A231009 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.

1, 2, 4, 5, 7, 10, 11, 12, 13, 16, 19, 21, 24, 25, 27, 28, 30, 33, 36, 39, 42, 44, 45, 47, 48, 51, 53, 56, 59, 61, 62, 65, 70, 71, 73, 74, 79, 82, 83, 85, 88, 91, 93, 96, 97, 99, 100, 102, 105, 108, 111, 116, 119, 120, 125, 128, 131, 133, 134, 137, 139, 142, 143, 145, 146, 148

Offset: 1

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

• nonn
• easy

Comments

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.

Assuming this fact, this sequence is periodic with a period of 4800.

This is the complement of A231006.

Links

• Time And Date, Repeating Months
• Time And Date, Gregorian Calendar

Crossrefs

Cf. A230995-A231014.

Cf. A231014 (Julian calendar).

Programs

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", ")))