A210885 Number of days after Jan 01 1000 such that the date written in the format DDMMYYYY is palindromic.

374, 749, 1124, 3652, 4027, 4402, 4778, 7305, 7681, 8056, 10959, 11334, 11709, 14612, 14988, 15363, 18266, 18641, 19016, 21919, 22295, 22670, 25573, 25948, 26323, 29226, 29602, 29977, 32880, 33255, 33630, 37202, 37577, 37952, 40480, 40855, 41231, 44134, 44509

Offset

1,1

Comments

There are exactly 335 palindromic dates between Jan 1 1000 and Dec 31 9999 (see b-file for the complete list).

See A210886 for the corresponding dates.

Because of the Gregorian Calendar Reform of 1582 these terms are fictitious, strictly speaking. They are valid only then, if we adopt the reform for dates before 1582, subsequently. To get the real number of days after 'Jan 1 1000' for years after 1582 subtract 5 from terms representing a date after 1582 (these are the terms with n higher than 61). The value 5 comes from the 10 skipped "Gregorian" days between Oct 4 and Oct 15 1582 minus the 5 additional Julian leap days in the years 1000, 1100, 1300, 1400 and 1500 which are not Gregorian leap years.

To determine the real number of days after 'Jan 01 2000' just subtract 365242 from each term. The first palindromic date after 'Jan 01 2000' is 10022001 which is the 62nd term of the sequence A210886. The resulting number of days after Jan 01 2000 are 406, 781, 3684, 4059, 4434, 7337, 7713, 8088, 10991, 11366, 11741, 14644, 15020, 15395, 18298, ... for the 62nd, 63rd, ... 76th ... palindromic dates (see A210886).

Links

Hieronymus Fischer, Table of n, a(n) for n = 1..335

Index entries for sequences related to calendars

Formula

n-th number of days after 'Jan 01 2000' = a(61+n)-365242.

Example

The first palindromic date in DDMMYYYY format after 'Jan 01 1000' is A210886(1)=10011001 (= 'Jan 10 1001' = 'Jan 01 1000' + 374 days);
The fourth palindromic date in DDMMYYYY format after 'Jan 01 1000' is A210886(4)=1011010=01011010 (= 'Jan 01 1010' = 'Jan 01 1000' + 3652 days);
The 66th palindromic date in DDMMYYYY format after 'Jan 01 1000' is A210886(66)=21022012 (= 'Feb 21 2012' = 'Jan 01 1000' + 369676 days = 'Jan 1 2000' + 4434 days)
The last (335th) palindromic date in DDMMYYYY format before the year 10000 is A210886(335)=29099092 (= 'Sep 29 9092' = 'Jan 01 1000' + 2955814 days).

Crossrefs

Cf. A210883 - A210884, A210886 - A210895, A106605, A107273, A107275.

Sequence in context: A025336 A025328 A045284 * A364009 A093237 A269113

Adjacent sequences: A210882 A210883 A210884 * A210886 A210887 A210888

Keyword

nonn,base

Author

Hieronymus Fischer, Apr 01 2012

Status

approved