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%I #30 Dec 25 2022 14:52:43
%S 640,672,705,738,770,803,835,868,900,933,966,998,1031,1063,1096,1129,
%T 1161,1194,1226,1259,1291,1324,1357,1389,1422,1454,1487,1520,1552,
%U 1585,1617,1650,1682,1715,1748,1780,1813,1845,1878,1911,1943,1976,2008,2041
%N Gregorian years containing two Islamic New Year Days.
%C Gregorian years containing "blue" Islamic New Year Days. The boundary of a calendrical period is hereby called "blue" w.r.t. a similarly named period in another calendar when the shorter one does not contain the boundaries of the longer one. Gregorian calendar prior to 1582 is proleptic, extrapolated according to the calculator in the links.
%C The ratio of Gregorian to Islamic Year is 365.2425/354.36666... = 438291/425240. The interesting approximating continuous fractions are 403/391, 638/619, 1041/1010 and a very long sequence of (1041+403*n)/(1010+391*n), ending with 7489/7266, so the 403/391 pattern will remain for thousands of years.
%C Because 4382910 Islamic years = 1553157207 days = 4252400 Gregorian years, each cycle contains 130510 blue Gregorian years and therefore a(n + 130510*k) = a(n) + 425400*k for k >= 0. - _Robert B Fowler_, Dec 06 2022
%C Note that the unique "crosspoint" year of the two calendars (20874) is also a blue year, with the first Islamic New Year falling on January 3, i.e., 01/01/20874 (Islamic) = 01/03/20874 (Gregorian). - _Robert B Fowler_, Dec 06 2022
%H John Walker, <a href="http://www.fourmilab.ch/documents/calendar/">Calendar Converter</a> (Warning: as of 12/7/2022, the first leap years in the 30-year Islamic cycle are listed as 1, 5, 7, ..., but are actually 2, 5, 7, ...; the calendar calculations, however, are correct. - _Robert B Fowler_, Dec 08 2022)
%F I(i) = chronological Julian Day Number (JDN) of New Year day of Islamic year i
%F = 1948086 + floor((10631*i+3)/30) (see A350539)
%F G(g) = JDN of New Year day of Gregorian year g+1 (not year g)
%F = 1721426 + floor(1461*g/4) - floor(g/100) + floor(g/400)
%F initialize values: i=1, g=621, n=0
%F repeat forever:
%F increment: i = i+1, g = g+1
%F IF I(i) < G(g) THEN n = n+1, a(n) = g, i = i+1 - _Robert B Fowler_, Dec 06 2022
%e 1396-1-1 A.H. = 1976-01-03 C.E.
%e 1397-1-1 A.H. = 1976-12-23 C.E. therefore 1976 is listed. (Corrected by _Robert B Fowler_, Mar 03 2022)
%Y Cf. A350539.
%K nonn
%O 1,1
%A _Leonid Broukhis_, Sep 15 2004
%E New name by _Robert B Fowler_, Mar 03 2022
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