OFFSET
1,2
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
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 1..10000
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)
EXAMPLE
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.
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 *)
PROG
(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
(PARI) a(n) = sumdiv(n, d, d*(bitor(n, d)==n)); \\ Michel Marcus, Sep 07 2014
CROSSREFS
KEYWORD
nonn,base
AUTHOR
N. J. A. Sloane, Sep 06 2014
STATUS
approved