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.
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
Examples
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.
Links
- 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
Programs
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Haskell
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 -
Mathematica
<< 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 *) -
VBA
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
Formula
a(2n+1) = a(2n + 1377), a(2n) = a(2n + 1376) + 400. - Charles R Greathouse IV, May 17 2011
Comments