A193910 Leap centuries in the revised Julian calendar.
2, 6, 11, 15, 20, 24, 29, 33, 38, 42, 47, 51, 56, 60, 65, 69, 74, 78, 83, 87, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 141, 146, 150, 155, 159, 164, 168, 173, 177, 182, 186, 191, 195, 200, 204, 209, 213, 218, 222, 227, 231, 236, 240, 245, 249
Offset: 1
Examples
20 mod 9 is 2; 2000 was a leap year in the revised Julian calendar. 24 mod 9 is 6; 2400 and 2000 also happen to be Gregorian leap years. 28 is the first integer greater than 16 only contained in A008586. 29 is the first integer greater than 16 not contained in A008586.
Links
- M. Milankovitch, Das Ende des julianischen Kalenders und der neue Kalender der orientalischen Kirchen, Astronomische Nachrichten, volume 220 (1924), 379-384.
- Claus Tøndering, Frequently Asked Questions about Calendars, Don't the Greeks do it differently?
- Wikipedia, Revised Julian calendar
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Crossrefs
Programs
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Mathematica
Table[1/4 (18 m - (-1)^m - 11), {m, 56}] (* Farideh Firoozbakht, Oct 08 2014 *) -
PARI
a(n)=(9*n-5)\2 \\ Charles R Greathouse IV, Aug 23 2011
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Rexx
do C = 0 to 250; J = C // 9; if J = 2 | J = 6 then say C; end C
Formula
a(n) = a(n-2) + 9. - Charles R Greathouse IV, Aug 09 2011
a(n) = 2 or 6 (mod 9).
For all positive integers n, a(n) = (1/4)*(18*n-17*(-1)^n-11), which implies a(2*n-1) = 9*n-3 and a(2*n) = 9*n-7. - Farideh Firoozbakht, Oct 08 2014
G.f.: x*(2 + 4*x + 3*x^2)/((1 + x)*(1 - x)^2). - Philippe Deléham, Nov 30 2016
Comments