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 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
LINKS
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.
FORMULA
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
EXAMPLE
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.
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
(Python)
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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Adam M. Kalman (mocha(AT)clarityconnect.com), Dec 22 2004
STATUS
approved