%I
%S 1010100,10110100,101000100,101101000,1010101100,1011101000,
%T 10110001000,10111010000,101001000100,101101011000,101111010000,
%U 1010111001100,1011100010000,1011110100000,10101001010100,10110010001000,10111010110000,10111110100000,101001001000100,101010010110100
%N Binary numbers with string structure 10s00, s in {0,1}*, such that it results in a nonpalindromic cycle of length 4 in the Reverse and Add! procedure in base 2.
%C If the binary representation of the binary string 10s00 is in the sequence, so is 101s000.
%C For decimal representation see A306514.
%C This sequence is a subset of A066059.
%C These regular patterns can be represented by the contextfree grammar with production rules:
%C S > S_a  S_b  S_c  S_d
%C S_a > 10 T_a 00, T_a > 1 T_a 0  T_a0,
%C S_b > 11 T_b 01, T_b > 0 T_b 1  T_b0,
%C S_c > 10 T_c 000, T_c > 1 T_c 0  T_c0,
%C S_d > 11 T_d 101, T_d > 0 T_d 1  T_d0,
%C where T_a0, T_b0, T_c0 and T_d0 are some terminating strings.
%C Numbers in this sequence are generated by choosing S_a or S_c from the starting symbol S.
%C From the fact that all strings derived from S_b have prefix 11 and suffix 00, it can be proved that all strings derived from S_a must have prefix 111 (i.e., 1 is prefix of s, with s as in the name of this sequence). Similarly, from the fact that all strings derived from S_d have prefix 11 and suffix 000, it can be proved that all strings derived from S_c must have prefix 111 (i.e., again, 1 is prefix of s, with s as in the name of this sequence). In the later case, 11 is a prefix of s, which is even stronger. I believe additional stronger conditions can be observed and proved, so I challenge others to take a look at it too.
%H A.H.M. Smeets, <a href="/A306515/b306515.txt">Table of n, a(n) for n = 1..6976</a>
%Y Cf. A306514, A306516, A306517.
%K nonn,base
%O 1,1
%A _A.H.M. Smeets_, Feb 21 2019
