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A231001 Number of years after which an entire year can have the same calendar, in the Julian calendar.

%I A231001 #11 Jun 19 2023 11:57:10
%S A231001 0,6,11,17,22,28,34,39,45,50,56,62,67,73,78,84,90,95,101,106,112,118,
%T A231001 123,129,134,140,146,151,157,162,168,174,179,185,190,196,202,207,213,
%U A231001 218,224,230,235,241,246,252,258,263,269,274,280,286,291,297,302,308,314,319,325
%N A231001 Number of years after which an entire year can have the same calendar, in the Julian calendar.
%C A231001 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 A231001 Assuming this fact, this sequence is periodic with a period of 28.
%C A231001 This is a subsequence of A231000.
%H A231001 Colin Barker, <a href="/A231001/b231001.txt">Table of n, a(n) for n = 0..1000</a>
%H A231001 Time And Date, <a href="http://www.timeanddate.com/calendar/repeating.html">Repeating Calendar</a>
%H A231001 Time And Date, <a href="http://www.timeanddate.com/calendar/julian-calendar.html">Julian Calendar</a>
%H A231001 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F A231001 From _Colin Barker_, Oct 17 2019: (Start)
%F A231001 G.f.: x*(2 + 3*x + 2*x^2)*(3 - 2*x + 3*x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
%F A231001 a(n) = a(n-1) + a(n-5) - a(n-6) for n>5.
%F A231001 (End)
%t A231001 LinearRecurrence[{1,0,0,0,1,-1},{0,6,11,17,22,28},60] (* _Harvey P. Dale_, Jun 19 2023 *)
%o A231001 (PARI) for(i=0,420,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),print1(i", ");break)))
%o A231001 (PARI) concat(0, Vec(x*(2 + 3*x + 2*x^2)*(3 - 2*x + 3*x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^40))) \\ _Colin Barker_, Oct 17 2019
%Y A231001 Cf. A230995-A231014.
%Y A231001 Cf. A230996 (Gregorian calendar).
%K A231001 nonn,easy
%O A231001 0,2
%A A231001 _Aswini Vaidyanathan_, Nov 02 2013