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A008685
Lengths of months in the Gregorian calendar.
10
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

  NODES
orte 1
see 1
Story 1