## co.combinatorics – Is the order of the Coxeter matrix even when the number of points is odd?

Let $$P$$ be a finite connected poset.
The Cartan matrix $$C_P$$ of $$P$$ is defined as the matrix with entries $$c_{i,j}=1$$ if $$i leq j$$ and $$c_{i,j}=0$$ else for $$i,j in P$$.
The Coxeter matrix of $$P$$ is defined as $$M_P:=-C_P^{-1}C_P^T$$. The order of a matrix $$M$$ is defined as the smallest $$k$$ such that $$M^k=id$$.

Question 1: When the poset $$P$$ has $$n$$ points for $$n$$ odd, does $$M_P$$ have even order when the order is finite?

This is true for all posets with at most 7 points and also for $$n=9$$ when $$P$$ is a lattice.

Question 2: Is there a good (maybe even optimal) bound for the maximal (finite) order of the Coxeter matrix of a poset with $$n geq 2$$ points? Is it true that the order is bounded by $$n^2$$?

For $$n=2,…,7$$ the sequence of largest periods starts with 3,6,6,12,15,30.

Question 3: Is it more generally true that the Coxeter matrix of a finite dimensional algebra $$A$$ with finite global dimension has even order when $$A$$ has an odd number of simple modules?

This has a positive answer for all Nakayama algebras with at most 7 simples.

## set theory – Odd Steinhaus problem for finite sets

Call a finite subset $$S$$ of the plane with an even number of points an odd Jackson set, if there is an $$Asubset mathbb R^2$$ such that $$A$$ meets every congruent copy of $$S$$ in an odd number of points.

Are there any odd Jackson sets?

For a bit of history of related questions (and an explanation of the nomenclature), see here.
For my motivation, see here.

## Probalitity of selecting even or odd numbers from a random sample?

Regarding a lottery, the winning numbers are determined by selecting 6 different numbers from the first 17 positive integers at random. What is the probability that there will be more even numbers than odd numbers drawn?

## encryption – Odd, worrying alert. What if I use same AppleID with different MacOS logins? Safe (in terms of security and data corruption) to do?

So I have an install of Catalina that isn’t my main installation, and doesn’t have a copy of my main login on it, hence I’m logged in using another admin-level account (shown) with a different password. I’m getting this alert in System Preferences..Apple ID.

Safe (in terms of security and data corruption) to use same AppleID with different MacOS logins? I’m wondering / thinking of how this could get messed up… Not to mention what about the case of same userid and Apple ID, but different passwords? So strange. Haven’t found a good explanation of what Apple is doing here…

## nt.number theory – On a variant of the nearest-square function and the quantity \$m^2 – p^k\$ where \$p^k m^2\$ is an odd perfect number

This question has been cross-posted from this MSE question and is an offshoot of this other MSE question.

(Note that MSE user mathlove has posted an answer in MSE, which I could not completely understand. I have therefore cross-posted this question in MO, hoping the sages here would be able to give some enlightenment. I hope this is okay, and that the question is research-level.)

Let $$n = p^k m^2$$ be an odd perfect number with special prime $$p$$ satisfying $$p equiv k equiv 1 pmod 4$$ and $$gcd(p,m)=1$$.

It was conjectured in Dris (2018) and Dris (2012) that the inequality $$p^k < m$$ holds.

Brown (2016) showed that the Dris Conjecture (that $$p^k < m$$) holds in many cases.

It is trivial to show that $$m^2 – p^k equiv 0 pmod 4$$. This means that $$m^2 – p^k = 4z$$, where it is known that $$4z geq {10}^{375}$$. (See this MSE question and answer, where the case $$m < p^k$$ is considered.) Note that if $$p^k < m$$, then $$m^2 – p^k > m^2 – m = m(m – 1),$$
and that
$${10}^{1500} < n = p^k m^2 < m^3$$
where the lower bound for the magnitude of the odd perfect number $$n$$ is due to Ochem and Rao (2012). This results in a larger lower bound for $$m^2 – p^k$$. Therefore, unconditionally, we have
$$m^2 – p^k geq {10}^{375}.$$
We now endeavor to disprove the Dris Conjecture.

