A324392 -id:A324392 - cp's OEIS frontend

A380844 The number of divisors of n that have the same binary weight as n.

1, 2, 1, 3, 1, 2, 1, 4, 2, 2, 1, 3, 1, 2, 1, 5, 1, 4, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 2, 2, 2, 6, 1, 2, 1, 4, 1, 4, 1, 3, 2, 2, 1, 5, 2, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 2, 4, 1, 3, 1, 4, 1, 8, 1, 2, 2, 3, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1

Offset: 1

Amiram Eldar, Feb 05 2025

First differs from A325565 at n = 133: a(133) = 3 while A325565(133) = 2.

The sum of these divisors is A380845(n).

			a(6) = 2 because 6 = 110_2 has binary weight 2, and 2 of its divisors, 3 = 11_2 and 6, have the same binary weight.

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

Cf. A000005, A000043, A000120, A000265, A000396, A007814, A324392, A325565, A380845.

a[n_] := Module[{h = DigitCount[n, 2, 1]}, DivisorSum[n, 1 &, DigitCount[#, 2, 1] == h &]]; Array[a, 100]

a(n) = {my(h = hammingweight(n)); sumdiv(n, d, hammingweight(d) == h);}

a(n) = Sum_{d|n} [A000120(d) = A000120(n)], where [ ] is the Iverson bracket.
a(2^n) = n+1.
a(n) <= A000005(n) with equality if and only if n is a power of 2.
a(n) = a(A000265(n)) * (A007814(n)+1), or equivalently, a(k*2^n) = a(k)*(n+1) for k odd and n >= 0.
In particular, since a(p) = 1 for an odd prime p, a(p*2^n) = n+1 for an odd prime p and n >= 0.
a(A000396(n)) = A000043(n), assuming that odd perfect numbers do no exist.

A324393 a(n) is the number of such divisors d of n that A000120(d) does not divide n, where A000120(d) gives the binary weight of d.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 2, 3, 0, 1, 0, 1, 0, 1, 2, 1, 0, 2, 2, 3, 3, 1, 2, 1, 0, 2, 0, 3, 0, 1, 2, 2, 0, 1, 0, 1, 3, 5, 2, 1, 0, 2, 2, 3, 3, 1, 2, 2, 4, 2, 2, 1, 0, 1, 2, 3, 0, 3, 0, 1, 0, 2, 4, 1, 0, 1, 2, 4, 3, 3, 2, 1, 0, 3, 2, 1, 0, 3, 2, 3, 4, 1, 4, 3, 0, 3, 2, 3, 0, 1, 4, 4, 3, 1, 2, 1, 4, 4

Offset: 1

Antti Karttunen, Mar 05 2019

Number of such positive integers k that divide n but A000120(k) [the Hamming weight of k] does not divide n.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences related to binary expansion of n

Cf. A000005, A000120, A324392, A306263 (positions of zeros).

Cf. also A093653, A192895, A266342, A292257.

a[n_] := DivisorSum[n, 1 &, !Divisible[n, DigitCount[#, 2, 1]] &]; Array[a, 100] (* Amiram Eldar, Dec 04 2020 *)

A324393(n) = sumdiv(n, d, !!(n%hammingweight(d)));

a(n) = Sum_{d|n} [A000120(d) does not divide n], where [ ] is the Iverson bracket.
a(n) = A000005(n) - A324392(n).
a(p) = 1 for all odd primes p.

cp's OEIS Frontend

A380844 The number of divisors of n that have the same binary weight as n.

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