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A127376
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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.
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1
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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.
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LINKS
| Ian Duff, Table of n, a(n) for n = 1..400
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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.
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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
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CROSSREFS
| Cf. A090651.
Sequence in context: A137332 A084047 A145077 * A086146 A082618 A083758
Adjacent sequences: A127373 A127374 A127375 * A127377 A127378 A127379
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KEYWORD
| easy,nonn
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AUTHOR
| Ian Duff (ianfduff(AT)yahoo.co.uk), Mar 30 2007
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