%I #11 Sep 08 2022 08:46:06
%S 0,5,6,11,17,22,23,28,33,34,39,45,50,51,56,61,62,67,73,78,79,84,89,90,
%T 95,101,106,107,112,117,118,123,129,134,135,140,145,146,151,157,162,
%U 163,168,173,174,179,185,190,191,196,201,202,207,213,218,219,224,229,230,235
%N Number of years after which a date can fall on the same day of the week, 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.
%H Colin Barker, <a href="/A231000/b231000.txt">Table of n, a(n) for n = 0..1000</a>
%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_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,1,-1).
%F From _Colin Barker_, Oct 17 2019: (Start)
%F G.f.: x*(1 - x + x^2)*(5 + 6*x + 6*x^2 + 6*x^3 + 5*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).
%F a(n) = a(n-1) + a(n-7) - a(n-8) for n>7.
%F (End)
%o (PARI) for(i=0,420,for(y=0,420,if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7),print1(i", ");break)))
%o (PARI) concat(0, Vec(x*(1 - x + x^2)*(5 + 6*x + 6*x^2 + 6*x^3 + 5*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)) + O(x^40))) \\ _Colin Barker_, Oct 17 2019
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( x*(1-x+x^2)*(5+6*x+6*x^2+6*x^3+5*x^4)/((1-x)^2*(1+x+x^2+x^3+x^4+x^5+x^6)) )); // _Marius A. Burtea_, Oct 17 2019
%Y Cf. A230995-A231014.
%Y Cf. A230995 (Gregorian calendar).
%K nonn,easy
%O 0,2
%A _Aswini Vaidyanathan_, Nov 02 2013