1901,1

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

(*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

Sequence in context: A105609 A101872 A069929 this_sequence A154263 A035942 A036989

Adjacent sequences: A101309 A101310 A101311 this_sequence A101313 A101314 A101315

nonn

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