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

Author

Amiram Eldar, Feb 05 2025

Keywords

• nonn
• base
• easy

Comments

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

The sum of these divisors is A380845(n).

Examples

			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.
		

Links

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

Crossrefs

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

Programs

Mathematica

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

PARI

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

Formula

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.