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

1,1

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.

Sequence in context: A066354 A220289 A133219 * A306517 A043643 A269564

Adjacent sequences: A306512 A306513 A306514 * A306516 A306517 A306518

Keyword

nonn,base

Author

A.H.M. Smeets, Feb 21 2019

Status

approved