cp's OEIS Frontend

Base-2 analog of A063048 (base 10) and A075421 (base 4); subsequence of A066059. - For the trajectory of 22 (cf. A061561) and the trajectory of 77 (cf. A075253) it has been proved that they do not contain a palindrome. A similar proof can be given for most terms of this sequence, but there are a few terms (4262, 17498, 33378, 33898, ...) whose trajectory does not show the kind of regularity that can be utilized for the construction of a proof. - If the trajectory of an integer k joins the trajectory of a smaller integer which is a term of the present sequence, then this occurs after very few 'Reverse and Add!' steps (at most 84 for numbers < 20000). On the other hand, the trajectories of the terms of this sequence do not join the trajectory of any smaller term within at least 1000 steps.

From A.H.M. Smeets, Feb 12 2019: (Start)

Most terms in this sequence eventually give rise to a regular binary pattern. These regular patterns can be represented by contextfree grammars:

S_a -> 10 T_a 00, T_a -> 1 T_a 0 | A_a(n);

S_b -> 11 T_b 01, T_b -> 0 T_b 1 | B_a(n);

S_c -> 10 T_c 000, T_c -> 1 T_c 0 | C_a(n) and

S_d -> 11 T_d 101, T_d -> 0 T_d 1 | D_a(n).

A_22 = 1101, B_22 = 1000, C_22 = 1101, D_22 = 0010 (see also A058042);

A_77 = 1100010, B_77 = 0000101, C_77 = 1101011, D_77 = 0100000 (see also A075253)

Decimal representations for 10 A_a(n) 00 are given by A306514(n).

Binary representations for 10 A_a(n) 00 are given by A306515(n).

(End)

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.

It is conjectured that all terms in this sequence are derived from S -> S_a -> 10 T_a 00 -> 10 T_a0 00 (in decimal representation here) when we represent the cyclic patterns 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;

T_a0, T_b0, T_c0 and T_d0 being some terminating strings.

It is observed that numbers of the form 81*2^k + 2^floor((k+6)/2) + 2^6*(2^floor((k-8)/2) - 1) + 4 for k >= 11 (i.e. numbers lower(k) as defined in A306514) are all included in this sequence. By assuming this conjecture is true, it is easily to see that the number of seeds for the Reverse and Add! are infinite in base 2.

From A.H.M. Smeets, May 30 2019: (Start)

Several subsets of this sequence can be defined, each of them proving that there exists an infinite number of seeds in the Reverse and Add! procedure in base 2:

In the string representations, + stands for concatenation and ^ stands for repeated concatenation.

Example 1:

With f_1(n) = 81*2^n + 12*2^floor(n/2) - 60, {f_1(n) | n >= 11} is a proper subset.

String representation for the binary representation of f_1(n): 1010001 + (0)^floor((n-11)/2) + 0010 + (1)^floor((n-10)/2) + 1000100 for n >= 11.

f_1(n) = 3*f_1(n-1) - 6*f_1(n-3) + 4*f_1(n-4) for n > 3, f_1(0) = 33, f_1(1) = 114, f_1(2) = 288, f_1(3) = 612 (from Colin Barker).

f_1(n) = (-60 + 81*2^n + 3*2^((1+n)/2)*(1+(-1)^(n+1) + sqrt(2) + (-1)^n*sqrt(2))) (from Colin Barker).

G.f. for f_1: 3*(11 + 5*x - 18*x^2 - 18*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)) (from Colin Barker).

Example 2:

With f_2(n) = 32*8^n + 64*(8^n - 8)/7 + 68, {f_2(n) | n >= 2} is a proper subset.

String representation for the binary representation of f_2(n): 10 + (100)^(n-2) + 1000100 for n >= 2.

f_2(n) = 8*f_2(n-1) + 36 for n > 0, a(0) = 0.

f_2(n) = 36*(-1 + 8^(n+1))/7 (from Colin Barker).

f_2(n) = 9*f_2(n-1) - 8*f_2(n-2) for n > 1, a(0) = 0, a(1) = 36 (from Colin Barker).

G.f. for f_2: 36 / ((1 - x)*(1 - 8*x)) (from Colin Barker).

Example 3:

f_3(n) = 369*2^n - 24*2^floor(n/2) +132, {f_3(n) | n >= 12} is a proper subset.

String representation for the binary representation of f_3(n): 101110000 + (1)^ceiling((n-12)/2) + 101 + (0)^floor((n-12)/2) + 010000100 for n >= 12.

Example 4:

f_4(n) = 21*2^n + 6*2^floor(n/2) - 12, {f_4(n) | n >= 12} is a proper subset.

String representation for the binary representation of f_4(n): 10101 + (0)^floor((n-9)/2) + 0010 + (1)^ceiling((n-9)/2) + 10100 for n >= 12.

f_4(n) = 2*f_4(n-1) + 12*(-1)^n for n >= 4. (End)