OFFSET
0,1
COMMENTS
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.
LINKS
Claus Tondering, Frequently Asked Questions about Calendars.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
FORMULA
a(n) = (11*(A074805(n)-1)) mod 30, but replacing every 0 result with 30. See program and link.
G.f.: (18*x^18 +7*x^17 +26*x^16 +15*x^15 +4*x^14 +23*x^13 +12*x^12 +x^11 +20*x^10 +9*x^9 +28*x^8 +17*x^7 +6*x^6 +25*x^5 +14*x^4 +3*x^3 +22*x^2 +11*x +30)/(-x^19 +1). - Colin Barker, Jul 18 2013
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18}, 78] (* Ray Chandler, Aug 27 2015 *)
PadRight[{}, 120, {30, 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, 1, 12, 23, 4, 15, 26, 7, 18}] (* Harvey P. Dale, Sep 14 2019 *)
PROG
(PARI) a(n)= JE=(11*(n%19))%30; if(JE==0, 30, JE)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rick L. Shepherd, Jul 14 2007
STATUS
approved