cp's OEIS Frontend

A231014 Number of months after which it is not possible to have the same calendar for the entire month with the same number of days, in the Julian calendar.

%I A231014 #4 Nov 02 2013 23:41:06
%S A231014 1,2,4,5,7,10,11,12,13,16,19,21,24,25,27,28,30,33,36,39,41,42,44,45,
%T A231014 47,48,49,50,51,53,56,58,59,61,62,65,66,67,70,71,73,74,76,79,82,83,84,
%U A231014 85,87,88,90,91,93,96,97,99,100,102,104,105,107,108,111,113,116,119,120
%N A231014 Number of months after which it is not possible to have the same calendar for the entire month with the same number of days, in the Julian calendar.
%C A231014 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 A231014 Assuming this fact, this sequence is periodic with a period of 336.
%C A231014 This is the complement of A231011.
%H A231014 Time And Date, <a href="http://www.timeanddate.com/calendar/repeating-month.html">Repeating Months</a>
%H A231014 Time And Date, <a href="http://www.timeanddate.com/calendar/julian-calendar.html">Julian Calendar</a>
%o A231014 (PARI) m=[0,3,3,6,1,4,6,2,5,0,3,5];n=[31,28,31,30,31,30,31,31,30,31,30,31];y=vector(336,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-!((i-1)%48)-!((i-2)%48))))%7);x=vector(336,i,n[((i-1)%12)+1]+!((i-2)%48));for(p=0,336,j=0;for(q=0,336,if(y[(q%336)+1]==y[((q+p)%336)+1]&&x[(q%336)+1]==x[((q+p)%336)+1],j=1;break));if(j==0,print1(p", ")))
%Y A231014 Cf. A230995-A231014.
%Y A231014 Cf. A231009 (Gregorian calendar).
%K A231014 nonn,easy
%O A231014 1,2
%A A231014 _Aswini Vaidyanathan_, Nov 02 2013