For odd numbers $n$ let:

$$a(n) = \sum_{d^2|n} d \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$$

where $\sigma^*(k) = $ sum of unitary ($\gcd(d,k/d)=1$) divisors of $k$ and $\omega$ counts the prime divisors.

Is this function multiplicative for odd numbers, and if so what is its value at odd prime powers?

Edit: Sorry for not giving any context when asking the question, which I will do now:

Let $n$ be a natural number. Let $U_n = \{d \in \mathbb{N}| d|n \text{ and } \gcd(d,n/d)=1 \}$ be the set of unitary divisors, $D_n$ be the set of divisors and $S_n=\{d \in \mathbb{N}|d^2 | n\}$ be the set of square divisors of $n$.

The set $U_n$ is a group with $a\oplus b := \frac{ab}{\gcd(a,b)^2}$. It operates on $D_n$ via:

$$ u \oplus d := \frac{ud}{\gcd(u,d)^2}$$

The orbits of this operation "seem" to be

$$ U_n \oplus d = d \cdot U_{\frac{n}{d^2}} \text{ for each } d \in S_n$$

From this conjecture it follows (also one can prove this directly since both sides are multiplicative and equal on prime powers):

$$\sigma(n) = \sum_{d\in S_n} d\sigma^*(\frac{n}{d^2})$$

where $\sigma^*$ denotes the sum of unitary divisors.

Since $\sigma^*(k)$ is divisible by $2^{\omega(k)}$ if $k$ is odd, where $\omega=$ counts the number of distinct prime divisors of $k$, for an odd perfect number $n$ we get (Let now $n$ be an odd perfect number):

$$2n = \sigma(n) = \sum_{d \in S_n} d \sigma^*(\frac{n}{d^2}) = \sum_{d \in S_n} d 2^{\omega(n/d^2)} k_d $$

where $k_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$ are natural numbers. Let $\hat{d}$ be the largest square divisor of $n$. Then: $\omega(n/d^2)\ge \omega(n/\hat{d}^2)$.

Hence we get:

$$2n = 2^{\omega(n/\hat{d}^2)} \sum_{d \in S_n} d l_d$$ for some natural numbers $l_d$.

If the prime $2$ divides not the prime power $2^{\omega(n/\hat{d}^2})$, we must have $\omega(n/\hat{d}^2)=0$ hence $n=\hat{d}^2$ is a square number, which is in contradiction to Eulers theorem on odd perfect numbers.

So the prime $2$ must divide the prime power $2^{\omega(n/\hat{d}^2})$ and we get:

$$n = 2^{\omega(n/\hat{d}^2)-1} \sum_{d \in S_n} d l_d$$

with $l_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$. Hence the odd perfect number, satisifies:

$$n = \sum_{d^2|n} d \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}=a(n)$$

Hence an odd perfect number satisifies:

$$n = a(n)$$

So my idea was to study the function on the right hand side and what properties it has to maybe derive insights into odd perfect numbers.

The real question is if it ever can happen that an odd number $n$ satisfies: $n=a(n)$? (checked for $n=2k+1$ and $1 \le k \le 10^7$)

  • $\begingroup$ Yes, it is multiplicative for odd numbers. How did you come up with this function? $\endgroup$ – Somos Aug 28 '20 at 22:39
  • $\begingroup$ @Somos: I updated the question with some context. $\endgroup$ – user276611 Aug 29 '20 at 6:14

The function $a(n)$ is multiplicative. If we define \begin{align*} f(m) &= \begin{cases} \sqrt m, &\text{if $m$ is a perfect square}, \\ 0, &\text{otherwise}, \end{cases} \\ g(m) &= \frac{\sigma^*(m)}{2^{\omega(m)}}, \end{align*} then it's reasonably easy to see that both $f$ and $g$ are multiplicative functions and that $$ a(n) = \sum_{k\mid n} f(k) g(n/k). $$ In other words $a=f*g$ is a Dirichlet convolution of two multiplicative functions and is therefore multiplicative. Its value at the prime power $p^j$ is $$ \sum_{d^2\mid p^j} d \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}} = \sum_{i=0}^{\lfloor j/2\rfloor} p^i \frac{\sigma^*(p^{j-2i})}{2^{\omega(p^{j-2i})}}. $$ If $j$ is odd then this is $$ \frac12 \sum_{i=0}^{\lfloor j/2\rfloor} p^i \sigma^*(p^{j-2i}) = \frac12 \sum_{i=0}^{\lfloor j/2\rfloor} p^i (p^{j-2i}+1) = \frac12\sum_{i=0}^{(j-1)/2} (p^{j-i}+p^i), $$ which can be evaluated via finite geometric series; a similar calculation will cover the case where $j$ is even (careful with the $2^\omega$ term when $i=\frac j2$).

  • $\begingroup$ Thanks @GregMartin for your answer! I did a similar calculation and come to the same result. $\endgroup$ – user276611 Aug 29 '20 at 8:27

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