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

1,2

Comments

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

The sum of these divisors is A380845(n).

Links

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

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.

Example

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.

Mathematica

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

Prog

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

Crossrefs

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

Sequence in context: A242923 A375040 A065704 * A325565 A286552 A324826

Adjacent sequences: A380841 A380842 A380843 * A380845 A380846 A380847

Keyword

nonn,base,easy,new

Author

Amiram Eldar, Feb 05 2025

Status

approved