A189916 Sequence for finding the day of the week for the first day of the month in leap years.

0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6

Offset: 1

Author

Wolfdieter Lang, May 02 2011

Keywords

• nonn
• easy
• fini
• full

Comments

The days of the week, starting with Sunday, have indices 0,1,..,6. The months of the year, starting with January, are numbered 1,2,...,12. The following pattern holds for leap years. Remember that the year 2000 was a leap year. See A189915 for non-leap years.

If Jan 01 falls on a day of the week with index I, then Feb 01 falls on the day with index I+3 (mod 7), Mar 01 falls on the day with index I+4 (mod 7), Apr 01 is again on the day with index I, etc.

If one uses 0->A, 1->B, 2->C, 3->D, 4->E, 5->F and 6->G then the sequence becomes A, D, E, A, C, F, A, D, G, B, E, G.

Examples

			In the year 2008 Jan 01 has index 2 (Tuesday). Therefore, Feb 01 has index 2+3 = 5 (mod 7) (Friday), Mar 01 has index 2+4 = 6 (mod 7) (Saturday), Apr 01 falls again on a Tuesday, May 01 has index 2+2 = 4 (Thursday), Jun 01 has index 2+5= 0 (mod 7) (Sunday). Jul 01 falls again on Tuesday, etc.
For leap years in which Jan 01 has index 0 (Sunday) the pattern for the first days of the months is Sun, Wed, Thu, Sun, Tue, Fri, Sun, Wed, Sat, Mon, Thu, Sat.
		

References

• See A189915.

Crossrefs

Cf. A061251, A178054 (index pattern starting with the year 2000), A189915.

Formula

I(n) = I + a(n) (mod 7), n=1,2,..,12, with I the index of the day of Jan 01 in a leap year, and I(n) the index of the first day the n-th month in this year.

a(n) = A061251(35+n) - 1095 mod(7), n=1..12.

a(n) = A178054(144+n), n=1,...,12 (indices for the year 2012).