%I #8 Jun 04 2024 10:51:55
%S 1,2,3,4,5,7,8,9,10,12,13,14,15,16,18,19,20,21,23,24,25,26,27,29,30,
%T 31,32,33,35,36,37,38,40,41,42,43,44,46,47,48,49,51,52,53,54,55,57,58,
%U 59,60,61,63,64,65,66,68,69,70,71,72,74,75,76,77,79,80,81,82,83,85,86,87
%N Number of years after which it is not possible to have the same calendar for the entire year, in the Julian calendar.
%C In the Julian calendar, a year is a leap year if and only if it is a multiple of 4 and all century years are leap years.
%C Assuming this fact, this sequence is periodic with a period of 28.
%C This is the complement of A231001.
%H Time And Date, <a href="http://www.timeanddate.com/calendar/repeating.html">Repeating Calendar</a>
%H Time And Date, <a href="http://www.timeanddate.com/calendar/julian-calendar.html">Julian Calendar</a>
%H <a href="/index/Rec#order_24">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).
%F From _Chai Wah Wu_, Jun 04 2024: (Start)
%F a(n) = a(n-1) + a(n-23) - a(n-24) for n > 24.
%F G.f.: x*(x^23 + x^22 + x^21 + x^20 + x^19 + 2*x^18 + x^17 + x^16 + x^15 + 2*x^14 + x^13 + x^12 + x^11 + x^10 + 2*x^9 + x^8 + x^7 + x^6 + 2*x^5 + x^4 + x^3 + x^2 + x + 1)/(x^24 - x^23 - x + 1). (End)
%o (PARI) for(i=0,420,j=0;for(y=0,420,if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7)&&((5*(y\4)+(y%4)-!(y%4))%7)==((5*((y+i)\4)+((y+i)%4)-!((y+i)%4))%7),j=1));if(j==0,print1(i", ")))
%Y Cf. A230995-A231014.
%Y Cf. A230999 (Gregorian calendar).
%K nonn,easy
%O 1,2
%A _Aswini Vaidyanathan_, Nov 02 2013