A127376 Perpetual calendar, giving the calendar for successive years of the 400-year cycle starting in year xy01, where xy is a 2-digit multiple of 4. Calendars 1 to 7 are for normal years starting on Monday, Tuesday, ..., Sunday; 8 to 14 are for leap years likewise.

1, 2, 3, 11, 6, 7, 1, 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1, 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1, 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1, 9, 4, 5, 6, 14, 2

Offset: 1

Author

Ian Duff, Mar 30 2007

Keywords

• easy
• nonn

Comments

A090651 gives the 14-year repeating cycle applying between century years not a multiple of 400; this sequence extends that to the complete 400-year cycle.

Examples

			2001 is an ordinary year starting on a Monday, as a(1)=1; 2004 is a leap year starting on a Thursday, as a(4)=11; 2400 is a leap year starting on Saturday, as a(400)=13.
		

Links

• Ian Duff, Table of n, a(n) for n = 1..400
• Index entries for sequences related to calendars
• Index entries for linear recurrences with constant coefficients, order 400.

Crossrefs

Cf. A090651.

Programs

BASIC

a(0)=13
for n=1 to 400
if (n mod 4 =0 and n mod 100 <>0) or n mod 400 =0 then
a(n)=a(n-1)+8 : if a(n)>14 then a(n)=a(n)-7
else
if (n-1 mod 4 =0 and n-1 mod 100 <>0) or n-1 mod 400 =0 then
a(n)=a(n-1)-5
else
a(n)=a(n-1)+1
end if
if a(n)>7 then a(n)=a(n)-7
end if
next n