|
|
A116369
|
|
Day of the week corresponding to Jan 01 of a given year (n=0 for the year 2000).
|
|
7
|
|
|
7, 2, 3, 4, 5, 7, 1, 2, 3, 5, 6, 7, 1, 3, 4, 5, 6, 1, 2, 3, 4, 6, 7, 1, 2, 4, 5, 6, 7, 2, 3, 4, 5, 7, 1, 2, 3, 5, 6, 7, 1, 3, 4, 5, 6, 1, 2, 3, 4, 6, 7, 1, 2, 4, 5, 6, 7, 2, 3, 4, 5, 7, 1, 2, 3, 5, 6, 7, 1, 3, 4, 5, 6, 1, 2, 3, 4, 6, 7, 1, 2, 4, 5, 6, 7, 2, 3, 4, 5, 7, 1, 2, 3, 5, 6, 7, 1, 3, 4, 5, 6, 7, 1, 2, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
The number of days in the 400 year cycle of the Gregorian calendar is 365 * 400 + 100 (leap year every 4 years) - 4 (no leap year in centuries) + 1 (leap year every 400 years) = 146097 days. Since 146097 is (coincidentally) divisible by 7 (7 * 20871), the cycle repeats exactly every 400 years. As a consequence, the probability of Jan 01 of a given year being any given weekday is not 1/7. Sunday, Tuesday and Friday have the highest probability (14.50%); Wednesday and Thursday 14.25%; Monday and Saturday 14.00%.
|
|
REFERENCES
|
N. Dershowitz and E. M. Reingold, Calendrical Calculations, Cambridge University Press, 1997.
|
|
LINKS
|
|
|
FORMULA
|
1 = Sunday, 2 = Monday, 3 = Tuesday, 4 = Wednesday, 5 = Thursday, 6 = Friday and 7 = Saturday. a(n+400) = a(n) since the cycle repeats every 400 years.
|
|
EXAMPLE
|
a(6) = 1 because Jan 01 2006 was a Sunday.
|
|
MATHEMATICA
|
(* first do *) Needs["Miscellaneous`Calendar`"] (* then *) Table[DayOfWeek[{2000 + n, 1, 1}], {n, 0, 104}] /. {Sunday -> 1, Monday -> 2, Tuesday -> 3, Wednesday -> 4, Thursday -> 5, Friday -> 6, Saturday -> 7} (* Robert G. Wilson v, Apr 04 2006 *)
|
|
PROG
|
(Python)
from datetime import date
def a(n): return (date(2000+n, 1, 1).isoweekday())%7 + 1
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|