A188528 - OEIS

%I #34 Feb 16 2025 08:33:14

%S 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,

%T 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,

%U 6,2,8,10,4,9,6,2,5,1,4,6,2,8,5,1,9

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

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

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

%H Reinhard Zumkeller, <a href="/A188528/b188528.txt">Table of n, a(n) for n = 1901..3000</a>

%H J. R. Stockton, <a href="https://web.archive.org/web/20140111181708/http://www.merlyn.demon.co.uk/zel-86px.htm">Rektor Chr. Zeller's 1886 Paper "Kalender-Formeln"</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Triskaidekaphobia.html">Triskaidekaphobia</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Triskaidekaphobia">Triskaidekaphobia</a>

%H <a href="/index/Ca#calendar">Index entries for sequences related to calendars</a>

%e Number of times all months occur on "Friday the 13th" in the past century and the current one:

%e           | Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

%e ----------+------------------------------------------------

%e 1991-2000 |  14  14   4  10  14  14   0  11  15   4   0   0

%e 2000-2100 |  14  15   3  11  14  15   0  11  14   3   0   0

%e 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.

%t << Calendar`;

%t Table[Select[ Table[ {yr,n,13},{n,12}],DayOfWeek[#]==Friday&,1][[1,2]],{yr,1901,2011}] (* _Harvey P. Dale_, Oct 26 2011 *)

%o (Haskell)

%o import Data.List (findIndex)

%o import Data.Maybe (fromJust)

%o a188528 n = succ $ fromJust $

%o   findIndex (\m -> h n m 13 == 6) [1..12] where

%o     h year month day  -- cf. Zeller reference.

%o       | month <= 2 = h  (year - 1)  (month + 12)  day

%o       | otherwise  = (day + 26 * (month + 1) `div` 10 + y + y `div` 4

%o                      + century `div` 4 - 2 * century) `mod` 7

%o         where (century, y) = divMod year 100

%o b188528 = bFileFun "A188528" a188528 1991 3000

%o -- For statistics (see example) ...

%o ff13_perMonth ys m = length $ filter (== m) (map a188528 ys)

%o century20 = map (ff13_perMonth [1901..2000]) [1..12]

%o century21 = map (ff13_perMonth [2001..2100]) [1..12]

%o (Python)

%o from datetime import date

%o def a(n):

%o     for month in range(1, 13):

%o         if date.isoweekday(date(n, month, 13)) == 5: return month

%o print([a(n) for n in range(1901, 1986)]) # _Michael S. Branicky_, Sep 06 2021

%Y Cf. A101312, A157962.

%K nonn,changed

%O 1901,1

%A _Reinhard Zumkeller_, May 16 2011