cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A097105 Gregorian years containing two Islamic New Year Days.

Original entry on oeis.org

640, 672, 705, 738, 770, 803, 835, 868, 900, 933, 966, 998, 1031, 1063, 1096, 1129, 1161, 1194, 1226, 1259, 1291, 1324, 1357, 1389, 1422, 1454, 1487, 1520, 1552, 1585, 1617, 1650, 1682, 1715, 1748, 1780, 1813, 1845, 1878, 1911, 1943, 1976, 2008, 2041
Offset: 1

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Author

Leonid Broukhis, Sep 15 2004

Keywords

Comments

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.
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.
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
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

Examples

			1396-1-1 A.H. = 1976-01-03 C.E.
1397-1-1 A.H. = 1976-12-23 C.E. therefore 1976 is listed. (Corrected by _Robert B Fowler_, Mar 03 2022)
		

Crossrefs

Cf. A350539.

Formula

I(i) = chronological Julian Day Number (JDN) of New Year day of Islamic year i
= 1948086 + floor((10631*i+3)/30) (see A350539)
G(g) = JDN of New Year day of Gregorian year g+1 (not year g)
= 1721426 + floor(1461*g/4) - floor(g/100) + floor(g/400)
initialize values: i=1, g=621, n=0
repeat forever:
increment: i = i+1, g = g+1
IF I(i) < G(g) THEN n = n+1, a(n) = g, i = i+1 - Robert B Fowler, Dec 06 2022

Extensions

New name by Robert B Fowler, Mar 03 2022