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A057891 In base 2, neither a palindrome nor becomes a palindrome if trailing 0's are omitted.

%I A057891 #27 Oct 13 2017 16:41:34
%S A057891 11,13,19,22,23,25,26,29,35,37,38,39,41,43,44,46,47,49,50,52,53,55,57,
%T A057891 58,59,61,67,69,70,71,74,75,76,77,78,79,81,82,83,86,87,88,89,91,92,94,
%U A057891 95,97,98,100,101,103,104,105,106,109,110,111,113,114,115,116,117,118
%N A057891 In base 2, neither a palindrome nor becomes a palindrome if trailing 0's are omitted.
%C A057891 These could be called "asymmetric bit strings".
%C A057891 Fixed pairs of A057889, complement of A057890.
%C A057891 If these numbers are converted to their binary polynomial, one of the roots of that polynomial will have absolute values other than 1 or 0. For example 11 = 2^3 + 2^1 + 2^0, the absolute values of the roots of x^3 + x + 1 are 0.682328... and 1.21061... which are not 1 or 0, so 11 is in the sequence. The first number with this property which is not a term is A057890(53) = 107. - _Benedict W. J. Irwin_, Sep 07 2017 and _Andrey Zabolotskiy_, Oct 13 2017
%H A057891 Reinhard Zumkeller, <a href="/A057891/b057891.txt">Table of n, a(n) for n = 1..10000</a>
%F A057891 A030101(A030101(n)) != A030101(n). - David Wilson, Jun 09 2009, Jun 18 2009
%F A057891 A178225(A000265(a(n))) = 0. - _Reinhard Zumkeller_, Oct 21 2011
%e A057891 11 is included because 1011 is asymmetrical, but 12 is not because 001100 is a palindrome.
%o A057891 (Haskell)
%o A057891 a057891 n = a057891_list !! (n-1)
%o A057891 a057891_list = filter ((== 0) . a178225 . a000265) [1..]
%o A057891 -- _Reinhard Zumkeller_, Oct 21 2011
%Y A057891 Cf. A061917, A006995. Complement of A057890.
%Y A057891 Cf. A030101, A000265, A006519, A057889.
%K A057891 easy,nonn,base
%O A057891 1,1
%A A057891 _Marc LeBrun_, Sep 25 2000
%E A057891 Edited by _N. J. A. Sloane_, Jun 09 2009 at the suggestion of _Ray Chandler_
%E A057891 A-numbers in formula corrected by _R. J. Mathar_, Jun 18 2009