Do all the zeroes of this arithmetic function occur at even integers?

Question

Let $a(n)$ be the number of natural numbers $\le n$ which have an odd number of distinct prime factors. Let $b(n)$ be the number of natural numbers $\le n$ which have an even number of distinct prime factors.

Conjecture: If $a(n) = b(n)$ then $n$ is even.

I have verified this conjecture for each of $8883$ zeroes of $s(n) = a(n)-b(n)$ for $n \le 1.25 \times 10^8$. Is this true in general?

Sage code

n = 1
s = f = 0
target = set = 10^6
while True:
    s = s + (-1)^(len(prime_divisors(n)))
    if s == 0:
        f = f + 1
        if n%2 == 1:
            print "Odd zero", n
            break
    if n > target:
        print n,f
        target = target + set
    n = n + 1

Yes, $$s(n) = -\sum_{k = 1}^{n} (-1)^{\omega(k)},,$$ so $s(n) \equiv n \pmod{2}$. — Daniel Fischer, Dec 26 '19 at 12:38

score 7 · Accepted Answer · answered Dec 26 '19 at 12:38

7

Since $a(n)+b(n)=n$ the claim is clear.

answered Dec 26 '19 at 12:38

lulu

70,402

Do all the zeroes of this arithmetic function occur at even integers?

1 Answers1