A354722 Composite numbers whose divisors have distinct binary weights (A000120).

25, 39, 55, 57, 69, 87, 95, 111, 115, 119, 121, 123, 125, 141, 145, 159, 169, 177, 183, 185, 187, 201, 203, 205, 213, 215, 219, 221, 235, 237, 249, 253, 265, 289, 291, 299, 301, 303, 305, 319, 321, 323, 329, 335, 339, 355, 361, 365, 371, 377, 391, 393, 411, 413

Offset: 1

Author

Amiram Eldar, Jun 04 2022

Keywords

• nonn
• base

Comments

Without the restriction of composite numbers, 1 and all the odd primes would have been terms of this sequence.

Since 1 and 2 have the same binary weight, all the terms are odd.

Examples

			25 is a term since its divisors, 1, 5 and 25, have binary weights 1, 2 and 3, respectively.
55 is a term since its divisors, 1, 5, 11 and 55, have binary weights 1, 2, 3 and 5, respectively.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000120, A027750.

Subsequences: A255401 and A354724.

Programs

Mathematica

bw[n_] := DigitCount[n, 2, 1]; q[n_] := CompositeQ[n] && UnsameQ @@ (bw /@ Divisors[n]); Select[Range[1, 400, 2], q]

PARI

isok(c) = {if ((c>1) && !isprime(c), my(d=divisors(c)); #Set(apply(hammingweight, d)) == #d;);} \\ Michel Marcus, Jun 04 2022

Python

from sympy import divisors
def binwt(n): return bin(n).count("1")
def ok(n):
    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 divs > 2
print([k for k in range(415) if ok(k)]) # Michael S. Branicky, Jun 04 2022