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On comparing two almost injective divisor maps

Edit 2018.08.08 This answer https://mathoverflow.net/a/307881 will be updated to give recent information about S, especially a forthcoming preprint. End Edit 2018.08.08 In an introductory post on ...
Gerhard Paseman's user avatar
12 votes
1 answer
2k views

Power of primes

$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ is the $n^\text{th}$ prime number and $\sigma(n)$ is the sum of the divisors of $n$. Consider the expression: $...
Craw Craw's user avatar
10 votes
7 answers
2k views

Getting a bound on the coefficients of the factor polynomial

Suppose $f(x):=a_0+a_1x+\cdots+a_nx^n$ is a polynomial in $\mathbb{Z}[x]$ and $|a_i|\leq M$ for each $i=0,\ldots ,n.$ Now suppose $g(x)$ is a factor of $f(x)$ in $\mathbb{Z}[x]$, then is it possible ...
Pritam Majumder's user avatar
3 votes
1 answer
356 views

Primitive sequence $a_i$ attaining Pillai's bound on $\sum_{i} 1/a_i$

A primitive sequence $1<a_1<\ldots<a_k\leq n$ is a sequence of integers no one of which divides any other, investigated by Erdos, Behrend and others, over the last 80 years. In fact, $\max k=\...
kodlu's user avatar
  • 10.4k
9 votes
0 answers
695 views

Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$): $$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
Perfect Number's user avatar
8 votes
2 answers
722 views

Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?

This question is related to the last question about van der Pol's identity for the sum of divisors. In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following ...
mathoverflowUser's user avatar
5 votes
2 answers
1k views

A truncated divisor function sum

Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$. The following estimate is well known $$ \sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal ...
kodlu's user avatar
  • 10.4k
5 votes
1 answer
960 views

There at least 4 divisors of $n-1$ which do not divide $\phi(n)$ if $n$ is a composite of the form $6k+1$

If $n$ is composite then $\phi(n) < n-1$ (Euler's totient function) hence there must be one or more divisors of $n-1$ which do not divide $\phi(n)$. For lack of a better terminology, let us call ...
Nilotpal Kanti Sinha's user avatar
2 votes
2 answers
643 views

On 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 ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
345 views

On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$

Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$. We could prove that: (1) $2^r t > 2n$. (We can modestly ...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
751 views

Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
928 views

Is there a formula that can predict the primes in the sequence of ratios of consecutive superior highly composite numbers? : $2, 3, 2, 5, 2, 3, 7,...$

This is the sequence of prime numbers which are the elementary building blocks for the superior highly composite numbers: $2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 2, 23, ...$ The $n^{th}$ ...
user50746's user avatar
  • 341
1 vote
1 answer
321 views

On odd perfect numbers and a GCD - Part III

Let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^...
Jose Arnaldo Bebita's user avatar
49 votes
4 answers
4k views

Strange (or stupid) arithmetic derivation

Let us consider the following operation on positive integers: $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \qquad f(n):= \prod_{i=1}^{k}\alpha_ip_i^{\alpha_i-1}$$ (Is it true that if we apply this operation to ...
Daniel Soltész's user avatar
36 votes
2 answers
7k views

Why do primes dislike dividing the sum of all the preceding primes?

I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I ...
Nilotpal Kanti Sinha's user avatar
10 votes
4 answers
4k views

Sum of the sum-of-divisors function

I was looking at the abstract of a paper 1 which claims that [2] and [3] prove $$ \sum_{n\le x}\sigma(n)-\frac{\pi^2}{12}x^2=\Omega(x\log\log x). $$ But I cannot find the above—or indeed, ...
Charles's user avatar
  • 9,114
9 votes
1 answer
1k views

Sum of divisors below threshold

Let $\sigma(n)$ denote the sum of divisors of $n$, that is, $$ \sigma(n) = \sum_{d | n} d. $$ It is known that $\sigma$ can have values as large as order $n \log \log n$. However, obviously the sum is ...
Kurisuto Asutora's user avatar
9 votes
1 answer
558 views

Is the divisor counting function equidistributed mod $p$?

Let $\sigma_0(n)$ be the divisor counting function: $$\sigma_0(n) = \sum_{d \vert n} 1.$$ I ran some numerical experiments that showed when $p$ is prime, the function $\sigma_0(n)$ is equidistributed ...
Adithya Chakravarthy's user avatar
8 votes
1 answer
306 views

Identity?: $\frac{\varphi(2^n-1)}{n}=\frac{2^{\varphi(2^n-1)}-1 \bmod (2^n-1)^2}{2^n-1}$

The computer found this. Let $n$ be a positive integer. Up to $n=200$ we have: $$\frac{\varphi(2^n-1)}{n}=\frac{2^{\varphi(2^n-1)}-1 \bmod (2^n-1)^2}{2^n-1}. \tag{1}\label{483144_1}$$ Q1 Is \eqref{...
joro's user avatar
  • 25.4k
6 votes
0 answers
506 views

Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions?

This question was posted in MSE in early August 2020. It did garner several upvotes, but did not receive any responses. I have therefore cross-posted it here, hoping that it gets answered. Let $\...
Jose Arnaldo Bebita's user avatar
6 votes
0 answers
222 views

Growth rate of signed sum of $N \sigma_0(n)-\sigma_1(n)$

Let $\sigma_k(n)=\sum_{d|n} d^k,$ for a positive integer $n$ and $k\geq 0$. A lot is known about the averages for the functions $\sigma_k(n),$ such as the estimates $$ \sum_{n\leq x} \sigma_0(n)=x \...
kodlu's user avatar
  • 10.4k
6 votes
0 answers
535 views

When is $ \sigma(n!-1) $ a perfect square?

