A093653 Total number of 1's in binary expansion of all divisors of n.

1, 2, 3, 3, 3, 6, 4, 4, 5, 6, 4, 9, 4, 8, 9, 5, 3, 10, 4, 9, 9, 8, 5, 12, 6, 8, 9, 12, 5, 18, 6, 6, 8, 6, 9, 15, 4, 8, 10, 12, 4, 18, 5, 12, 15, 10, 6, 15, 7, 12, 9, 12, 5, 18, 11, 16, 10, 10, 6, 27, 6, 12, 17, 7, 8, 16, 4, 9, 10, 18, 5, 20, 4, 8, 16, 12, 11, 20, 6, 15, 12, 8, 5, 27, 9, 10, 12

Offset: 1

Author

Jason Earls, May 16 2004

Keywords

• base
• easy
• nonn

Examples

			a(8) = 4 because the divisors of 8 are [1, 2, 4, 8] and in binary: 1, 10, 100, 1000, so four 1's.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384 (first 500 terms from Jaroslav Krizek)
• Maxwell Schneider and Robert Schneider, Digit sums and generating functions, arXiv:1807.06710 [math.NT], 2018. See (22) p. 6.
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000120, A093687, A192895, A292257.

Cf. A226590 (number of 0's in binary expansion of all divisors of n).

Cf. A182627 (number of digits in binary expansion of all divisors of n).

Cf. A034690 (a decimal equivalent).

Programs

Maple

a:= n-> add(add(i, i=Bits[Split](d)), d=numtheory[divisors](n)):
seq(a(n), n=1..100);  # Alois P. Heinz, May 17 2022

Mathematica

Table[Plus@@DigitCount[Divisors[n], 2, 1], {n, 75}] (* Alonso del Arte, Sep 01 2013 *)

PARI

A093653(n) = sumdiv(n,d,hammingweight(d)); \\ Antti Karttunen, Dec 14 2017

PARI

a(n) = {my(v = valuation(n, 2), n = (n>>v)); sumdiv(n, d, hammingweight(d)) * (v + 1)} \\ David A. Corneth, Feb 15 2023

Python

from sympy import divisors
def a(n): return sum(bin(d).count("1") for d in divisors(n))
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Apr 20 2022

Python

from sympy import divisors
def A093653(n): return sum(d.bit_count() for d in divisors(n, generator=True))
print([A093653(n) for n in range(1, 88)]) # Michael S. Branicky, Feb 15 2023

Formula

a(n) = Sum_{k = 0..n} if(mod(n, k) = 0, A000120(k), 0). - Paul Barry, Jan 14 2005

a(n) = A182627(n) - A226590(n). - Jaroslav Krizek, Sep 01 2013

a(n) = A292257(n) + A000120(n). - Antti Karttunen, Dec 14 2017

From Bernard Schott, May 16 2022: (Start)

If prime p = A000043(n), then a(2^p-1) = a(A000668(n)) = p+1 = A050475(n).

a(2^n) = n+1 (End)