A101312 -id:A101312 - cp's OEIS frontend

A190651 Years with exactly one "Friday the 13th", starting from 1901.

1902, 1904, 1909, 1910, 1913, 1915, 1916, 1919, 1921, 1924, 1926, 1927, 1930, 1932, 1937, 1938, 1941, 1943, 1944, 1947, 1949, 1952, 1954, 1955, 1958, 1960, 1965, 1966, 1969, 1971, 1972, 1975, 1977, 1980, 1982, 1983, 1986, 1988, 1993, 1994, 1997, 1999, 2000

Offset: 1

Reinhard Zumkeller, May 16 2011

nonn

			1902 is a term, since only Jun 13 1902 fell on a Friday.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Index entries for sequences related to calendars

Cf. A101312, A190652, A190653.

a190651 n = a190651_list !! (n-1)
a190651_list = filter ((== 1) . a101312) [1901..]

Select[Range[1901,2001],Count[Table[DayName@{#,m,13},{m,12}],Friday]==1&] (* Giorgos Kalogeropoulos, Sep 12 2021 *)

from datetime import date
def ok(n): return sum(date.isoweekday(date(n, m, 13)) == 5 for m in range(1, 13)) == 1
print(list(filter(ok, range(1901, 2001)))) # Michael S. Branicky, Sep 12 2021

A101312(a(n)) = 1, 1 <= A101312(n) <= 3.

A190652 Years with exactly two "Friday the 13ths", starting from 1901.

Original entry on oeis.org

1901, 1905, 1906, 1907, 1908, 1911, 1912, 1917, 1918, 1920, 1922, 1923, 1929, 1933, 1934, 1935, 1936, 1939, 1940, 1945, 1946, 1948, 1950, 1951, 1957, 1961, 1962, 1963, 1964, 1967, 1968, 1973, 1974, 1976, 1978, 1979, 1985, 1989, 1990, 1991, 1992, 1995, 1996

Offset: 1

Reinhard Zumkeller, May 16 2011

nonn

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Index entries for sequences related to calendars

Cf. A101312, A190651, A190653.

a190652 n = a190652_list !! (n-1)
a190652_list = filter ((== 2) . a101312) [1901..]

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

from datetime import date
def ok(n): return sum(date.isoweekday(date(n, m, 13)) == 5 for m in range(1, 13)) == 2
print(list(filter(ok, range(1901, 2000)))) # Michael S. Branicky, Sep 12 2021

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

A190653 Years with exactly three "Friday the 13ths", starting from 1901.

Original entry on oeis.org

1903, 1914, 1925, 1928, 1931, 1942, 1953, 1956, 1959, 1970, 1981, 1984, 1987, 1998, 2009, 2012, 2015, 2026, 2037, 2040, 2043, 2054, 2065, 2068, 2071, 2082, 2093, 2096, 2099, 2105, 2108, 2111, 2122, 2133, 2136, 2139, 2150, 2161, 2164, 2167, 2178, 2189, 2192

Offset: 1

Reinhard Zumkeller, May 16 2011

nonn

			2012 is a term, since only Jan 13 2012, Apr 13 2012 and Jul 13 2012 fell on Fridays.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Index entries for sequences related to calendars

Cf. A101312, A190651, A190652.

a190653 n = a190653_list !! (n-1)
a190653_list = filter ((== 3) . a101312) [1901..]

Module[{mos={#[[1]],Length[#]}&/@(Flatten[Take[#,1]&/@DateSelect[ DateRange[ {1900,1,1},{2200,12,1}],#Day==13&&#DayName== Friday&]]// Split)},Select[ mos,#[[2]]>2&][[All,1]]] (* Requires Mathematica version 12 or later *) (* Harvey P. Dale, Jan 17 2021 *)

from datetime import date
def ok(n): return sum(date.isoweekday(date(n, m, 13)) == 5 for m in range(1, 13)) == 3
print(list(filter(ok, range(1901, 2193)))) # Michael S. Branicky, Sep 12 2021

A101312(a(n)) = 3, 1 <= A101312(n) <= 3.

A157962 Table read by rows: Occurrences of Friday the 13th, T(n,1)=month (1 for January), T(n,2)=year-2000, starting year 2000.

Original entry on oeis.org

10, 0, 4, 1, 7, 1, 9, 2, 12, 2, 6, 3, 2, 4, 8, 4, 5, 5, 1, 6, 10, 6, 4, 7, 7, 7, 6, 8, 2, 9, 3, 9, 11, 9, 8, 10, 5, 11, 1, 12, 4, 12, 7, 12, 9, 13, 12, 13, 6, 14, 2, 15, 3, 15, 11, 15, 5, 16, 1, 17, 10, 17, 4, 18, 7, 18, 9, 19, 12, 19, 3, 20, 11, 20, 8, 21, 5, 22, 1, 23, 10, 23, 9, 24, 12, 24

Offset: 1

Pierre CAMI, Mar 10 2009

Uses the Gregorian calendar.

			a(39) = 1, a(40) = 12 -> Jan 2012,
a(41) = 4, a(42) = 12 -> Apr 2012,
a(43) = 7, a(44) = 12 -> Jul 2012,
a(45) = 9, a(46) = 13 -> Sep 2013.

