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%I #25 Jul 28 2022 12:10:26
%S 0,0,1,1,2,0,1,1,2,0,2,2,3,1,2,0,1,1,3,1,2,0,1,1,2,0,2,2,3,1,3,1,2,0,
%T 1,1,2,0,2,2,3,1,3,1,2,0,1,1,2,0,2,2,3,1,2,0,1,1,3,1,2,0,2,2,3,1,3,1,
%U 2,0,1,1,2,0,2,2,3,1,2,0,1,1,3,1,2,0,2,2,3,1,3,1,2,0,1,1,2,0,2,2,3,1,2
%N Hamming weight of A354783(n).
%C Conjecture: This sequence appears to have a simple structure. Encode it by making the following substitutions, in this order:
%C Replace the initial 28 terms 0011201120223120113120112022 by S (as usual, the start is irregular), then map:
%C 3 1 3 -> 7
%C 3 1 2 -> 6
%C 1 2 0 1 1 2 0 2 2 -> 9
%C 0 1 1 -> 2
%C 0 2 2 -> 4
%C Then it appears that the encoded sequence is the concatenation of the following blocks:
%C S
%C 79
%C 79(6264)^1
%C 79(6264)^1
%C 79(6264)^3
%C 79(6264)^3
%C 79(6264)^15
%C 79(6264)^15
%C 79(6264)^31
%C 79(6264)^31
%C 79(6264)^63
%C 79(6264)^63
%C 79(6264)^127
%C 79(6264)^127
%C ...
%C This is probably not the most efficient encoding, but I was happy to find any one that revealed the structure.
%C From _Michel Dekking_, Jul 23 2022: (Start)
%C The following is another way to present the conjecture above, which shows the close connection with sequence A355150.
%C Conjecture: It appears that this sequence is almost a periodic sequence, with period 12. Let x:=A354789.
%C If n > 28, n == 5 (mod 12) is not an element of x then (written as words)
%C a(n)a(n+1)...a(n+11) = 312011312022.
%C If n > 28, n == 5 (mod 12) is an element of x then
%C a(n)a(n+1)...a(n+11) = 313120112022.
%C (End)
%H N. J. A. Sloane, <a href="/A354793/b354793.txt">Table of n, a(n) for n = 1..4900</a>
%Y Cf. A354169, A354757, A354783, A355150.
%K nonn
%O 1,5
%A _N. J. A. Sloane_, Jul 19 2022