A354767 Indices of terms in A354169 that have Hamming weight 1.

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

Offset: 1

Author

N. J. A. Sloane, Jun 21 2022

Keywords

• nonn

Comments

The terms of A354169 only have Hamming weights 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 generating function for the sequence. See link below. (End)

For proofs of these statements see De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..3296
• Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, The Binary Two-Up Sequence, arXiv:2209.04108 [math.CO], Sep 11 2022.
• N. J. A. Sloane, A conjectured generating function for A354169.

Crossrefs

Cf. A354169, A354680, A354779, A354788, A354798.

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)

The conjecture is now known to be true. - N. J. A. Sloane, Aug 29 2022