cp's OEIS Frontend

A097105 Gregorian years containing two Islamic New Year Days.

%I A097105 #30 Dec 25 2022 14:52:43
%S A097105 640,672,705,738,770,803,835,868,900,933,966,998,1031,1063,1096,1129,
%T A097105 1161,1194,1226,1259,1291,1324,1357,1389,1422,1454,1487,1520,1552,
%U A097105 1585,1617,1650,1682,1715,1748,1780,1813,1845,1878,1911,1943,1976,2008,2041
%N A097105 Gregorian years containing two Islamic New Year Days.
%C A097105 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 A097105 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 A097105 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 A097105 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 A097105 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 A097105 I(i) = chronological Julian Day Number (JDN) of New Year day of Islamic year i
%F A097105      = 1948086 + floor((10631*i+3)/30)     (see A350539)
%F A097105 G(g) = JDN of New Year day of Gregorian year g+1   (not year g)
%F A097105      = 1721426 + floor(1461*g/4) - floor(g/100) + floor(g/400)
%F A097105 initialize values: i=1, g=621, n=0
%F A097105 repeat forever:
%F A097105      increment: i = i+1, g = g+1
%F A097105      IF I(i) < G(g) THEN n = n+1, a(n) = g, i = i+1 - _Robert B Fowler_, Dec 06 2022
%e A097105 1396-1-1 A.H. = 1976-01-03 C.E.
%e A097105 1397-1-1 A.H. = 1976-12-23 C.E. therefore 1976 is listed. (Corrected by _Robert B Fowler_, Mar 03 2022)
%Y A097105 Cf. A350539.
%K A097105 nonn
%O A097105 1,1
%A A097105 _Leonid Broukhis_, Sep 15 2004
%E A097105 New name by _Robert B Fowler_, Mar 03 2022