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?

Odd, worrying alert:
Use not usual username's pass for sensitive data in iCloud??

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.

(related question https://apple.stackexchange.com/questions/394399/why-am-i-being-asked-for-local-password-in-update-apple-id-settings )

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.
SEMrush

 

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.