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Let $p_n$ be the $n$th prime number and all variables, unless otherwise specified, are natural numbers.

Conjecture: For all square-free $n \geq 2$, the following function evaluates to a positive integer: $$F(n) = -1 + \sum_{d \mid n} \mu(d) \sum_{a^2 = 1\mod d, \ \\ 0 \leq a \lt d}\left\lfloor\frac{\frac{12}{5} p_{\text{max}}(n)-a}{d} \right\rfloor$$ where $p_{\text{max}}(n)$ is the maximum prime divisor of $n$. And this conjecture is tight with respect to the constant $C = \frac{12}{5}$. Any deviation to a lower value will yield a false conjecture.

Verification Code:

from sympy import *

N = 5000
epsilon =  0.00000000    # Try setting to 0.000000001


for n in range(2, N):
   if all(j == 1 for j in factorint(n).values()):  # n is square-free
      m = ((12 / 5) - epsilon)* max(primefactors(n))
      S = -1
      for d in divisors(n):
         for a in range(0, d):
            if (a ** 2 - 1) % d == 0:
               S += (-1) ** primeomega(d) * floor((m - a)/d)

      assert S > 0
      print(m,n, S)

What is interesting is that $\{ \bar{a} \in \Bbb{Z}/d : a^2 = 1\mod d\} \leqslant (\Bbb{Z}/d)^{\times}$ is a subgroup of the units modulo $d$ and you can't change that either unless you adjust $C$.

Anyway, was wanting some information about sums of this type. There's not much data on the web that I could find about strictly sums over divisors of square-free integers. When you remove the square-free requirement, the summation goes chaotic, i.e. takes on negative or positive values. Are there any good papers out there about this?

Attempt. This doesn't attempt to answer the question but instead partially prove the result for when $n = q$ is a prime number. Let $[z]_q =$ the usual (least, non-negative) residue of $z$ modulo $q$, for any $z \in \Bbb{Q}$.

$$\begin{align} F(q) &= -1 + \sum_{d \in \{1,q\}}\mu(d) \sum_{a^2 = 1 \pmod d} \lfloor\frac{\frac{12}{5} q - a}{d} \rfloor \\ &= -1 + \frac{12}{5}q - \lfloor\frac{\frac{12}{5} q - 1}{q} \rfloor - \lfloor\frac{\frac{12}{5} q - (q-1)}{q} \rfloor \\ &= -1 + \frac{12}{5}q - \frac{\frac{12}{5}q - 1 - [\frac{12}{5}q - 1]_q}{q} - \frac{\frac{12}{5}q - (q-1) - [\frac{12}{5}q - (q-1)]_q}{q} \\ &= -1 + \frac{12}{5} q - \frac{12}{5} +\frac{1}{q} +\frac{[\frac{12}{5}q - 1]_q}{q} -\frac{12}{5} +1 -\frac{1}{q} + \frac{[\frac{12}{5}q + 1]_q}{q} \\ &= \frac{12}{5}q - \frac{24}{5} + \frac{[\frac{12}{5} q - 1]_q}{q} + \frac{[\frac{12}{5} q + 1]_q}{q} \end{align}$$

Now plug in $q = 2$. We have:

$$ \frac{24}{5} - \frac{24}{5} + \text{ two numbers that can't both be zero} \gt 0 $$

since the function can only (clearly) take on integer values, we have that $F(2) \geq 1$. By induction, for larger primes, the value is clearly also $\geq 1$. $\blacksquare$

What remains to be proven is that it's true for $F(n)$ with $\Omega(n) \gt 1$, $n$ square-free.

Shaun
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Daniel Donnelly
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    What is meant by " tight with respect to a constant " ? – mick Nov 26 '23 at 19:16
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    @mick that is explained in the post. But basically, if you change the constant in a certain direction, then the conjecture no longer holds, by any change amount that is. But if you keep the constant as stated, the conjecture holds. But this is not a good formula to use if you're starting out (I saw your other post). By Sieve of Eratosthenes, the largest number we can plugin in place of $12/5 p_n$ is $p_{n+1}^2 - 2$. That's for $k = 1$ sep primes. For general $k$, it's $p_{n + 1}^2 - -(k + 1)$. The form I posted about in your answer puts in $n$ in place of $p_{n + 1}^2 - (k+1)$ ... – Daniel Donnelly Nov 26 '23 at 19:21
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    $n$ and so then we take the primorial of $\sqrt{n + k}$ in the summation bound. That simply means $p_1 \cdot p_2 \cdot p_3 \cdots p_{\pi\sqrt{n + k}}$. Where $\pi$ is the prime counting function. So that is in English the product of all primes $\leq \sqrt{n + k}$. – Daniel Donnelly Nov 26 '23 at 19:26
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    The above post I made is just noticing that we can take $b$ much smaller than $p_{n + 1}^2 - (k + 1)$ namely down to $12/5 p_n$ at least. Since it is definitely smaller than $p_{n + 1}^2 - (k+ 1)$, it would imply the higher $b$ version. Reason is by construction of the formula. When you derive it you see that it's an inclusion-exclusion set counting function. – Daniel Donnelly Nov 26 '23 at 19:30

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