Why is this alternating summation involving reciprocals of divisors always positive? A conjecture.

Question

Conjecture. For all $n,m \in \Bbb{N}$,

$$ f(n, m) := \sum_{c\mid d\mid n\# \\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\omega(d)}}{d} $$

is greater than $0$.

Example verification code:

from sympy import *

def reciprocal_sum(n, m):
    s = 0
    for d in divisors(n):
        t = 0
        for c in divisors(d):
             if gcd(c, m) == 1:
                  t += 1/d
        s += (-1)**primeomega(d) * t
    return s

N = 30

for m in range(1, N):
     for n in range(1, N):
         n = primorial(n, nth=False)   # Note primorial of n used
         m = 2*m
         F = reciprocal_sum(n, m)
         assert(F > 0)
         print(F)

As you can see, the conjecture is water-tight. How would we prove such a thing?

Note that you cannot replace $n\#$ with general $n$ as the you will get negative or $0$ values.

Also you can't change $2m$ in the $\gcd$ to just $m$ or you will likewise get negative values. But with this combination of things, the sum always seems to be positive.

---

The code prints:

1.0
0.5
0.16666666666666669
0.16666666666666669
0.1
0.1
0.07142857142857151
0.07142857142857151
0.07142857142857151
0.07142857142857151
0.05844155844155851
0.05844155844155851
0.04945054945054941
0.04945054945054941
0.04945054945054941
0.04945054945054941
0.043632837750484876
0.043632837750484876
0.039039907460960095
0.039039907460960095
0.039039907460960095
0.039039907460960095
0.035645132899137504
0.035645132899137504

until it gets too spatially complex to compute further primorials. Anyway, take away any of the ingredients, and this special summation fails to be always positive.

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Attempt. It's true for $\Omega(n\#) =: n' = 0$ and for all $m \in \Bbb{N}$. Exercise. Prove this.

Now assume that it's true for all $0 \leq \Omega(n\#) \lt r$, then in the inductive case that $\Omega((n+i)\#) = \Omega(qn\#) = r, \ q$ some prime, we have that:

$$ f(n + i,m) = \sum_{c \mid d \mid (qn\#)\\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\Omega(d)}}{d} \\ = \sum_{c \mid d \mid n\# \\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\Omega(d)}}{d} + \sum_{d \mid n\#} \sum_{c \mid (qd) \\ \gcd(c, 2m) = 1}\dfrac{(-1)^{\omega(qd)}}{qd} \\ = \sum_{c \mid d \mid n\# \\ \gcd(c, 2m)=1}\dfrac{(-1)^{\Omega(d)}}{d} + \sum_{d \mid n\#} \left(\sum_{c \mid d \\ \gcd(c,2m)=1}\dfrac{(-1)^{\Omega(qd)}}{qd} + \sum_{c \mid d \\ \gcd(qc,2m)=1}\dfrac{(-1)^{\Omega(qd)}}{qd} \right) \\ = \sum_{c \mid d \mid n\# \\ \gcd(c, 2m)=1}\dfrac{(-1)^{\Omega(d)}}{d} - \dfrac{1}{q}\sum_{d \mid n\#} \left(\sum_{c \mid d \\ \gcd(c,2m)=1}\dfrac{(-1)^{\Omega(d)}}{d} + \sum_{c \mid d \\ \gcd(qc,2m)=1}\dfrac{(-1)^{\Omega(d)}}{d} \right) $$

That is the previous case times a factor of $(1-\dfrac{1}{q})$ (which is always positive), minus $\dfrac{1}{q}\sum_{c \mid d \mid n\# \\ \gcd(qc, 2m) = 1} \dfrac{(-1)^{\Omega(d)}}{d}$. There are two cases: $q \mid m$ and $q \nmid m$. When $q \mid m$ clearly, the last spoken of summation term is $0$ and we're done. So assume that $q\nmid m$. Then $\{c \mid d : \gcd(qc, 2m) = 1\} = \{c \mid d: \gcd(c, 2m) = 1\}$. In which case the summation is $\dfrac{-1}{q} \times$ the previous case so that, we have the total summation is $\text{previous case} \times (1 - \dfrac{2}{q})$ which positive if and only if $1 - \dfrac{2}{q} \gt 0 \iff q \gt 2$. If on the other hand $q = 2$, then similarly in the case above, we have a summation value of $0$ in this last summation term.

In more mathematical terms we have that:

$$ f(n, m) = \prod_{q \mid n\# \\ q \text{ prime}} \left(\begin{cases}(1 - \dfrac{1}{q}) \text{ if } \gcd(q, 2m) \neq 1 \\ (1 - \dfrac{2}{q}) \text{ if } \gcd(q, 2m) = 1 \end{cases}\right) $$

which is always positive because if $q = 2$ then it falls to the first case, and if $q \gt 2$ then it falls to either case which will still be a positive number in the second case, since $q \gt 2$.

I think that's a QED. Please confirm. I started this as an attempt, but it ended in an answer.