A383364 -id:A383364 - cp's OEIS frontend

A383363 Composite numbers k all of whose proper divisors have binary weights that are not equal to the binary weight of k.

15, 25, 27, 39, 51, 55, 57, 63, 69, 77, 81, 85, 87, 91, 95, 99, 111, 115, 117, 119, 121, 123, 125, 141, 143, 145, 147, 159, 169, 171, 175, 177, 183, 185, 187, 201, 203, 205, 207, 209, 213, 215, 219, 221, 231, 235, 237, 243, 245, 247, 249, 253, 255, 261, 265, 275

Offset: 1

Amiram Eldar, Apr 24 2025

First differs from A325571 at n = 56: A325571(56) = 267 is not a term of this sequence. Differs from A325571 by having the terms 16849, 35235, 101265, 268357, 295717, ..., and not having the terms 267, 295, 327, 387, 395, ... .
Composite numbers k such that A380844(k) = 1.
All the odd primes p have A380844(p) = 1.
All the terms are odd numbers since for an even number k, A000120(k) = A000120(k/2).

			15 = 3 * 5 is a term since it is composite, and its binary weight, A000120(15) = 4 is different from the binary weights of its proper divisors: A000120(1) = 1, A000120(3) = 2, and A000120(5) = 2.

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

Cf. A000120, A325571, A380844, A383364, A383365.

q[k_] := CompositeQ[k] && DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; Select[Range[1, 300, 2], q]

isok(k) = if(k == 1 || isprime(k), 0, my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) == 1);

A383365 Numbers k with a record number of proper divisors, where all of them have binary weights that are different from the binary weight of k.

Original entry on oeis.org

1, 3, 15, 63, 231, 405, 495, 1485, 2475, 4095, 14175, 21735, 24255, 31185, 79695, 190575, 218295, 239085, 294525, 904365, 1276275, 2789325, 3586275, 4937625, 6912675, 10072755, 17342325, 17972955, 26801775, 46621575, 80405325, 192567375, 326351025, 333107775, 654729075

Offset: 1

Amiram Eldar, Apr 24 2025

The corresponding record values are 0, 1, 3, 5, 7, 9, 11, 15, 17, ... (see the link for more values).

All terms are odd as an even number k has proper divisor k/2 with the same binary weight. - David A. Corneth, Apr 24 2025

			a(1) = 1 since 1 has no proper divisors.
a(2) = 3 since 3 has one proper divisor, 1, and A000120(1) = 1 != A000120(3) = 2, while 2 also has one proper divisor, 1, but A000120(2) = A000120(1) = 1.
a(3) = 15 since 15 has 3 proper divisors, 1, 3 and 5, and A000120(1) = 1 and A000120(3) = A000120(5) = 2 are different from A000120(15) = 4. All the numbers below 15 have fewer proper divisors with this property.

David A. Corneth, Table of n, a(n) for n = 1..91 (first 46 terms from Amiram Eldar)
Amiram Eldar, Table of n, a(n), A032741(a(n)) for n = 1..46.

Cf. A000120, A032741, A380844, A383363, A383364.

q[k_] := DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; seq[kmax_] := Module[{s = {}, d, dm = 0}, Do[d = DivisorSigma[0, k]; If[d > dm && q[k], dm = d; AppendTo[s, k]], {k, 1, kmax}]; s]; seq[10^5]

is1(k) = {my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) ==  1};
list(kmax) = {my(d, dm = 0); for(k = 1, kmax, d = numdiv(k); if(d > dm && is1(k), dm = d; print1(k, ", ")));}

cp's OEIS Frontend

A383363 Composite numbers k all of whose proper divisors have binary weights that are not equal to the binary weight of k.

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