A246601 Sum of divisors d of n with property that the binary representation of d can be obtained from the binary representation of n by changing any number of 1's to 0's.

1, 2, 4, 4, 6, 8, 8, 8, 10, 12, 12, 16, 14, 16, 24, 16, 18, 20, 20, 24, 22, 24, 24, 32, 26, 28, 40, 32, 30, 48, 32, 32, 34, 36, 36, 40, 38, 40, 43, 48, 42, 44, 44, 48, 60, 48, 48, 64, 50, 52, 72, 56, 54, 80, 61, 64, 58, 60, 60, 96, 62, 64, 104, 64, 66, 68, 68, 72, 70, 72

Offset: 1

Author

N. J. A. Sloane, Sep 06 2014

Keywords

• nonn
• base

Comments

Equivalently, the sum of the divisors d of n such that the bitwise OR of d and n is equal to n. - Chai Wah Wu, Sep 06 2014

Equivalently, the sum of the divisors d of n such that the bitwise AND of n and d is equal to d. - Amiram Eldar, Dec 15 2022

Examples

			12 = 1100_2; only the divisors 4 = 0100_2 and 12 = 1100_2 satisfy the condition, so(12) = 4+12 = 16.
15 = 1111_2; all divisors 1,3,5,15 satisfy the condition, so a(15)=24.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000203, A075708, A246600.

Programs

Maple

with(numtheory);
sd:=proc(n) local a,d,s,t,i,sw;
s:=convert(n,base,2);
a:=0;
for d in divisors(n) do
sw:=-1;
t:=convert(d,base,2);
for i from 1 to nops(t) do if t[i]>s[i] then sw:=1; fi; od:
if sw<0 then a:=a+d; fi;
od;
a;
end;
[seq(sd(n),n=1..100)];

Mathematica

a[n_] := DivisorSum[n, #*Boole[BitOr[#, n] == n] &]; Array[a, 100] (* Jean-François Alcover, Dec 02 2015, adapted from PARI *)

PARI

a(n) = sumdiv(n, d, d*(bitor(n,d)==n)); \\ Michel Marcus, Sep 07 2014

Python

from sympy import divisors
def A246601(n):
    return sum(d for d in divisors(n) if n|d == n)
# Chai Wah Wu, Sep 06 2014

Formula

a(2^i) = 2^i.

a(odd prime p) = p+1.

From Amiram Eldar, Dec 15 2022: (Start)

a(2*n) = 2*a(n), and therefore a(m*2^k) = 2^k*a(m) for m odd and k>=0.

a(2^n-1) = sigma(2^n-1) = A075708(n). (End)

a(n) = Sum_{d|n} d*(binomial(n,d) mod 2). - Ridouane Oudra, Apr 09 2025