A306516 a(n) is the decimal representation of the binary number 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 and is not in the trajectory of any m < a(n).

84, 324, 2628, 10836, 21060, 21684, 23556, 43188, 43668, 86388, 86964, 166212, 168516, 172404, 173364, 332484, 344820, 345972, 348564, 664260, 666180, 688884, 690804, 695220, 1328580, 1330884, 1348164, 1362564, 1377780, 1380084, 1385268, 1387092, 1510020, 2655684, 2659524, 2726532, 2754036

Offset: 1

Author

A.H.M. Smeets, Feb 21 2019

Keywords

• nonn
• base

Comments

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)

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..1701