OFFSET
0,6
COMMENTS
All the following conjectures are now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022
Conjecture: It appears that this sequence may be computed by a fast algorithm:
We begin with an initial sequence 0,1,1,1,1,2. Let n be the index of the last element added. Then extend by the rules:
If a(n) = 2, a((n-3)/2) = 1, and a((n-1)/2) = 2 extend this sequence by 1,2.
If a(n) = 2, a((n-3)/2) = 2, a((n-1)/2) = 1, and a(n-2) = 1, extend this sequence by 1,2.
In all other cases extend this sequence by 1,1,1,2.
This conjecture was verified for n = 0..2^16 against the b-file provided by Michael De Vlieger. - Thomas Scheuerle, Jul 14 2022
[Typos corrected by N. J. A. Sloane, Jul 10 2022 at the suggestion of Michel Dekking.]
From Michel Dekking, Jul 12 2022: (Start)
Conjecture: It appears that this sequence is almost a periodic sequence, with period 6. Let x be the sequence defined below.
If n > 25, n == 2 (mod 6) is not an element of x then (written as words)
a(n)a(n+1)...a(n+5) = 111212.
If n > 25, n == 2 (mod 6) is an element of x then
a(n)a(n+1)...a(n+5) = 121112.
The sequence x = {32, 44, 68, 92, 140, 188, 284, ...} is a sparse sequence defined via the sequence A007283, given by A007283(n)=3*2^n, which has also been encountered in A354169. In fact, x(1) = 32, and
x(2n+2) - x(2n+1) = 3*2^(n+2) for n=0,1,2,....
x(2n+1) - x(2n) = 3*2^(n+2) for n=1,2,.... (End)
From Michel Dekking, Jul 23 2022: (Start)
Extending the sequence x to the right with the four numbers 5,8,14,21 we obtain sequence A354789.
So the sparse positions are given by 9*2^k - 4 for k even, and by 12*2^k - 4 for k odd, for k = 2,3,... (End)
LINKS
Rémy Sigrist, Table of n, a(n) for n = 0..20000 (first 4941 terms from N. J. A. Sloane)
Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, The Binary Two-Up Sequence, arXiv:2209.04108 [math.CO], Sep 11 2022.
Rémy Sigrist, C++ program
N. J. A. Sloane, Blog post about the Two-Up sequence, June 13 2022. Mentions A354169.
PROG
(C++) See Links section.
(MATLAB)
function a = A355150( max_n ) % Note: a(0) is omitted here because
% a(1) will be a(1) in the sequence.
a = [1 1 1 1 2];
m = length(a);
while length(a) < max_n
if (((a((m-3)/2) == 2)&&(a((m-1)/2) == 1)&&(a(m-2) == 1)) ...
||((a((m-3)/2) == 1)&&(a((m-1)/2) == 2)))
a(m+1:m+2) = [1 2];
m = m+2;
else
a(m+1:m+4) = [1 1 1 2];
m = m+4;
end
end
end
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Scheuerle, Jun 21 2022
EXTENSIONS
Edited by N. J. A. Sloane, Jul 10 2022
STATUS
approved