A354789 -id:A354789 - cp's OEIS frontend

A354798 Indices of terms in A354169 that are not powers of 2.

0, 5, 9, 13, 15, 19, 21, 25, 29, 31, 33, 37, 41, 43, 45, 49, 53, 55, 59, 61, 65, 67, 69, 73, 77, 79, 83, 85, 89, 91, 93, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 141, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175, 179

Offset: 1

Rémy Sigrist and N. J. A. Sloane, Jun 06 2022

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, The Binary Two-Up Sequence, arXiv:2209.04108 [math.CO] [math.CO], Sep 11 2022.
Rémy Sigrist, PARI program

Cf. A057716, A136252, A354169, A354680 (corresponding terms), A354788, A354798.

PARI
```
See Links section.
```

from itertools import count, islice
from collections import deque
from functools import reduce
from operator import or_
def A354798_gen(): # generator of terms
    aset, aqueue, b, f, i = {0,1,2}, deque([2]), 2, False, 2
    yield 0
    while True:
        for k in count(1):
            m, j, j2, r, s = 0, 0, 1, b, k
            while r > 0:
                r, q = divmod(r,2)
                if not q:
                    s, y = divmod(s,2)
                    m += y*j2
                j += 1
                j2 *= 2
            if s > 0:
                m += s*2**b.bit_length()
            if m not in aset:
                i += 1
                if m.bit_count() > 1:
                    yield i
                aset.add(m)
                aqueue.append(m)
                if f: aqueue.popleft()
                b = reduce(or_,aqueue)
                f = not f
                break
A354798_list = list(islice(A354798_gen(),30)) # Chai Wah Wu, Jun 06 2022

Conjecture from N. J. A. Sloane, Jul 15 2022: (Start)
The following is a conjectured explicit formula for a(n).
Define the "fence posts" by F(0) = 1, F(2i+1) = 2^(i+4) - 3 for i >= 0, F(2i) = 3*2^(i+2) - 3 for i >= 1.
The F(i) sequence begins 1, 13, 21, 29, 45, 61, 93, 125, 189, 253, 381, ... (cf. A136252 or A354788)
The value of a(n) at n = F(i) is V(i) = 0 if i = 0, V(i) = 3*F(i)+2 if i >= 1.
The V(i) sequence begins 0, 41, 65, 89, 137, 185, 281, 377, 569, 761, ... (cf. A354789).
The first 12 terms of the sequence are irregular, so we simply define a(n) for F(0) = 1 <= n <= 12 to be the n-th term of
  [0, 5, 9, 13, 15, 19, 21, 25, 29, 31, 33, 37]
Assume now that n >= F(1) = 13, and define i and j by F(i) <= n < F(i+1), n = F(i) + j.
Then we conjecture that a(n) = V(i) + f(j) where f(0) .. f(3) are 0,2,4,8, and for j >= 4, f(j) = 3*j if j is even, f(j) = 3*j-1 if j is odd.
The f(i), i >= 0, sequence is independent of n (to find a(n) we use only an initial segment of f(n)), and begins:
0, 2, 4, 8, 12, 14, 18, 20, 24, 26, 30, 32, 36, 38, 42, 44, 48, 50, 54, 56, ...
The conjecture has been checked for the first 5000 terms.
(End)
The conjecture is now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022

A355150 The Hamming weight of A354169, a(n) = A000120(A354169(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1

Offset: 0

Thomas Scheuerle, Jun 21 2022

nonn

All the following conjectures are now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022
Conjecture: It appears that this sequence may be computed by a fast algorithm:
  We begin with an initial sequence 0,1,1,1,1,2. Let n be the index of the last element added. Then extend by the rules:
  If a(n) = 2, a((n-3)/2) = 1, and a((n-1)/2) = 2 extend this sequence by 1,2.
  If a(n) = 2, a((n-3)/2) = 2, a((n-1)/2) = 1, and a(n-2) = 1, extend this sequence by 1,2.
  In all other cases extend this sequence by 1,1,1,2.
This conjecture was verified for n = 0..2^16 against the b-file provided by Michael De Vlieger. - Thomas Scheuerle, Jul 14 2022
[Typos corrected by N. J. A. Sloane, Jul 10 2022 at the suggestion of Michel Dekking.]
From Michel Dekking, Jul 12 2022: (Start)
Conjecture: It appears that this sequence is almost a periodic sequence, with period 6. Let x be the sequence defined below.
If n > 25, n == 2 (mod 6) is not an element of x then (written as words)
      a(n)a(n+1)...a(n+5) = 111212.
If n > 25, n == 2 (mod 6) is an element of x then
      a(n)a(n+1)...a(n+5) = 121112.
The sequence x = {32, 44, 68, 92, 140, 188, 284, ...} is a sparse sequence defined via the sequence A007283, given by A007283(n)=3*2^n, which has also been encountered in A354169. In fact, x(1) = 32, and
      x(2n+2) - x(2n+1) = 3*2^(n+2)    for n=0,1,2,....
      x(2n+1) - x(2n)   = 3*2^(n+2)    for n=1,2,.... (End)
From Michel Dekking, Jul 23 2022: (Start)
Extending the sequence x to the right with the four numbers 5,8,14,21 we obtain sequence A354789.
So the sparse positions are given by 9*2^k - 4 for k even, and by 12*2^k - 4 for k odd, for k = 2,3,... (End)

Rémy Sigrist, Table of n, a(n) for n = 0..20000 (first 4941 terms from N. J. A. Sloane)
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.
Rémy Sigrist, C++ program
N. J. A. Sloane, Blog post about the Two-Up sequence, June 13 2022. Mentions A354169.

Cf. A000120, A354169, A354767, A354798.

function a = A355150( max_n ) % Note: a(0) is omitted here because
                              % a(1) will be a(1) in the sequence.
    a = [1 1 1 1 2];
    m = length(a);
    while length(a) < max_n
        if (((a((m-3)/2) == 2)&&(a((m-1)/2) == 1)&&(a(m-2) == 1)) ...
            ||((a((m-3)/2) == 1)&&(a((m-1)/2) == 2)))
            a(m+1:m+2) = [1 2];
            m = m+2;
        else
            a(m+1:m+4) = [1 1 1 2];
            m = m+4;
        end
    end
end

a(A354767(n)) = 1.

a(A354798(n+1)) != 2.

Edited by N. J. A. Sloane, Jul 10 2022

A354793 Hamming weight of A354783(n).

Original entry on oeis.org

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

N. J. A. Sloane, Jul 19 2022

nonn

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)

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

Cf. A354169, A354757, A354783, A355150.

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A354798 Indices of terms in A354169 that are not powers of 2.

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A355150 The Hamming weight of A354169, a(n) = A000120(A354169(n)).

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A354793 Hamming weight of A354783(n).

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