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A354793
Hamming weight of A354783(n).
2
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, 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, 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
OFFSET
1,5
COMMENTS
Conjecture: This sequence appears to have a simple structure. Encode it by making the following substitutions, in this order:
Replace the initial 28 terms 0011201120223120113120112022 by S (as usual, the start is irregular), then map:
3 1 3 -> 7
3 1 2 -> 6
1 2 0 1 1 2 0 2 2 -> 9
0 1 1 -> 2
0 2 2 -> 4
Then it appears that the encoded sequence is the concatenation of the following blocks:
S
79
79(6264)^1
79(6264)^1
79(6264)^3
79(6264)^3
79(6264)^15
79(6264)^15
79(6264)^31
79(6264)^31
79(6264)^63
79(6264)^63
79(6264)^127
79(6264)^127
...
This is probably not the most efficient encoding, but I was happy to find any one that revealed the structure.
From Michel Dekking, Jul 23 2022: (Start)
The following is another way to present the conjecture above, which shows the close connection with sequence A355150.
Conjecture: It appears that this sequence is almost a periodic sequence, with period 12. Let x:=A354789.
If n > 28, n == 5 (mod 12) is not an element of x then (written as words)
a(n)a(n+1)...a(n+11) = 312011312022.
If n > 28, n == 5 (mod 12) is an element of x then
a(n)a(n+1)...a(n+11) = 313120112022.
(End)
LINKS
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jul 19 2022
STATUS
approved