A339024 a(1) = 1, a(n) is the least m not already in the sequence whose binary expansion begins with the binary expansion of the binary weight of a(n-1).

1, 2, 3, 4, 5, 8, 6, 9, 10, 11, 7, 12, 16, 13, 14, 15, 17, 18, 19, 24, 20, 21, 25, 26, 27, 32, 22, 28, 29, 33, 23, 34, 35, 30, 36, 37, 31, 40, 38, 48, 39, 64, 41, 49, 50, 51, 65, 42, 52, 53, 66, 43, 67, 54, 68, 44, 55, 45, 69, 56, 57, 70, 58, 71, 72, 46, 73, 59

Offset: 1

Author

Michael De Vlieger, Dec 16 2020

Keywords

• nonn
• base
• easy

Comments

We define binary weight wt(n) = A000120(n) as the number of 1s in n_2, the number n expressed in binary. Let w = wt(a(n-1)) the binary weight of the previous term, where w_2 is w expressed in binary, and let interval I(j) = 2^j <= n <= (2^(j+1)-1).'

Likely a permutation of the natural numbers.

The plot (n, a(n)) is organized into streaky clouds that pertains to a "family" M(i) <= m < M(i+1) whose binary expansion begins with an odd "prefix" m/2^v, where v is the 2-adic valuation of m. There are thus 2^v numbers in this range.

The numbers in this range accommodate the binary weights wt(a(n-1)) = w with 1 <= w <= ceiling(log_2 a(n-1)) such that w_2 appears in part or all of the binary expansion of the prefix m/2^v, and perhaps an additional bit in m after the prefix.

Small values of w, for instance w = 1, may appear in any family, but large w require the entire prefix and potentially more (if even).

The w that cannot be found in a particular family are found in a different family that has M(i+1) as its least member.

The families M(i) belong in turn to classes according to odd prefixes. Thus, for example, we may find w = 1, 2, 4, and 9 in class 9, since "1", "10", "100", and "1001" can be found in numbers m that begin, "1001...".

For w in interval I(j), we have values 1 <= k <= j - 1 distributed binomially.

Permutation of the natural numbers. We can always find w in a number m in family M(i) that pertains to a class C of numbers that in binary start with the binary expansion of an odd number c.

Numbers m that begin with numbers that are formed of left-trimmed bits of c exhaust the numbers in M(i) before moving to M(i+1) in the same class C.

When we have recordsetting odd w, a new class C opens up based on the binary expansion of a larger odd number c.

A permutation of the integers since n appears at or before index 2^n - 1, the first number with binary weight n. - Michael S. Branicky, Dec 16 2020

Examples

			Let wt(n) = A000120(n).
a(2) = 2 since wt(a(1)) = wt(1) = 1, and we find "1" at the beginning of the binary expansion of the yet unused 2 = "10"_2.
a(3) = 3 since wt(2) = 1, we find "1" as first bit of yet unused 3 = "11"_2.
a(4) = 4 since wt(3) = 2 = "10"_2, we find "10" as first bits of yet unused 4 = "100"_2.
a(5) = 5 since wt(4) = 1, and yet unused 5 = "101"_2 starts with 1.
a(6) = 8 since wt(5) = 2 = "10"_2; we see that the yet unused 6 and 7 start with "11"_2, and it isn't until 8 that we have a number that when expressed in binary starts with "10"_2.
a(7) = 6 since wt(8) = 1, we can now apply the yet unused 6 = "110"_2 because it starts with 1, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..16384
• Michael De Vlieger, Plot (n, a(n)) for 1 <= n <= 2^10 color-coded to show wt(a(n-1)), with the first term in the family indicated.
• Michael De Vlieger, Plot (n, a(n)/A007814(a(n))) for 1 <= n <= 2^11, color-coded to show wt(a(n-1)).
• Wikipedia, Hamming weight.
• Wolfram Research, Numbers in Pascal's triangle.
• Index entries for sequences related to binary expansion of n.

Crossrefs

Cf. A000120, A338209, A339607.

Programs

Mathematica

Nest[Append[#, Block[{k = 1, r = IntegerDigits[DigitCount[#[[-1]], 2, 1], 2]}, While[Nand[FreeQ[#, k], Take[IntegerDigits[k, 2], Length@ r] == r], k++]; k]] & @@ {#, Length@ #} &, {1}, 2^7]

Python

def aupto(n):
  alst, used = [1], {1}
  for i in range(2, n+1):
    binprev = bin(alst[-1])[2:]
    binwt = binprev.count("1")
    lsbs, extra = 0, 0
    while binwt + extra in used:
      lsbs += 1
      binwt *= 2
      for extra in range(2**lsbs):
        if binwt + extra not in used: break
    alst.append(binwt+extra); used.add(binwt+extra)
  return alst    # use alst[n-1] for a(n)
print(aupto(68)) # Michael S. Branicky, Dec 16 2020