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Search: id:A131773
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| A131773 |
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Epact in Julian calendar for a year n with Golden Number (n mod 19) + 1 = A074805(n). |
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+0
1
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| 30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18, 30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18, 30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18, 30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18, 30, 11
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Periodic with period 19 as there are 19 Golden Numbers. Basis for calculating Gregorian Epact. See sections about Epacts, Golden Numbers, and discussion of the 19-year Metonic cycle (in Chapter 1) of the Calendar FAQ link. The FAQ also discusses in detail in which years the different calendars have been adopted by different countries and that there was no year 0 (unless considering, say, a "proleptic" Gregorian calendar) -- so the first term here (and in A074805) is actually for 1 BC (1 BCE) of the Julian calendar.
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LINKS
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Claus Tondering, Frequently Asked Questions about Calendars.
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FORMULA
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a(n) = (11*(A074805(n)-1)) mod 30, but replacing every 0 result with 30. See program and link.
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EXAMPLE
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a(2007)=12 as the Julian Epact for the year 2007 is (11*(2007 mod 19)) mod 30 = (11*12) mod 30 = 12. ((2007 mod 19)+1 = 12+1 = 13 = A074805(2007) is the corresponding Golden Number for 2007).
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PROGRAM
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(PARI) a(n)= JE=(11*(n%19))%30; if(JE==0, 30, JE)
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CROSSREFS
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Cf. A074805.
Sequence in context: A040876 A073401 A040875 this_sequence A091746 A040874 A033350
Adjacent sequences: A131770 A131771 A131772 this_sequence A131774 A131775 A131776
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KEYWORD
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nonn
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AUTHOR
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Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jul 14 2007
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