I am looking for pairs of positive integers $(m,n)$ such that $ \sigma(n!-1) =m^2$, where $\sigma$ is the sum of divisors function. Examples occur with $(m,n)=(12,5),(1,2)$. Question: Are there ...
user avatar
5 votes
2 answers
403 views

Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots,x\}.$

I apologise for the long-windedness of this question. Let $a$ be a positive real constant and let $d(n)$ denote the number of divisors of $n.$ Define $$ S_a(x)=\sum_{n\leq x} d(n)^a. $$ For $a=1,$ ...
kodlu's user avatar
  • 10.4k
5 votes
0 answers
171 views

Touchard / van der Pol's identity for the sum of divisors and an elliptic curve for perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$, satisifies the following recurrence relation ($n>1$): $$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
Perfect Number's user avatar
5 votes
0 answers
772 views

The Grimm Machine(s): A Collatz Conjecture Rival?

Edit 2018.08.08 This answer https://mathoverflow.net/a/307881 will be updated to give recent information about S, especially a forthcoming preprint. End Edit 2018.08.08 Just as the Collatz ...
Gerhard Paseman's user avatar
3 votes
0 answers
180 views

Does this Theorem 2 from Dandapat et al. imply that $\gcd(\sigma(p^k),\sigma(a^2)) > 1$?

Write the odd perfect number $m=p^k a^2$ as a product of primes $$m = p^k {p_1}^{2a_1} \cdots {p_v}^{2a_v}.$$ (Note that it is known that $v \geq 9$ by work of Nielsen.) Let $N(m)$ be the number of ...
Jose Arnaldo Bebita's user avatar
3 votes
1 answer
318 views

Problem related to Frobenius coin problem

Let's say that the linear form $ax+by$ represents $n$ if $ax+by=n$ for some positive integer $x$ and $y$. Call a pair $(a,b)\in\Bbb N\times\Bbb N$ with $\mathsf{gcd}(a,b)=1$ good if, for any $r,s,u,...
Turbo's user avatar
  • 13.9k
3 votes
0 answers
82 views

Estimating from below positive moments of "clipped" divisor function on subsets of $\{1,2,\ldots,x\}$ with positive density

The question here about estimating positive moments of the divisor function on sets of nonzero density $A\subset \{1,2,\ldots,x\}$ was answered giving $$ S_a(x):=\sum_{n \in A} d(n)^a \geq |A|(\ln x)^{...
kodlu's user avatar
  • 10.4k
3 votes
4 answers
1k views

A conjecture regarding odd perfect numbers

(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.) Let $\sigma(z)$ denote the sum of ...
Jose Arnaldo Bebita's user avatar
3 votes
2 answers
795 views

Estimate about primes

Can anyone give an estimate (upper bound or lower bound) for the number of divisors $d\mid P_r$ such that $\frac{\sqrt{P_r}}{2}< d < \sqrt{P_r}$, where $P_r$ is the product of the $r$ smallest ...
Farzad Aryan's user avatar
2 votes
1 answer
326 views

Does there exist an integer that is both solitary and almost perfect?

This question is an offshoot from the following MSE post. I hope that it is appropriate for this site. Let $\sigma(x)$ be the sum of the divisors of $x$. An integer $a$ is said to be solitary if ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
167 views

On "Euclidean" odd perfect numbers

In what follows, we let $N = r^s u^2$ be an odd perfect number given in Eulerian form, i.e. $r$ is the special prime satisfying $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,u)=1$. In this preprint, ...
Jose Arnaldo Bebita's user avatar
1 vote
2 answers
387 views

Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Define the abundancy index $$I(x)=\frac{\sigma(x)}{x}$$ where $\sigma(x)$ ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
106 views

Lower bound on a Truncated Divisor Sum

Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$. I am interested in estimating, the following sum $$ A(a,x)=\sum_{n\leq x} \min[ d(n), M]^a $$ ...
kodlu's user avatar
  • 10.4k
1 vote
1 answer
466 views

Some divisibility constraints in Frobenius coin problem

Let's say that the linear form $ax+by$ represents $n$ if $ax+by=n$ for some positive integer $x$ and $y$. Call a pair $(a,b)\in\Bbb N\times\Bbb N$ with $\mathsf{gcd}(a,b)=1$ excellent if linear form ...
Turbo's user avatar
  • 13.9k
0 votes
1 answer
146 views

On $\mathsf{LCM}$ of a set of integers

For integers $a,b$ define $$\mathcal R(a,b)=\{q\in\mathbb Z\cap[1,\min(a,b)]: a\equiv b\bmod q\}$$ and $\mathsf{LCM}(\mathcal R(a,b))$ to be $\mathsf{LCM}$ of all entries in $\mathcal R(a,b)$. How ...
VS.'s user avatar
  • 1,836
0 votes
0 answers
759 views

On sets of coprime integers in intervals

Briefly, Question: Is it "good enough" to use least prime factor in choosing a maximal set of coprime integers in an interval? The post title comes from a 1993 paper of Erdos and Sarkozy. They ...
Gerhard Paseman's user avatar
-10 votes
1 answer
556 views

Arithmetic billiards, prime numbers and the Goldbach conjecture

I've edited the following post on Mathematics Stack Exchange, (now closed, at that date I'm suspended) with identifier 4510963, please let me to know if you've some doubt or I can improve the post. On ...
user142929's user avatar