A231006 Number of months after which a date can fall on the same day of the week, and the two months can have the same number of days, in the Gregorian calendar.

0, 3, 6, 8, 9, 14, 15, 17, 18, 20, 22, 23, 26, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 46, 49, 50, 52, 54, 55, 57, 58, 60, 63, 64, 66, 67, 68, 69, 72, 75, 76, 77, 78, 80, 81, 84, 86, 87, 89, 90, 92, 94, 95, 98, 101, 103, 104, 106, 107, 109, 110, 112, 113, 114, 115, 117, 118

Offset: 0

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 a subsequence of A231005.

Examples

			3 belongs to this sequence because December 1, 2011 falls on the same day as March 1, 2012 and both December and March have 31 days each.
6 belongs to this sequence because January 1, 2012 falls on the same day as July 1, 2012 and both January and July have 31 days each.
8 belongs to this sequence because May 1, 2013 falls on the same day as January 1, 2014 and both May and January have 31 days each.
9 belongs to this sequence because January 1, 2013 falls on the same day as October 1, 2013 and both January and October have 31 days each.
		

Links

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

Crossrefs

Cf. A230995-A231014.

Cf. A231011 (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,for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1],print1(p", ");break)))