Reinhard Zumkeller, Table of n, a(n) for n = 1..3442
J. R. Stockton, Rektor Chr. Zeller's 1886 Paper "Kalender-Formeln" [Wayback Machine link from _Felix Fröhlich_, Sep 23 2019]
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Chr. Zeller, Kalender-Formeln, Acta Mathematica, 9 (1886), 131-136.
Index entries for linear recurrences with constant coefficients, order 1376.
Index entries for sequences related to calendars

Cf. A101312, A188528.

a157962 n = a157962_list !! (n-1)
a157962_list = concat $ map (t 1 {- January -}) [0..] where
   t 13 _                       = []
   t m n | h (n+2000) m 13 == 6 = m : n : t (succ m) n
         | otherwise            = t (succ m) n
   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
-- Reinhard Zumkeller, May 17 2011

<< Calendar`; {y, m} = {2000, 1}; A157962 = {}; Do[ If[ m == 12, y = y + 1; m = 1, m = m + 1]; If[ DayOfWeek[ {y, m, 13}] == Friday, AppendTo[ A157962, {m , y - 2000}]] , {300}]; Flatten[A157962] (* Jean-François Alcover, Dec 16 2011 *)
Flatten[Table[If[DateValue[{2000+n,m,13},"DayName"]==Friday,{m,n},{}], {n,0,25},{m,12}]/.{}->Nothing] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 10 2017 *)

jo = 6 For a = 2000 To 2399 For b = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And b = 13 Then ll = ll + 1: Cells(ll, 1) = 1: Cells(ll, 2) = a - 2000 Next b If a - 4 * Int(a / 4) = 0 Then GoTo 10 For c = 1 To 28 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And c = 13 Then ll = ll + 1: Cells(ll, 1) = 2: Cells(ll, 2) = a - 2000 Next c GoTo 20 10 For c = 1 To 29 jo = jo + 1: If jo = 7
Then jo = 0 If jo = 6 And c = 13 Then ll = ll + 1: Cells(ll, 1) = 2: Cells(ll, 2) = a - 2000 Next c 20 For d = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And d = 13 Then ll = ll + 1: Cells(ll, 1) = 3: Cells(ll, 2) = a - 2000 Next d For e = 1 To 30 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And e = 13 Then ll = ll + 1: Cells(ll, 1) = 4: Cells(ll, 2) = a - 2000 Next e For f = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And f = 13 Then ll = ll + 1: Cells(ll, 1) = 5: Cells(ll, 2) = a - 2000 Next f For g = 1 To 30 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And g = 13 Then ll = ll + 1: Cells(ll, 1) = 6: Cells(ll, 2) = a - 2000 Next g For h = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And h = 13 Then ll = ll + 1: Cells(ll, 1) = 7: Cells(ll, 2) = a - 2000 Next h For i = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And i = 13
Then ll = ll + 1: Cells(ll, 1) = 8: Cells(ll, 2) = a - 2000 Next i For j = 1 To 30 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And j = 13 Then ll = ll + 1: Cells(ll, 1) = 9: Cells(ll, 2) = a - 2000 Next j For k = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And k = 13 Then ll = ll + 1: Cells(ll, 1) = 10: Cells(ll, 2) = a - 2000 Next k For l = 1 To 30 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And l = 13 Then ll = ll + 1: Cells(ll, 1) = 11: Cells(ll, 2) = a - 2000 Next l For m = 1 To 31 jo = jo + 1: If jo = 7 Then jo = 0 If jo = 6 And m = 13 Then ll = ll + 1: Cells(ll, 1) = 12: Cells(ll, 2) = a - 2000 Next m Next a End Sub

a(2n+1) = a(2n + 1377), a(2n) = a(2n + 1376) + 400. - Charles R Greathouse IV, May 17 2011

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

Original entry on oeis.org

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

Reinhard Zumkeller, May 16 2011

nonn

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

			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.

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

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

Cf. A101312, A157962.

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]

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

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

A116386 Number of calendar weeks in the year n (starting at n=0 for the year 2000).

Original entry on oeis.org

54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53

Offset: 0

Sergio Pimentel, Mar 15 2006

Since 365/7 = 52.14 > 52, every year has at least 53 weeks (although the first and / or the last calendar weeks might not be complete and belong to two different years). Only if a leap year begins in a Saturday (the last day of the calendar week), a year can have 54 different calendar weeks (being the first and last of only one day). Years with 54 calendar weeks are: 2000, 2028, 2056, 2084, 2124, 2152, etc. It happens 13 times in a 400 year cycle.

			E.g. a(0)=54 because the year 2000 had 54 calendar weeks (since Jan 01 2000 was a Saturday and Dec 31 2000 was a Sunday)

Index entries for sequences related to calendars

Cf. A060512, A061251, A003786, A090651, A101312.

cp's OEIS Frontend

A190651 Years with exactly one "Friday the 13th", starting from 1901.

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A190652 Years with exactly two "Friday the 13ths", starting from 1901.

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Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Python

Formula

A190653 Years with exactly three "Friday the 13ths", starting from 1901.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Python

Formula

A157962 Table read by rows: Occurrences of Friday the 13th, T(n,1)=month (1 for January), T(n,2)=year-2000, starting year 2000.

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

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Haskell

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Python

A116386 Number of calendar weeks in the year n (starting at n=0 for the year 2000).

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