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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

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Last modified August 17 12:30 EDT 2022. Contains 356189 sequences. (Running on oeis4.)