1,1

The starting point for the sequence is explained by the fact that the Gregorian calendar was only introduced in 1582.

The complete Easter cycle lasts 5700000 years. In this cycle, Mar 25 occurs 110200 times and Apr 25 occurs 42000 times for a total of 152200 times. This reduces to 761 occurrences every 28500 years (~2.67%). - Hans Havermann (pxp(AT)rogers.com), Jan 27 2008

Author?, Calculation of Easter.

M. Montes, Frequency of the Date of Easter over one complete Gregorian Easter Cycle.

The formula is based on the algorithm of Oudin (1940) taken from the link.

(* first do *) Needs["Miscellaneous`Calendar`"] (* then *) Select[ Range[1582, 2941], EasterSunday[ # ][[3]] == 25 &] (from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 06 2005)

(PARI) edate(yr1, yr2, day) = { local(flag=1, d, y, y2, ct, dt); for(d=day, day, ct=0; for(y=yr1, yr2, dt=oudin(y); if(eval(mid(dt, 4, 2))==d, if(flag, y2=y; flag=0); ct++; \ print(ct" "dt" "y-y2); print1(y", "); if(y2<>y, y2=y); ); ); \ print1(ct", "); ) } oudin(y) = \This is based on the algorithm of Oudin (1940) { local(c, n, k, i1, i2, i3, a1, a2, m, d, l, dt, dat=""); c=floor(y/100); n=y-19*floor(y/19); k=floor((c-17)/25); i1=c-floor(c/4)-floor((c-k)/3)+19*n+15; i2=i1-30*floor(i1/30); i3=i2-floor(i2/28)*(1-floor(i2/28)*floor(29/(i2+1))*floor((21-n)/11)); a1=y+floor(y/4)+i3+2-c+floor(c/4); a2=a1-7*floor(a1/7); l=i3-a2; m=3+floor((l+40)/44); d=l+28-31*floor(m/4); dat = concat(dat, right(Str(m+100), 2)); dat = concat(dat, " "); dat = concat(dat, right(Str(d+100), 2)); dat = concat(dat, " "); dat = concat(dat, Str(y)); return(dat); }

Cf. A104034.

Sequence in context: A154075 A054809 A163273 this_sequence A054810 A068744 A054811

Adjacent sequences: A104016 A104017 A104018 this_sequence A104020 A104021 A104022

nonn

Cino Hilliard (hillcino368(AT)gmail.com), Mar 31 2005

More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 06 2005.