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Let $e_d$ be the $d$-th standard-basis vector in the Hilbert space $H=l_2(\mathbb{N})$. Let $h(n) = J_2(n)$ be the second Jordan totient function. Define:

$$\phi(n) = \frac{1}{n} \sum_{d|n}\sqrt{h(d)} e_d$$.

Then we have:

$$ \left < \phi(a),\phi(b) \right > = \frac{\gcd(a,b)^2}{ab}=:k(a,b)$$

The vectors $\phi(a_i)$ are linearly independent for each finite set $a_1,\cdots,a_n$ of natural numbers, since

$$\det(G_n) = \prod_{i=1}^n \frac{h(a_i)}{a_i^2} $$ is not zero, where $G_n$ denotes the Gram matrix.

Define:

$$\hat{\phi}(n) := \sum_{d|n} \phi(d) = \frac{1}{n} \sum_{d|n} \sigma(\frac{n}{d})\sqrt{h(d)} e_d$$

Then we have:

$n$ is an odd perfect numbers, if and only if:

$$\left < \hat{\phi}(n),\phi(2) \right > = 1$$

By the triangle inequality we have:

$$|\hat{\phi}(n)| \le \tau(n)$$

where $\tau$ counts the number of divisors of $n$.

Geometric intuition: Since the vectors $\phi(d), d|n$ are almost orthogonal and have norm $1$, we should have by Pythagoras:

$$|\hat{\phi}(n)|^2 \approx \sum_{d|n} |\phi(d)|^2 = \tau(n)$$

A more concrete claim, which I have not been able to prove yet is: $$|\hat{\phi}(n)|^2 \ge \tau(n)$$ for all $n$?

Let $\alpha$ be the angle between $\phi(2)$ and $\hat{\phi}(n)$, where $n$ is an OPN. Then, by Jordans inequality for the $\sin$-e we get after some algebraic manipulation (and using the last claim), the following upper and lower bound for $\tau(n)$ for the OPN $n$:

$$\frac{1}{\sqrt{1-\frac{4\alpha^2}{\pi^2}}} \le \tau(n) \le \frac{1}{1-\alpha^2}$$

However it seems that numerical experiments suggest, that the last inequality can hold only for $n=1$ or $n=$ a prime, which would contradict the OPN property.

My question is, if one can prove the claim.

Also asked on MSE, since it may not be research level: https://math.stackexchange.com/questions/3854989/a-geometric-approach-to-the-odd-perfect-number-problem

Here are some notes with more details of the claims I have written above.

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    $\begingroup$ The question links to a document whose author has the same name as a deleted MathOverflow account, so the downvote may reflect that this new account might also be deleted for similar reasons. $\endgroup$ – Matt F. Oct 7 at 10:01
  • $\begingroup$ @MattF: Ok I understand. $\endgroup$ – stackExchangeUser Oct 7 at 10:09
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    $\begingroup$ I don't understand why some people judge the OP based on social considerations rather than the question based on mathematical criteria. $\endgroup$ – Sylvain JULIEN Oct 8 at 10:22
  • $\begingroup$ @SylvainJULIEN: Thank you for your nice words. $\endgroup$ – stackExchangeUser Oct 8 at 10:34
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We have for all $n$:

$$|\hat{\phi}(n)|^2 = |\sum_{d|n} \phi(d)|^2 = \left < \sum_{d|n} \phi(d),\sum_{d|n} \phi(d)\right >$$ $$= \sum_{d|n} |\phi(d)|^2 + 2 \sum_{d_1 < d_2,d_1,d_2|n} \left < \phi(d_1),\phi(d_2)\right >$$ $$= \tau(n) + 2 \sum_{d_1 < d_2} \frac{\gcd(d_1,d_2)^2}{d_1 d_2}$$ $$\ge \tau(n) + 2\sum_{d_1 < d_2} \frac{1}{d_1 d_2}$$ $$\ge \tau(n) + 2\sum_{d_1 < d_2} \frac{1}{n^2}$$ $$= \tau(n) + \frac{2}{n^2} \frac{\tau(n)(\tau(n)-1)}{2}$$ $$= \tau(n) + \frac{\tau(n)(\tau(n)-1)}{n^2}$$

hence:

$$|\hat{\phi}(n)| \ge \sqrt{\tau(n)(1+\frac{\tau(n)-1}{n^2})}\ge \sqrt{\tau(n)}$$

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