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A101312
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Number of "Friday the 13ths" in year n (starting at 1901).
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7
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1901,1
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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, 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]
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REFERENCES
| Chr. Zeller, Kalender-Formeln, Acta mathematica, 9 (1886), 131-136.
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LINKS
| 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
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EXAMPLE
| a(2004) = 2, since there were 2 "Friday the 13ths" that year: Feb 13, 2004 and Aug 13, 2004 and both fell on a Friday.
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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
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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
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CROSSREFS
| Cf. A157962, A188528.
Sequence in context: A101872 A174892 A069929 * A154263 A182850 A035942
Adjacent sequences: A101309 A101310 A101311 * A101313 A101314 A101315
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KEYWORD
| nonn
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AUTHOR
| Adam M. Kalman (mocha(AT)clarityconnect.com), Dec 22 2004
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