A080024 Number of divisors d of n such that in binary representation d and n/d have the same number of 1's as n.

1, 2, 0, 3, 0, 0, 0, 4, 1, 0, 0, 0, 0, 0, 0, 5, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0

Offset

1,2

Comments

a(n)<=A000005(n), a(n)=A000005(n) iff n=2^k (A000079).

Not multiplicative. Counterexample: 441=3^2*7^2, a(441)=0, but a(3^2) = a(7^2) = 1. - Christian G. Bower, May 16 2005

Links

G. C. Greubel, Table of n, a(n) for n = 1..5000

Example

Divisors of 36 = {1,2,3,4,6,9,12,18,36}, 36->100100 and also 3,6,12 have two 1's in binary notation with 36 = 3*12 = 6*6, therefore a(36)=3 (18->10010 doesn't count, as 36/18 = 2 -> 10 has only one binary 1).

Mathematica

h[n_] := Total[IntegerDigits[n, 2]]; w[n_] := DigitCount[n, 2, 1]; a[n_] := With[{hn = h[n]}, DivisorSum[ n, Boole[h[#] == hn && w[n/#] == hn]&]]; Array[a, 105] (* Jean-François Alcover, Dec 06 2015, adapted from PARI *)

Prog

(PARI) a(n) = my(hn=hammingweight(n)); sumdiv(n, d, (hammingweight(d) == hn) && (hammingweight(n/d) == hn)); \\ Michel Marcus, Feb 16 2015

Crossrefs

Cf. A007088, A000120, a(A080025(n))>0, a(A080026(n))=1.

Sequence in context: A085199 A338504 A085200 * A348223 A035199 A035229

Adjacent sequences: A080021 A080022 A080023 * A080025 A080026 A080027

Keyword

nonn,base

Author

Reinhard Zumkeller, Jan 21 2003

Status

approved