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A008685
Lengths of months in the Gregorian calendar.
9
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31
OFFSET
1,1
COMMENTS
To make the definition unambiguous, consider these as starting from January 2001 (or, equivalently, AD 1 in the proleptic Gregorian calendar). [Charles R Greathouse IV, Sep 28 2011]
FORMULA
a(n) = a(n-4800). [Charles R Greathouse IV, Sep 28 2011]
MATHEMATICA
year1 = 2001; year2 = 2100;
PreviousDate[#, "Day"][[1, 3]]& /@ Flatten[Table[{#, m, 1}, {m, Append[ Range[2, 12], 1]}]& /@ Range[year1, year2], 1] (* Jean-François Alcover, Aug 01 2018 *)
PROG
(PARI) v=[]; for(n=1, 50, v=concat(v, [31, if(n%4||(n%100==0&&n%400), 28, 29), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31])); v
(Haskell)
a008685 n = a008685_list !! (n-1)
a008685_list = concatMap t [1..] where
t y = [31, 28 + leap, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
where leap = if mod y 4 == 0 &&
(mod y 100 > 0 || mod y 400 == 0) then 1 else 0
-- Reinhard Zumkeller, Jun 19 2015
(Python)
def a(n):
y, m = 1 + (n-1)//12, (n-1)%12
leap = int((y%4 == 0 and y%100 != 0) or y%400 == 0)
return [31, 28+leap, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31][m]
print([a(n) for n in range(1, 64)]) # Michael S. Branicky, Feb 04 2024
CROSSREFS
Cf. A061251.
Sequence in context: A361315 A291471 A276992 * A162156 A141529 A022987
KEYWORD
nonn,easy
AUTHOR
STATUS
approved