A144179 Table in which row n lists the digits of n, the digits of the number of days in year n and if n is a leap year the digits of the number of days in February of that year.

1, 3, 6, 5, 2, 3, 6, 5, 3, 3, 6, 5, 4, 3, 6, 6, 2, 9, 5, 3, 6, 5, 6, 3, 6, 5, 7, 3, 6, 5, 8, 3, 6, 6, 2, 9, 9, 3, 6, 5, 1, 0, 3, 6, 5, 1, 1, 3, 6, 5, 1, 2, 3, 6, 6, 2, 9, 1, 3, 3, 6, 5, 1, 4, 3, 6, 5, 1, 5, 3, 6, 5, 1, 6, 3, 6, 6, 2, 9, 1, 7, 3, 6, 5, 1, 8, 3, 6, 5, 1, 9, 3, 6, 5, 2, 0, 3, 6, 6, 2, 9, 2, 1, 3, 6

Offset

1,2

Comments

This sequence uses the proleptic Gregorian calendar. - Charles R Greathouse IV, Oct 25 2022

Links

Table of n, a(n) for n=1..105.

Example

Array begins
  [1, 3, 6, 5]
  [2, 3, 6, 5]
  [3, 3, 6, 5]
  [4, 3, 6, 6, 2, 9]
  [5, 3, 6, 5]
  [6, 3, 6, 5]
  [7, 3, 6, 5]
  [8, 3, 6, 6, 2, 9]
  [9, 3, 6, 5]
  [1, 0, 3, 6, 5]
  ...
Row 9 is 9, 3, 6, 5 as we start with the year. Then year 9 has 365 days so follow up with the digits 365 to it giving 9, 3, 6, 5 (as 9 is not a leap year we do not include the digits of the number of days in February of year 9). - David A. Corneth, Oct 24 2022

Prog

(PARI) first(n) = my(res = []); for(i = 1, oo, res = concat(res, row(i)); if(#res >= n, return(res) ) )
row(n) = my(res = digits(n)); if(n % 400 == 0 || (n % 100 != 0 && n % 4 == 0), res = concat(res, [3, 6, 6, 2, 9]), res = concat(res, [3, 6, 5]) ); res \\ David A. Corneth, Oct 24 2022

Crossrefs

Cf. A144189.

Sequence in context: A296224 A135097 A154008 * A155540 A225451 A131955

Adjacent sequences: A144176 A144177 A144178 * A144180 A144181 A144182

Keyword

nonn,tabf,base,less

Author

Juri-Stepan Gerasimov, Nov 20 2008

Extensions

New name from and edited by David A. Corneth, Oct 24 2022

Status

approved