A354724 Numbers k whose ordered binary weights (A000120) of their divisors are the numbers 1 to A000005(k).

1, 3, 5, 17, 25, 39, 57, 69, 145, 201, 257, 265, 289, 291, 323, 393, 579, 1075, 1083, 2307, 2645, 2875, 4205, 4503, 5555, 5593, 7955, 8815, 9399, 9401, 9519, 11033, 11155, 11407, 12297, 12455, 12711, 12909, 13205, 13281, 13611, 13737, 14001, 14915, 15879, 16629

Offset: 1

Author

Amiram Eldar, Jun 04 2022

Keywords

• nonn
• base

Comments

The subsequence of terms whose divisors have binary weights in order is A255401.

All terms are odd. Proof: All binary weights must be distinct but the binary weights of k and 2*k are equal. A contradiction. - David A. Corneth, Jun 04 2022

Examples

			3 is a term since its divisors, 1 and 3, have binary weights 1 and 2, respectively.
69 is a term since its divisors, 1, 3, 23 and 69, have binary weights 1, 2, 4 and 3, respectively.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..2475

Crossrefs

Cf. A000005, A000120, A354722.

A255401 is a subsequence.

Programs

Mathematica

bw[n_] := DigitCount[n, 2, 1]; q[n_] := Module[{d = Divisors[n]}, Union[bw /@ d] == Range[Length[d]]]; Select[Range[1, 10^4, 2], q]

Python

from sympy import divisors
def binwt(n): return bin(n).count("1")
def ok(n):
    if n%2 == 0: return False
    binwts, divs = set(), 0
    for d in divisors(n, generator=True):
        b = binwt(d)
        if b in binwts: return False
        binwts.add(b)
        divs += 1
    return binwts == set(range(1, divs+1))
print([k for k in range(20000) if ok(k)]) # Michael S. Branicky, Jun 04 2022