A188528 First month in year n with a "Friday the 13th", starting from 1901.

9, 6, 2, 5, 1, 4, 9, 3, 8, 5, 1, 9, 6, 2, 8, 10, 4, 9, 6, 2, 5, 1, 4, 6, 2, 8, 5, 1, 9, 6, 2, 5, 1, 4, 9, 3, 8, 5, 1, 9, 6, 2, 8, 10, 4, 9, 6, 2, 5, 1, 4, 6, 2, 8, 5, 1, 9, 6, 2, 5, 1, 4, 9, 3, 8, 5, 1, 9, 6, 2, 8, 10, 4, 9, 6, 2, 5, 1, 4, 6, 2, 8, 5, 1, 9

Offset

1901,1

Comments

A101312(n) > 0, therefore a(n) is defined for all n.

References

Chr. Zeller, Kalender-Formeln, Acta mathematica, 9 (1886), 131-136.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1901..3000

J. R. Stockton, Rektor Chr. Zeller's 1886 Paper "Kalender-Formeln"

Eric Weisstein's World of Mathematics, Triskaidekaphobia

Wikipedia, Triskaidekaphobia

Index entries for sequences related to calendars

Example

Number of times all months occur on "Friday the 13th" in the past century and the current one:
          | Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
----------+------------------------------------------------
1991-2000 |  14  14   4  10  14  14   0  11  15   4   0   0
2000-2100 |  14  15   3  11  14  15   0  11  14   3   0   0
The table doesn't say that there are no "Friday the 13ths" in July, November, and December: just first occurrences are considered, e.g. Nov 13 2009 is on a Friday, but a(2009) = 2; Jul 13 2012 is on a Friday, but a(2012) = 1; and Dec 13 2024 is on a Friday, but a(2024) = 9.

Mathematica

{* First execute *) << Calendar`; (* then execute *) Table[Select[ Table[ {yr, n, 13}, {n, 12}], DayOfWeek[#]==Friday&, 1][[1, 2]], {yr, 1901, 2011}] (* Harvey P. Dale, Oct 26 2011 *)

Prog

(Haskell)
import Data.List (findIndex)
import Data.Maybe (fromJust)
a188528 n = succ $ fromJust $
  findIndex (\m -> h n m 13 == 6) [1..12] where
    h year month day  -- cf. Zeller reference.
      | month <= 2 = h  (year - 1)  (month + 12)  day
      | otherwise  = (day + 26 * (month + 1) `div` 10 + y + y `div` 4
                     + century `div` 4 - 2 * century) `mod` 7
        where (century, y) = divMod year 100
b188528 = bFileFun "A188528" a188528 1991 3000
-- For statistics (see example) ...
ff13_perMonth ys m = length $ filter (== m) (map a188528 ys)
century20 = map (ff13_perMonth [1901..2000]) [1..12]
century21 = map (ff13_perMonth [2001..2100]) [1..12]
(Python)
from datetime import date
def a(n):
    for month in range(1, 13):
        if date.isoweekday(date(n, month, 13)) == 5: return month
print([a(n) for n in range(1901, 1986)]) # Michael S. Branicky, Sep 06 2021

Crossrefs

Cf. A101312, A157962.

Sequence in context: A335563 A011219 A202543 * A243257 A194182 A019961

Adjacent sequences: A188525 A188526 A188527 * A188529 A188530 A188531

Keyword

nonn

Author

Reinhard Zumkeller, May 16 2011

Status

approved