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Leap years having 53 Fridays and Saturdays.
6

%I #17 Jul 21 2024 20:14:47

%S 1588,1616,1644,1672,1712,1740,1768,1796,1808,1836,1864,1892,1904,

%T 1932,1960,1988,2016,2044,2072,2112,2140,2168,2196,2208,2236,2264,

%U 2292,2304,2332,2360,2388,2416,2444,2472,2512,2540,2568,2596,2608,2636,2664,2692,2704,2732,2760,2788

%N Leap years having 53 Fridays and Saturdays.

%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 15 occurrences in 400 years.

%C Months having Friday the 13th: May.

%C February 29th falls on Monday.

%C 366 day leap year: 52 Sundays, 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 Thursdays, 53 Fridays, 53 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=2016">2016</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_16">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 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] == Friday && DaysBetween[{#, 1, 1}, {# + 1, 1, 1}, Calendar -> Gregorian] == 366 &, 50] (* _T. D. Noe_, Apr 22 2013 *)

%t ly53Q[y_]:=LeapYearQ[{y}]&&Count[DayRange[{y,1,1},{y,12,31}],_?(DayName[ #]==Friday&)]==53&&Count[DayRange[{y,1,1},{y,12,31}],_?(DayName[#] == Saturday&)] ==53; Select[Range[1570,2800],ly53Q] (* _Harvey P. Dale_, Oct 14 2021, v9 or later *)

%Y Cf. A224945, A224946, A224947, A224948, A224949, A224951.

%K nonn

%O 1,1

%A _Aswini Vaidyanathan_, Apr 21 2013