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Leap years having 53 Thursdays and Fridays.
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%I #15 Jul 21 2024 20:13:20

%S 1604,1632,1660,1688,1728,1756,1784,1824,1852,1880,1920,1948,1976,

%T 2004,2032,2060,2088,2128,2156,2184,2224,2252,2280,2320,2348,2376,

%U 2404,2432,2460,2488,2528,2556,2584,2624,2652,2680,2720,2748,2776

%N Leap years having 53 Thursdays and Fridays.

%C Gregorian calendar repeats after every 400 years because number of days in 400 years is 146097 which is a multiple of 7.

%C Non-century years are leap years if and only if they are multiples of 4 while century years are leap years if and only if they are multiples of 400.

%C 13 occurrences in 400 years.

%C Months having Friday the 13th: February and August.

%C February 29th falls on Sunday.

%C 366 day leap year: 52 Sundays, 52 Mondays, 52 Tuesdays, 52 Wednesdays, 53 Thursdays, 53 Fridays, 52 Saturdays.

%H Time and Date, <a href="http://www.timeanddate.com/calendar/gregorian-calendar.html">The Gregorian calendar</a>

%H Time and Date, <a href="http://www.timeanddate.com/calendar/?year=2004">2004</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Gregorian_calendar">Gregorian calendar</a>

%H <a href="/index/Ca#calendar">Index entries for sequences related to calendars</a>

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).

%t Needs["Calendar`"]; Select[Range[1583, 2800], DayOfWeek[{#, 1, 1}, Calendar -> Gregorian] == Thursday && DaysBetween[{#, 1, 1}, {# + 1, 1, 1}, Calendar -> Gregorian] == 366 &, 50] (* _T. D. Noe_, Apr 22 2013 *)

%t Select[Range[1583,3000],LeapYearQ[{#}]&&DayName[{#,1,1}]==Thursday&] (* _Ray Chandler_, Jul 26 2023, v9 or later, after _Harvey P.Dale_ at A224945 *)

%Y Cf. A224945, A224946, A224947, A224948, A224950, A224951.

%K nonn

%O 1,1

%A _Aswini Vaidyanathan_, Apr 21 2013