A354773 For terms of A354169 that are the sum of two distinct powers of 2, the exponent of the smaller power of 2.

0, 2, 0, 1, 2, 3, 8, 0, 1, 5, 12, 2, 3, 7, 16, 8, 9, 0, 10, 1, 4, 11, 24, 12, 13, 2, 14, 3, 6, 15, 32, 16, 17, 8, 18, 9, 19, 0, 20, 10, 21, 1, 22, 4, 5, 11, 48, 24, 25, 12, 26, 13, 27, 2, 28, 14, 29, 3, 30, 6, 7, 15, 64, 32, 33, 16, 34, 17, 35, 8, 36, 18, 37, 9, 38, 19, 39, 0, 40, 20, 41, 10, 42, 21, 43, 1, 44, 22, 45, 4, 46

Offset: 1

Author

N. J. A. Sloane, Jun 26 2022

Keywords

• base
• nonn

Links

• 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], Sep 11 2022.
• Rémy Sigrist, C++ program

Crossrefs

Cf. A354169, A354680, A354767, A354798, A354774, A354775.

Programs

Python

from itertools import count, islice
from collections import deque
from functools import reduce
from operator import or_
def A354773_gen(): # generator of terms
    aset, aqueue, b, f = {0,1,2}, deque([2]), 2, False
    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:
                if (s := bin(m)[:1:-1]).count('1') == 2:
                    yield s.index('1')
                aset.add(m)
                aqueue.append(m)
                if f: aqueue.popleft()
                b = reduce(or_,aqueue)
                f = not f
                break
A354773_list = list(islice(A354773_gen(),20)) # Chai Wah Wu, Jun 26 2022
(C++) See Links section.

Formula

Conjecture from N. J. A. Sloane, Jun 29 2022: (Start)

The following is a conjectured recurrence for a(n). Basically a(n) = a(n/2-1) if n is even, and a(n) = (n+1)/2 if n is odd, except that there are four types of n which have a different formula, and there are 19 exceptional values for small n. Note that a(n) does not depend on earlier values when n is odd.

Here is the formula, which agrees with the first 10000 terms.

There are exceptional values as far out as n=61, so we take care of them first.

Initial conditons:

If n is on the list

[1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 21, 22, 29, 30, 45, 61]

then a(n) is given by the n-th term of the following list:

[0, 2, 0, 1, 2, 3, 8, 0, 1, 5, 12, 2, 3, 7, 16, 8, 9, 0, 10,

1, 4, 11, 24, 12, 13, 2, 14, 3, 6, 15, 32, 16, 17, 8, 18, 9,

19, 0, 20, 10, 21, 1, 22, 4, 5, 11, 48, 24, 25, 12, 26, 13,

27, 2, 28, 14, 29, 3, 30, 6, 7].

Otherwise, if n is even, a(n) = a(n/2-1).

Otherwise n is odd and is not one of the exceptions.

(I) If n = 3*2^k-3, k >= 5, then a(n) = (n-1)/4.

(II) If n = 2^k-3, k >= 4 then a(n) = (n-1)/4.

(III) If n = 3*2^k-1, k >= 2 then a(n) = n+1.

(IV) If n = 2^k-1, k >= 3 then a(n) = n+1.

(V) Otherwise a(n) = (n+1)/2.

(End)

The conjecture is now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022