A380845 The sum of divisors of n that have the same binary weight as n.

1, 3, 3, 7, 5, 9, 7, 15, 12, 15, 11, 21, 13, 21, 15, 31, 17, 36, 19, 35, 28, 33, 23, 45, 25, 39, 27, 49, 29, 45, 31, 63, 36, 51, 42, 84, 37, 57, 39, 75, 41, 84, 43, 77, 60, 69, 47, 93, 56, 75, 51, 91, 53, 81, 55, 105, 57, 87, 59, 105, 61, 93, 63, 127, 70, 108

Offset: 1

Author

Amiram Eldar, Feb 05 2025

Keywords

• nonn
• base
• easy

Comments

The number of these divisors is A380844(n).

Examples

			a(6) = 9 because 6 = 110_2 has binary weight 2, 2 of its divisors, 3 = 11_2 and 6, have the same binary weight, and 3 + 6 = 9.
		

Links

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

Crossrefs

Cf. A000120, A000203, A000265, A000396, A000668, A007814, A038712, A380844.

Programs

Mathematica

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

PARI

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

Formula

a(n) = Sum_{d|n} d * [A000120(d) = A000120(n)], where [ ] is the Iverson bracket.

a(2^n) = 2^(n+1) - 1.

a(n) <= A000203(n) with equality if and only if n is a power of 2.

a(n) = a(A000265(n)) * (2^(A007814(n)+1)-1) = a(A000265(n)) * A038712(n), or equivalently, a(k*2^n) = a(k)*(2^(n+1)-1) for k odd and n >= 0.

In particular, since a(p) = p for an odd prime p, a(p*2^n) = p*(2^(n+1)-1) for an odd prime p and n >= 0.

a(A000396(n)) = A000668(n)^2, assuming that odd perfect numbers do no exist.