cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Amiram Eldar, Apr 24 2025

Keywords

Comments

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).

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    q[k_] := CompositeQ[k] && DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; Select[Range[1, 300, 2], q]
  • PARI
    isok(k) = if(k == 1 || isprime(k), 0, my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) == 1);

A383364 a(n) is the least number k with exactly n 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, 25, 15, 81, 63, 15625, 231, 1225, 405, 59049, 495, 531441, 5103, 2025, 1485, 33232930569601, 2475, 3814697265625, 6237, 18225, 295245, 31381059609, 4095, 1500625, 2657205, 81225, 25515, 22876792454961, 14175, 931322574615478515625, 21735, 31236921, 301327047
Offset: 0

Views

Author

Amiram Eldar, Apr 24 2025

Keywords

Examples

			a(0) = 1 since 1 has no proper divisors.
a(1) = 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.
		

Crossrefs

Programs

  • Mathematica
    q[k_] := DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; seq[len_] := Module[{s = Table[0, {len}], c = 0, k = 1, i}, While[c < len, i = DivisorSigma[0, k]; If[i <= len && s[[i]] == 0 && q[k], c++; s[[i]] = k]; k++]; s]; seq[16]
  • PARI
    is1(k) = {my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) ==  1};
    list(len) = {my(s = vector(len), c = 0, k = 1, i); while(c < len, i = numdiv(k); if(i <= len && s[i] == 0 && is1(k), c++; s[i] = k); k++); s;}
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