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

%I A210885 #17 Feb 06 2015 10:29:54
%S A210885 374,749,1124,3652,4027,4402,4778,7305,7681,8056,10959,11334,11709,
%T A210885 14612,14988,15363,18266,18641,19016,21919,22295,22670,25573,25948,
%U A210885 26323,29226,29602,29977,32880,33255,33630,37202,37577,37952,40480,40855,41231,44134,44509
%N A210885 Number of days after Jan 01 1000 such that the date written in the format DDMMYYYY is palindromic.
%C A210885 There are exactly 335 palindromic dates between Jan 1 1000 and Dec 31 9999 (see b-file for the complete list).
%C A210885 See A210886 for the corresponding dates.
%C A210885 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.
%C A210885 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).
%H A210885 Hieronymus Fischer, <a href="/A210885/b210885.txt">Table of n, a(n) for n = 1..335</a>
%H A210885 <a href="/index/Ca#calendar">Index entries for sequences related to calendars</a>
%F A210885 n-th number of days after 'Jan 01 2000' = a(61+n)-365242.
%e A210885 The first palindromic date in DDMMYYYY format after 'Jan 01 1000' is A210886(1)=10011001 (= 'Jan 10 1001' = 'Jan 01 1000' + 374 days);
%e A210885 The fourth palindromic date in DDMMYYYY format after 'Jan 01 1000' is A210886(4)=1011010=01011010 (= 'Jan 01 1010' = 'Jan 01 1000' + 3652 days);
%e A210885 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)
%e A210885 The last (335th) palindromic date in DDMMYYYY format before the year 10000 is A210886(335)=29099092 (= 'Sep 29 9092' = 'Jan 01 1000' + 2955814 days).
%Y A210885 Cf. A210883 - A210884, A210886 - A210895, A106605, A107273, A107275.
%K A210885 nonn,base
%O A210885 1,1
%A A210885 _Hieronymus Fischer_, Apr 01 2012