Interesting, John; it worked fine for me.

Actually, I had to create a bit of vb-based code a while ago for calculating Easter correctly, and then ported it into a SQL function too. The function DASUN has provided gave the same results as mine for all years between 1800 and 9999, so if you want an alternative, try my version as below:

CREATE FUNCTION [dbo].[fnEasterSunday]

(

@YearVal int

)

RETURNS datetime

AS

BEGIN

DECLARE @EasterDate datetime

declare @a int,

@b-2 int,

@C int,

@d int,

@e int,

@f int,

@g int,

@h int,

@j-2 int,

@m int,

@k int,

@mth int,

@dy int,

@easter datetime

set @a = @yearval - (floor(@yearval/19) * 19)

set @b-2 = floor(@yearval/100)

set @C = @yearval - (@b * 100)

set @d = floor(@b/4)

set @e = @b-2 - (@d * 4)

set @f = floor(@c/4)

set @g = @C - (@f*4)

set @h = floor(((8 * @b-2) + 13)/25)

set @j-2 = ((19 * @a) + (@b - @d - @h) + 15) - (floor(((19 * @a) + (@b - @d - @h) + 15)/30) * 30)

set @m = floor((@a + 11 * @j-2)/319)

set @k = ((2 * @e) + (2 * @f) - @g - @j-2 + @m + 32) - (floor(((2 * @e) + (2 * @f) - @g - @j-2 + @m + 32)/7) * 7)

set @mth = floor((@j - @m + @k + 90)/25)

set @dy = (@j - @m + @k + 19 + @mth) - (floor((@j - @m + @k + 19 + @mth) /32) * 32)

set @easter = convert(datetime, str(@yearval) + '-' + str(@mth) + '-' + str(@dy), 120)

if datepart(dw, @easter) = 1

begin

select @easterdate = @easter

end

else

begin

select @easterdate = dateadd(d, 8 - datepart(dw, @easter), @easter)

end

RETURN @EasterDate

END

And you might want to have a look at this website for the full details of the calculation.

Hope this helps