|
| |
|
|
A354767
|
|
Indices of terms in A354169 that have Hamming weight 1.
|
|
11
|
|
|
|
1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 16, 17, 18, 20, 22, 23, 24, 26, 27, 28, 30, 32, 34, 35, 36, 38, 39, 40, 42, 44, 46, 47, 48, 50, 51, 52, 54, 56, 57, 58, 60, 62, 63, 64, 66, 68, 70, 71, 72, 74, 75, 76, 78, 80, 81, 82, 84, 86, 87, 88, 90, 92, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 116
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
It is conjectured that the terms of A354169 only have Hamming weight 0, 1, or 2.
Comment from N. J. A. Sloane, Jun 26 2022: (Start)
Taking first differences, then applying the RUNS transform twice gives [1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 13, 1, 1, 1, 13, 1, 1, 1, 29, 1, 1, 1, 29, 1, 1, 1, 61, 1, 1, 1, 61, 1, 1, 1, 125, 1, 1, 1, 125, 1, 1, 1, 253, 1, 1, 1, 253, 1, 1, 1, 509, 1, 1, 1,...].
If the initial terms 1, 1, 1, 1, 5 are replaced by a single 1, this has an obvious regular structure, which can then be analyzed to give a conjectured generating function for the sequence. See link below. (End)
See also the more precise conjectured Formula below.
|
|
|
LINKS
|
N. J. A. Sloane, Table of n, a(n) for n = 1..3296
N. J. A. Sloane, A conjectured generating function for A354169.
|
|
|
FORMULA
|
Conjecture from N. J. A. Sloane, Jun 30 2022, modified Jul 30 2022: (Start)
We first define a sequence {b(n)} as follows.
Define "fence posts" by F(0) = 1, F(2i+1) = 2^(i+5) - 3 for i >= 0, F(2i) = 3*2^(i+3) - 3 for i >= 1.
The F(i) sequence begins 1, 29, 45, 61, 93, 125, 189, 253, 381, 509, ... (cf. A136252 or A354788).
The value of b(n) at n = F(i) is V(i) = 0 if i = 0, V(i) =(F(i)-7)/2 if i >= 1.
The V(i) sequence begins 0, 11, 19, 27, 43, 59, 91, 123, 187, 251, ...
The first 28 terms are irregular, and we simply define b(n) for F(0) = 1 <= n <= 28 to be the n-th term of
[0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9, 10, 10, 10, 11, 11].
Assume now that n >= F(1) = 29, and define i and j by F(i) <= n < F(i+1), n = F(i) + j.
Then b(n) = V(i) + f(j), where f(0) ... f(5) are 0,1,2,3,3,3, and for j >= 6, f(j) = 2 + 2*floor((j-2)/4) + epsilon(j), where epsilon(j) is 1 if j==1 mod 4 and is otherwise 0.
The f(i), i >= 0, sequence (A354779) is independent of n (to find b(n) we use only an initial segment of f(n)), and begins:
0, 1, 2, 3, 3, 3, 4, 4, 4, 5, 6, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 13, 14, ...
With b(n) defined in this way, we conjecture that a(n) = b(n) + n. This has been checked for the first 3296 terms.
(End)
|
|
|
CROSSREFS
|
Cf. A354169, A354680, A354779, A354788, A354798.
Sequence in context: A099619 A027564 A284941 * A351714 A248559 A186511
Adjacent sequences: A354764 A354765 A354766 * A354768 A354769 A354770
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
N. J. A. Sloane, Jun 21 2022
|
|
|
STATUS
|
approved
|
| |
|
|
