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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.
24

%I #4 Nov 02 2013 23:41:06

%S 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 47,48,49,50,51,53,56,58,59,61,62,65,66,67,70,71,73,74,76,79,82,83,84,

%U 85,87,88,90,91,93,96,97,99,100,102,104,105,107,108,111,113,116,119,120

%N 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 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 336.

%C This is the complement of A231011.

%H Time And Date, <a href="http://www.timeanddate.com/calendar/repeating-month.html">Repeating Months</a>

%H Time And Date, <a href="http://www.timeanddate.com/calendar/julian-calendar.html">Julian Calendar</a>

%o (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 Cf. A230995-A231014.

%Y Cf. A231009 (Gregorian calendar).

%K nonn,easy

%O 1,2

%A _Aswini Vaidyanathan_, Nov 02 2013