A306515 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.

1010100, 10110100, 101000100, 101101000, 1010101100, 1011101000, 10110001000, 10111010000, 101001000100, 101101011000, 101111010000, 1010111001100, 1011100010000, 1011110100000, 10101001010100, 10110010001000, 10111010110000, 10111110100000, 101001001000100, 101010010110100

Offset: 1

Author

A.H.M. Smeets, Feb 21 2019

Keywords

• nonn
• base

Comments

If the binary representation of the binary string 10s00 is in the sequence, so is 101s000.

For decimal representation see A306514.

This sequence is a subset of A066059.

These regular patterns can be represented by the context-free grammar with production rules:

S -> S_a | S_b | S_c | S_d

S_a -> 10 T_a 00, T_a -> 1 T_a 0 | T_a0,

S_b -> 11 T_b 01, T_b -> 0 T_b 1 | T_b0,

S_c -> 10 T_c 000, T_c -> 1 T_c 0 | T_c0,

S_d -> 11 T_d 101, T_d -> 0 T_d 1 | T_d0,

where T_a0, T_b0, T_c0 and T_d0 are some terminating strings.

Numbers in this sequence are generated by choosing S_a or S_c from the starting symbol S.

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.

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..6976

Crossrefs

Cf. A306514, A306516, A306517.