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%I #4 Nov 02 2013 23:40:25
%S 1,2,4,5,7,10,11,12,13,16,19,21,24,25,27,28,30,33,36,39,42,44,45,47,
%T 48,51,53,56,59,61,62,65,70,71,73,74,79,82,83,85,88,91,93,96,97,99,
%U 100,102,105,108,111,116,119,120,125,128,131,133,134,137,139,142,143,145,146,148
%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 Gregorian calendar.
%C In the Gregorian calendar, a non-century year is a leap year if and only if it is a multiple of 4 and a century year is a leap year if and only if it is a multiple of 400.
%C Assuming this fact, this sequence is periodic with a period of 4800.
%C This is the complement of A231006.
%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/gregorian-calendar.html">Gregorian 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(4800,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-((i-1)\1200)+((i-1)\4800)-!((i-1)%48)+!((i-1)%1200)-!((i-1)%4800)-!((i-2)%48)+!((i-2)%1200)-!((i-2)%4800))))%7);x=vector(4800,i,n[((i-1)%12)+1]+!((i-2)%48)-!((i-2)%1200)+!((i-2)%4800));for(p=0,4800,j=0;for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1],j=1;break));if(j==0,print1(p", ")))
%Y Cf. A230995-A231014.
%Y Cf. A231014 (Julian calendar).
%K nonn,easy
%O 1,2
%A _Aswini Vaidyanathan_, Nov 02 2013