%I #14 May 21 2021 13:46:06
%S 5,23,33,51,61,79,89,107,117,135,145,163,173,191,201,219,229,247,257,
%T 275,285,303,313,331,341,359,369,387,397,415,425,443,453,471,481,499,
%U 509,527,537,555,565,583,593,611,621,639,649,667,677,695,705,723,733,751,761,779,789
%N Number of years after which it is possible to have a date falling on same day of the week, but the entire year does not have the same calendar, 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 These are the terms of A231000 not in A231001.
%C The statement about the period is misleading: this is the sequence of (positive) numbers congruent to 5 or -5 (mod 28). It is strictly increasing, not periodic; the sequence a(n) - 28*floor(n/2) is 2-periodic. - _M. F. Hasler_, Apr 14 2015
%H Colin Barker, <a href="/A231002/b231002.txt">Table of n, a(n) for n = 1..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_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n+1) = a(n-1)+28, for all n > 1. - _M. F. Hasler_, Apr 14 2015
%F a(2n) = 28n-5 (n>0), a(2n+1) = 28n+5 (n>=0), a(n) = 28*floor(n/2)-5*(-1)^n. - _M. F. Hasler_, Apr 14 2015
%F From _Colin Barker_, Oct 15 2019: (Start)
%F G.f.: x*(5 + 18*x + 5*x^2) / ((1 - x)^2*(1 + x)).
%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
%F a(n) = -7 + 2*(-1)^n + 14*n.
%F (End)
%t LinearRecurrence[{1,1,-1},{5,23,33},70] (* _Harvey P. Dale_, May 21 2021 *)
%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),j=1;break));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=2;break));if(j==1,print1(i", ")))
%o (PARI) A231002(n) = n\2*28-5*(-1)^n \\ _M. F. Hasler_, Apr 14 2015
%o (PARI) Vec(x*(5 + 18*x + 5*x^2) / ((1 - x)^2*(1 + x)) + O(x^50)) \\ _Colin Barker_, Oct 15 2019
%Y Cf. A230995-A231014.
%Y Cf. A230997 (Gregorian calendar).
%K nonn,easy
%O 1,1
%A _Aswini Vaidyanathan_, Nov 02 2013