronomical Algorithms, Willmann-Bell, Richmond, Virginia. Second edition, 1998, chapter 9, pages 73-76.
- Edward M. Reingold and Nachum Dershowitz, Calendrical Calculations, Cambridge University, UK. 1st edition, 1997, chapter 6 and appendix B8. 4th edition, 2018. Chapter 7 and Appendix D7.
- Edward Graham Richards, Mapping Time, Oxford University, London, 1998. Chapter 15, pages 231-235, 311, 323-324.
- Paul Kenneth Seidelmann and Leroy Elsworth Doggett, Explanatory Supplement to the Astronomical Almanac, Mill Valley, 1992. Pages 589-591.
Links
- Robert Harry van Gent, Islamic Calendar Converter, Mathematical Institute, Utrecht University, December 2019. An excellent source. The "Origin" section clearly describes how the various leap year methods are constructed; the "Islamic calendar converter" section converts Gregorian/Julian dates simultaneously into eight Islamic dates (four leap year methods in two epochs); the "Other online converters" section indicates which of the eight possible Islamic dates are used by ten online calendar converters, most of which do not divulge their method.
- Dieter Koch & Alois Treindl, Astrodienst Planetary Positions for AD 622 July (page 272).
- E. G. Richards, The identification of days, Figure 22.1 in: Mapping Time: The Calendar and Its History.
- P. K. Seidelmann, The Islamic Calendar, Explanatory Supplement 1992.
- Wikipedia, Islamic calendar.
- Wikipedia, Tabular Islamic calendar.
- Index entries for sequences related to calendars
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).
Programs
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Mathematica
a[n_] := Floor[(n*10631 + 3)/30 + 1948086]; Array[a, 32] (* Amiram Eldar, Jan 04 2022 *) IslamicNewYear[n_] := Module[{}, date := DateObject[{n, 1, 1, 12}, CalendarType -> "Islamic", TimeZone -> "Europe/London"]; jl := CalendarConvert[date, "Julian"]; jd := JulianDate[jl]; MixedFractionParts[jd][[1]] ]; Table[IslamicNewYear[n], {n, 1, 32}] (* Peter Luschny, Feb 13 2022 *)
Formula
a(n) = floor((n*10631+c)/30) + 1948086.
c = 3 is used here; for other calendar methods, see comments section.
The epoch date of July 16 is assumed; for epoch July 15, subtract one from a(n).
Number of days in Islamic year n = a(n+1) - a(n).
Day of week of first day in year n = (a(n) mod 7) = 0 (Monday) to 6 (Sunday).
Julian day number of general Islamic date y,m,d = floor((y*10631+c)/30) + floor(m*59/2) + d + 1948056. Note that this single equation defines the entire Tabular Islamic calendar (for the four tabular methods mentioned in the comments).
Comments