A188528 -id:A188528 - cp's OEIS frontend

A101312 Number of "Friday the 13ths" in year n (starting at 1901).

2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1

Offset: 1901

Adam M. Kalman (mocha(AT)clarityconnect.com), Dec 22 2004

This sequence is basically periodic with period 28 [example: a(1901) = a(1929) = a(1957)], with "jumps" when it passes a non-leap-year century such as 2100 [all centuries which are not multiples of 400].
At these points [for example, a(2101)], the sequence simply "jumps" to a different point in the same pattern, "dropping back" 12 entries [or equivalently, "skipping ahead" 16 entries], but still continuing with the same repeating [period 28] pattern.
Every year has at least 1 "Friday the 13th," and no year has more than 3.
On average, 171 of every 400 years (42.75%) have 1 "Friday the 13th," 170 of every 400 years (42.5%) have 2 of them and only 59 in 400 years (14.75%) have 3 of them. [Corrected by Pontus von Brömssen, Sep 09 2021]
Conjecture: The same basic repeating pattern results if we seek the number of "Sunday the 22nds" or "Wednesday the 8ths" or anything else similar, with the only difference being that the sequence starts at a different point in the repeating pattern.
Periodic with period 400. (Because the number of days in 400 years is 400*365 + 97 = 146097, which happens to be divisible by 7.) - Pontus von Brömssen, Sep 09 2021

			a(1902) = 1, since only Jun 13 1902 fell on a Friday.
a(2004) = 2, since there were 2 "Friday the 13ths" that year: Feb 13 2004 and Aug 13 2004 each fell on a Friday.
a(2012) = 3, since Jan 13 2012, Apr 13 2012, and Jul 13 2012 are all Fridays.

Reinhard Zumkeller, Table of n, a(n) for n = 1901..3000
Andy Huchala, Sage program
J. R. Stockton, Rektor Chr. Zeller's 1886 Paper "Kalender-Formeln"
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Chr. Zeller, Kalender-Formeln, Acta Mathematica, 9 (1886), 131-136.
Index entries for sequences related to calendars
Index entries for linear recurrences with constant coefficients, order 400.

Cf. A011763, A157962, A188528, A190651, A190652, A190653.

a101312 n = f 1 {- January -} where
   f 13                = 0
   f m | h n m 13 == 6 = (f $ succ m) + 1
       | otherwise     = f $ succ m
   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 16 2011

Mathematica
```
(*Load <
				
```

from datetime import date
def a(n):
    return sum(date.isoweekday(date(n, m, 13)) == 5 for m in range(1, 13))
print([a(n) for n in range(1901, 2006)]) # Michael S. Branicky, Sep 09 2021

a(A190651(n)) = 1; a(A190652(n)) = 2; a(A190653(n)) = 3. - Reinhard Zumkeller, May 16 2011
1 <= a(n) <= 3. - Michael S. Branicky, Sep 09 2021
G.f.: x^1900 * p/q with p,q given in Sage file. - Andy Huchala, Mar 06 2022

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

cp's OEIS Frontend

A101312 Number of "Friday the 13ths" in year n (starting at 1901).

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