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Years with exactly two "Friday the 13ths", starting from 1901.
4

%I #18 Feb 16 2025 08:33:14

%S 1901,1905,1906,1907,1908,1911,1912,1917,1918,1920,1922,1923,1929,

%T 1933,1934,1935,1936,1939,1940,1945,1946,1948,1950,1951,1957,1961,

%U 1962,1963,1964,1967,1968,1973,1974,1976,1978,1979,1985,1989,1990,1991,1992,1995,1996

%N Years with exactly two "Friday the 13ths", starting from 1901.

%H Reinhard Zumkeller, <a href="/A190652/b190652.txt">Table of n, a(n) for n = 1..1000</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>

%F A101312(a(n)) = 2, 1 <= A101312(n) <= 3.

%e 2004 is a term, since only Feb 13 2004 and Aug 13 2004 fell on a Friday.

%t Select[Range[1901,2020],Count[Table[{#,m,13},{m,12}],_?(DayName[#] == Friday&)] == 2&] (* _Harvey P. Dale_, Oct 02 2018 *)

%o (Haskell)

%o a190652 n = a190652_list !! (n-1)

%o a190652_list = filter ((== 2) . a101312) [1901..]

%o (Python)

%o from datetime import date

%o def ok(n): return sum(date.isoweekday(date(n, m, 13)) == 5 for m in range(1, 13)) == 2

%o print(list(filter(ok, range(1901, 2000)))) # _Michael S. Branicky_, Sep 12 2021

%Y Cf. A101312, A190651, A190653.

%K nonn,changed

%O 1,1

%A _Reinhard Zumkeller_, May 16 2011