

A101312


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


7



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
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OFFSET

1901,1


COMMENTS

This sequence is basically periodic with period 28 [example: a(1901) = a(1929) = a(1957)], with "jumps" when it passes a nonleapyear 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, 3 of every 7 years (43%) have 1 "Friday the 13th," 3 of every 7 years (43%) have 2 of them and only 1 in 7 years (14%) has 3 of them.
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.
a(A190651(n)) = 1; a(A190652(n)) = 2; a(A190653(n)) = 3.  Reinhard Zumkeller, May 16 2011
Periodic with period 2800.  Charles R Greathouse IV, Jul 16 2012


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1901..3000
J. R. Stockton, Rektor Chr. Zeller's 1886 Paper "KalenderFormeln"
Eric Weisstein's World of Mathematics, Triskaidekaphobia
Wikipedia, Triskaidekaphobia
Chr. Zeller, KalenderFormeln, Acta Mathematica, 9 (1886), 131136.
Index entries for sequences related to calendars
Index entries for linear recurrences with constant coefficients, order 2800.


EXAMPLE

a(2004) = 2, since there were 2 "Friday the 13ths" that year: Feb 13 2004 and Aug 13 2004 each fell on a Friday.


MATHEMATICA

(*Load <<Miscellaneous`Calendar` package first*) s={}; For[n=1901, n<=2200, t=0; For[m=1, m<=12, If[DayOfWeek[{n, m, 13}]===Friday, t++ ]; m++ ]; AppendTo[s, t]; n++ ]; s


PROG

(Haskell)
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
(PARI) a(n)=[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][(n+(n\100n\40015)*12)%28+1] \\ Charles R Greathouse IV, Jul 16 2012


CROSSREFS

Cf. A157962, A188528.
Sequence in context: A320777 A069929 A304081 * A241273 A154263 A293435
Adjacent sequences: A101309 A101310 A101311 * A101313 A101314 A101315


KEYWORD

nonn,easy


AUTHOR

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


STATUS

approved



