A178055 Numbers representing the number of days in a month in the Gregorian calendar (modulus 7).

3, 1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 1, 3, 2, 3, 2, 3, 3, 2

Offset

1,1

Comments

Sequence first term represents January 2000. Sequence repeats after 4800 terms, representing 400 years in the Gregorian calendar system.

Actual number of days in a month can be determined by adding 28 to the value of the sequence term representing the month in question.

Links

Lyle P. Blosser, Table of n, a(n) for n = 1..4800

Index entries for sequences related to calendars

Example

a(1) = 3 -> January 2000 has 31 days (3+28), a(2) = 1 -> February 2000 has 29 days (1+28), a(3) = 3 -> March 2000 has 31 days (3+28).

Mathematica

dys[{y_, m_, 1}]:=If[m==12, DateDifference[{y, m, 1}, {y+1, 1, 1}], DateDifference[ {y, m, 1}, {y, m+1, 1}]][[1]]; Mod[#, 7]&/@(dys/@ Flatten[Table[{y, m, 1}, {y, 2000, 2010}, {m, 12}], 1])  (* Harvey P. Dale, Sep 04 2020 *)

Crossrefs

Cf. A178054. If a(n) is the n-th term in A178054 and b(n) is the n-th term in A178055, then a(n) + b(n) (modulus 7) = a(n+1)

Sequence in context: A229215 A123508 A117621 * A247856 A059660 A194003

Adjacent sequences: A178052 A178053 A178054 * A178056 A178057 A178058

Keyword

easy,nonn

Author

Lyle P. Blosser (lyleblosser(AT)att.net), May 18 2010

Status

approved