This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A355150 #114 Dec 10 2023 09:28:38
%S A355150 0,1,1,1,1,2,1,1,1,2,1,1,1,2,1,2,1,1,1,2,1,2,1,1,1,2,1,1,1,2,1,2,1,2,
%T A355150 1,1,1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,1,1,2,1,2,1,1,1,2,1,2,1,1,1,2,1,2,
%U A355150 1,2,1,1,1,2,1,1,1,2,1,2,1,1,1,2,1,2,1,1,1,2,1
%N A355150 The Hamming weight of A354169, a(n) = A000120(A354169(n)).
%C A355150 All the following conjectures are now known to be true. See De Vlieger et al. (2022). - _N. J. A. Sloane_, Aug 29 2022
%C A355150 Conjecture: It appears that this sequence may be computed by a fast algorithm:
%C A355150 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:
%C A355150 If a(n) = 2, a((n-3)/2) = 1, and a((n-1)/2) = 2 extend this sequence by 1,2.
%C A355150 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.
%C A355150 In all other cases extend this sequence by 1,1,1,2.
%C A355150 This conjecture was verified for n = 0..2^16 against the b-file provided by _Michael De Vlieger_. - _Thomas Scheuerle_, Jul 14 2022
%C A355150 [Typos corrected by _N. J. A. Sloane_, Jul 10 2022 at the suggestion of _Michel Dekking_.]
%C A355150 From _Michel Dekking_, Jul 12 2022: (Start)
%C A355150 Conjecture: It appears that this sequence is almost a periodic sequence, with period 6. Let x be the sequence defined below.
%C A355150 If n > 25, n == 2 (mod 6) is not an element of x then (written as words)
%C A355150 a(n)a(n+1)...a(n+5) = 111212.
%C A355150 If n > 25, n == 2 (mod 6) is an element of x then
%C A355150 a(n)a(n+1)...a(n+5) = 121112.
%C A355150 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
%C A355150 x(2n+2) - x(2n+1) = 3*2^(n+2) for n=0,1,2,....
%C A355150 x(2n+1) - x(2n) = 3*2^(n+2) for n=1,2,.... (End)
%C A355150 From _Michel Dekking_, Jul 23 2022: (Start)
%C A355150 Extending the sequence x to the right with the four numbers 5,8,14,21 we obtain sequence A354789.
%C A355150 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)
%H A355150 Rémy Sigrist, <a href="/A355150/b355150.txt">Table of n, a(n) for n = 0..20000</a> (first 4941 terms from N. J. A. Sloane)
%H A355150 Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, <a href="http://arxiv.org/abs/2209.04108">The Binary Two-Up Sequence</a>, arXiv:2209.04108 [math.CO], Sep 11 2022.
%H A355150 Rémy Sigrist, <a href="/A355150/a355150.txt">C++ program</a>
%H A355150 N. J. A. Sloane, <a href="https://njas.blog/2022/06/03/the-two-up-sequence-a090252/">Blog post about the Two-Up sequence</a>, June 13 2022. Mentions A354169.
%F A355150 a(A354767(n)) = 1.
%F A355150 a(A354798(n+1)) != 2.
%o A355150 (C++) See Links section.
%o A355150 (MATLAB)
%o A355150 function a = A355150( max_n ) % Note: a(0) is omitted here because
%o A355150 % a(1) will be a(1) in the sequence.
%o A355150 a = [1 1 1 1 2];
%o A355150 m = length(a);
%o A355150 while length(a) < max_n
%o A355150 if (((a((m-3)/2) == 2)&&(a((m-1)/2) == 1)&&(a(m-2) == 1)) ...
%o A355150 ||((a((m-3)/2) == 1)&&(a((m-1)/2) == 2)))
%o A355150 a(m+1:m+2) = [1 2];
%o A355150 m = m+2;
%o A355150 else
%o A355150 a(m+1:m+4) = [1 1 1 2];
%o A355150 m = m+4;
%o A355150 end
%o A355150 end
%o A355150 end
%Y A355150 Cf. A000120, A354169, A354767, A354798.
%K A355150 nonn
%O A355150 0,6
%A A355150 _Thomas Scheuerle_, Jun 21 2022
%E A355150 Edited by _N. J. A. Sloane_, Jul 10 2022