A354793 Hamming weight of A354783(n).

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

N. J. A. Sloane, Table of n, a(n) for n = 1..4900

Crossrefs

Cf. A354169, A354757, A354783, A355150.

Sequence in context: A083715 A037135 A202177 * A029360 A337541 A185279

Adjacent sequences: A354790 A354791 A354792 * A354794 A354795 A354796

Keyword

nonn

Author

N. J. A. Sloane, Jul 19 2022

Status

approved