A306515 - OEIS

%I #34 May 07 2019 18:36:11

%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 non-palindromic 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 context-free 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