A061251 -id:A061251 - cp's OEIS frontend

A008685 Lengths of months in the Gregorian calendar.

31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, 31, 28, 31

Offset: 1

N. J. A. Sloane

To make the definition unambiguous, consider these as starting from January 2001 (or, equivalently, AD 1 in the proleptic Gregorian calendar). [Charles R Greathouse IV, Sep 28 2011]

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Index entries for sequences related to calendars
Index entries for linear recurrences with constant coefficients, order 4800.

Cf. A061251.

a008685 n = a008685_list !! (n-1)
a008685_list = concatMap t [1..] where
   t y = [31, 28 + leap, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
         where leap = if mod y 4 == 0 &&
                         (mod y 100 > 0 || mod y 400 == 0) then 1 else 0
-- Reinhard Zumkeller, Jun 19 2015

year1 = 2001; year2 = 2100;
PreviousDate[#, "Day"][[1, 3]]& /@ Flatten[Table[{#, m, 1}, {m, Append[ Range[2, 12], 1]}]& /@ Range[year1, year2], 1] (* Jean-François Alcover, Aug 01 2018 *)

v=[];for(n=1,50,v=concat(v,[31, if(n%4||(n%100==0&&n%400),28,29), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]));v

def a(n):
    y, m = 1 + (n-1)//12, (n-1)%12
    leap = int((y%4 == 0 and y%100 != 0) or y%400 == 0)
    return [31, 28+leap, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31][m]
print([a(n) for n in range(1, 64)]) # Michael S. Branicky, Feb 04 2024

a(n) = a(n-4800). [Charles R Greathouse IV, Sep 28 2011]

A073304 Remaining days in non-leap year at end of n-th month.

Original entry on oeis.org

365, 334, 306, 275, 245, 214, 184, 153, 122, 92, 61, 31, 0

Offset: 0

Rick L. Shepherd, Jul 23 2002

			a(6)=184 because there are 184 days left in a (non-leap) year on Jun 30, the sixth month.

Cf. A073305 (remaining days in leap year), A061251 (elapsed days at end of n-th month beginning with non-leap year).

A073305 Remaining days in leap year at end of n-th month.

Original entry on oeis.org

366, 335, 306, 275, 245, 214, 184, 153, 122, 92, 61, 31, 0

Offset: 0

Rick L. Shepherd, Jul 23 2002

a(0)=A073304(0)+1 and a(1)=A073304(1)+1; these sequences otherwise are identical since there is only one leap day and it is in the second month.

			a(1)=335 because there are 335 days left in a leap year at the end of January, the first month.

Cf. A073304 (remaining days in non-leap year), A061251 (elapsed days at end of n-th month beginning with non-leap year).

Join[{366},(DateDifference[#,{2016,12,31}]&/@Table[Select[Table[ DatePlus[ {2016,m,1},{d,"Day"}],{d,30}],#[[2]]==m&][[-1]],{m,12}])[[All,1]]] (* Harvey P. Dale, Nov 29 2020 *)

A116386 Number of calendar weeks in the year n (starting at n=0 for the year 2000).

Original entry on oeis.org

54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53

Offset: 0

Sergio Pimentel, Mar 15 2006

Since 365/7 = 52.14 > 52, every year has at least 53 weeks (although the first and / or the last calendar weeks might not be complete and belong to two different years). Only if a leap year begins in a Saturday (the last day of the calendar week), a year can have 54 different calendar weeks (being the first and last of only one day). Years with 54 calendar weeks are: 2000, 2028, 2056, 2084, 2124, 2152, etc. It happens 13 times in a 400 year cycle.

			E.g. a(0)=54 because the year 2000 had 54 calendar weeks (since Jan 01 2000 was a Saturday and Dec 31 2000 was a Sunday)

Index entries for sequences related to calendars

Cf. A060512, A061251, A003786, A090651, A101312.

A189915 Sequence for finding the day of the week for the first day of the month in a common (non-leap) year.

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, May 02 2011

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 common (non-leap) years.
  See A189916 for leap years.
If Jan 01 falls on a day of the week with index I, then Feb 01 is on the day with index I+3 (mod 7), Mar 01 is also on the day with index I+3 (mod 7), Apr 01 is on the day with index I+6 (mod 7), etc.
If one uses 0->A, 1->B, 2->C, 3->D, 4->E, 5->F and 6->G the sequence becomes
  A, D, D, G, B, E, G, C, F, A, D, F.
  A mnemonic rhyme for this sequence is in English:
  At Dover dwells George Brown, Esquire,
  Good Christopher Fitch and David Frier.
  In German (attributed to Thomas Brown, Oxford):
  Allvater, der du gnaedig bist,
  Ein gesetzestreuer Christ
  Fordert Ablauf dieser Frist.
  See the L. Holford-Strevens reference pp. 106-7 (German translation).

			In the year 2011 Jan 01 has index 6 (Saturday). Therefore, Feb 01 has index 6+3 = 2 (mod 7) (Tuesday), Mar 01 also has index 2, Apr 01 has index 6+6 = 5 (mod 7) (Friday), etc.
In common years with Jan 01 on a Sunday (index 0) the sequence gives the index of the first day of the n-th month of this year. E.g., in the year 2006 (but not in the leap year 2012).

L. Holford-Strevens, The History of Time. A Very Short Introduction, Oxford University Press, 2005.
L. Holford-Strevens, Kleine Geschichte der Zeitrechnung und des Kalenders, Reclams Universalbibliothek Nr.18483, Stuttgart, 2008 (German translation).

Cf. A178054 (indices starting with Jan 01 2000), A061251.

I(n) = I + a(n) (mod 7), n=1,...,12, with I the index of January 01 in a common (non-leap) year, and I(n) the index of the day of the week of the first day of the n-th month in this year.
a(n) = A061251(n) (mod 7), n=0,..,11.
a(n) = A178054(72+n), n=1..12.

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

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, May 02 2011

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.

			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.

See A189915.

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

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).

A327846 Full days remaining in the month on the Gregorian calendar starting at n-th day of a non-leap year.

Original entry on oeis.org

30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6

Offset: 1

Andrew Pham, Sep 27 2019

			On March 27th, the 86th day of the year where n=86, there are 4 full days left in the month of March, a(86)=4.

Jinyuan Wang, Table of n, a(n) for n = 1..365

Cf. A061251, A073304.

Array[Range[# - 1, 0, -1] &@ Which[MemberQ[{4, 6, 9, 11}, #], 30, # == 2, 28, True, 31] &, 12] // Flatten (* Michael De Vlieger, Sep 30 2019 *)

cp's OEIS Frontend

A008685 Lengths of months in the Gregorian calendar.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Python

Formula

A073304 Remaining days in non-leap year at end of n-th month.

Views

Author

Keywords

Examples

Crossrefs

A073305 Remaining days in leap year at end of n-th month.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A116386 Number of calendar weeks in the year n (starting at n=0 for the year 2000).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A189915 Sequence for finding the day of the week for the first day of the month in a common (non-leap) year.

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Formula

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

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Formula

A327846 Full days remaining in the month on the Gregorian calendar starting at n-th day of a non-leap year.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica