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

1,2

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.

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.

Example

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.

Prog

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

Crossrefs

Cf. A090651.

Sequence in context: A354585 A282770 A145077 * A086146 A083758 A127494

Adjacent sequences: A127373 A127374 A127375 * A127377 A127378 A127379

Keyword

easy,nonn

Author

Ian Duff, Mar 30 2007

Status

approved