Consider the following sample proof arguments:

Theorem 1 If $$n = p^k m^2$$ is an odd perfect number satisfying $$m^2 – p^k = 8$$, then $$m < p^k$$.

Proof

Let $$p^k m^2$$ be an odd perfect number satisfying $$m^2 – p^k = 8$$.

Then $$(m + 3)(m – 3) = m^2 – 9 = p^k – 1.$$

This implies that $$(m + 3) mid (p^k – 1)$$, from which it follows that
$$m < m + 3 leq p^k – 1 < p^k.$$
We therefore conclude that $$m < p^k$$.

QED

Theorem 2 If $$n = p^k m^2$$ is an odd perfect number satisfying $$m^2 – p^k = 40$$, then $$m < p^k$$.

Proof

Let $$p^k m^2$$ be an odd perfect number satisfying $$m^2 – p^k = 40$$.

Then $$(m+7)(m-7)=m^2 – 49=p^k – 9,$$
from which it follows that
$$(m+7) mid (p^k – 9)$$
which implies that
$$m < m+7 leq p^k – 9 < p^k.$$

QED

Note that $$49$$ is not the nearest square to $$40$$ ($$36$$ is), but rather the nearest square larger than $$40$$.

With this minor adjustment in the logic, I would expect the general proof argument to work.

(Additionally, note that it is known that $$m^2 – p^k$$ is not a square, if $$p^k m^2$$ is an OPN with special prime $$p$$. See this MSE question and the answer contained therein.)

So now consider the equation $$m^2 – p^k = 4z$$. Following our proof strategy, we have:

Subtracting the smallest square that is larger than $$m^2 – p^k$$, we obtain

$$m^2 – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2 = p^k + Bigg(4z – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg).$$

So the only remaining question now is whether it could be proved that
$$Bigg(4z – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg) = -y < 0$$
for some positive integer $$y$$?

In other words, is it possible to prove that it is always the case that
$$Bigg((m^2 – p^k) – bigg(lfloor{sqrt{m^2 – p^k} + 1}rfloorbigg)^2Bigg) < 0,$$
if $$n = p^k m^2$$ is an odd perfect number with special prime $$p$$?

If so, it would follow that
$$Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg)Bigg(m – lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) = p^k – y$$
which would imply that
$$Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) mid (p^k – y)$$
from which it follows that
$$m < Bigg(m + lfloor{sqrt{m^2 – p^k} + 1}rfloorBigg) leq p^k – y < p^k.$$

## Assuming two integers are odd

Is there a way to assume both n and k are odd in Mathematica?
Assuming({n > 0, k > 0}, Exp(-I*((Pi)/2)*(n^k)) // FullSimplify)

## Odd 403 issue

A client of mine has a custom coded ecommerce site. Everything is running smooth, well for most customers at least.

Some customers can’t see any product images.

The site has a domain like domain.co.uk and the images are served from images.domain.co.uk. On images, there is no htaccess file, I checked in cpanel and hotlinking is allowed and no IP is banned. There is no CDN like cloudflare in use for images.domain.co.uk

When I managed to get screenshot with the developer console open, I saw that they get a 403 error on all of the product images.

What can cause this? The majority of customers have no issue.

Thanks for any help, suggestion or advice you can give.

## nt.number theory – A geometric approach to the odd perfect number problem?

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 $$n$$:

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

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 bold claim.

## number theory – When is \$b^2 – {b-1}_2\$ with odd \$b\$ a perfect power of \$2\$? (The bracket-notation explained below)

For the complete extraction of the factor $$p$$ and its powers from a natural number $$m$$
let’s define the notation $${m}_p := { m over p^{nu_p(m)}}$$
While looking at the question of existences of 2-step-cycles in the generalized Collatz-problem for $$mx+1$$ I arrived at the question for which odd $$b$$ the evaluation of $$f(b)$$
$$f(b) = b^2 – {b-1}_2 overset{?}=2^S tag 1$$
is a perfect power $$2^S$$.
I know, that $$b=3$$ and then $$S=3$$ is a solution (the famous $$3^2-1=2^3$$). Generating a list for sequential $$b$$ didn’t suggest any useful pattern so far, so this didn’t help me with a clue.