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2 votes
1 answer
62 views

On the "efficiency" of digit-sum divisibility tricks in other bases (or about the growth rate of the number of divisors function).

Out of the different divisibility tricks there's a really simple rule that works for more than one divisor: The digit-sum. Specifically, if the sum of the digits of a number is a multiple of $1,3$ or $...
Robert Lee's user avatar
  • 7,273
2 votes
1 answer
48 views

Solutions of equations of form $\tau(n+a_i)=\tau(x+a_i)$ where $\tau$ is the number of divisors function and $a_i$ is a diverging series

Consider the function $$\tau(n)=\sum_{d|n}1$$ which gives the number of divisors of a number. The question is: How much information does $\tau(n)$ contain about $n$? The answer is obviously: not very ...
xyz1234's user avatar
  • 103
1 vote
1 answer
144 views

On a conjecture involving multiplicative functions and the integers $1836$ and $137$

We denote the Euler's totient as $\varphi(x)$, the Dedekind psi function as $\psi(x)$ and the sum of divisors function as $\sigma(x)$. Are well-known arithmetic functions, see the corresponding ...
user759001's user avatar
4 votes
1 answer
107 views

On a conjecture involving multiplicative functions and the integers $1836$ and $136$

We denote the Euler's totient as $\varphi(x)$, the Dedekind psi function as $\psi(x)$ and the sum of divisors function as $\sigma(x)$. Are well-known arithmetic functions, see Wikipedia. I would like ...
user759001's user avatar
2 votes
0 answers
95 views

On Carmichael function and aliquot parts of odd perfect numbers

This post is cross-posted on MathOverflow with identifier 439563 and same title. We denote as $N$ an odd perfect number, and $d\mid N$ one of its divisors. We denote the Carmichael function as $\...
user759001's user avatar
0 votes
0 answers
94 views

Does $I = \gcd(n,\sigma(n^2)) = (\frac{n}{\sigma(q^k)/2})\cdot\gcd(\sigma(q^k)/2,n)$ imply that $\sigma(q^k)/2 \mid n$ holds?

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 GCDs: $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
116 views

Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. My question in the present post is closely related to this one in MO: QUESTION Does there exist a nontrivial ...
Jose Arnaldo Bebita Dris's user avatar
-1 votes
1 answer
161 views

On odd perfect numbers and a GCD - Part VII

(Pardon me for being somewhat stubborn, but this question will be the last for this week. This post is an offshoot of this one.) Let $N = q^k n^2$ be an odd perfect number be an odd perfect number ...
Jose Arnaldo Bebita Dris's user avatar
0 votes
1 answer
55 views

Does $G \mid I$ and $I \mid H$ still hold if $\sigma(q^k)/2$ is not squarefree, where $q^k n^2$ is an odd perfect number with special prime $q$?

This question is an offshoot of this post #1 and this post #2. Let $N = q^k n^2$ be an odd perfect number be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\...
Jose Arnaldo Bebita Dris's user avatar
0 votes
0 answers
47 views

Please check my proof: If $q^k n^2$ is an odd perfect number with special prime $q$, then $\sigma(q^k)/2 \mid n$ if and only if $n \mid \sigma(n^2)$.

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
74 views

On the reciprocal of $I(n^2) - \frac{2(q - 1)}{q}$, if $q^k n^2$ is a(n) (odd) perfect number with special prime $q$

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. A number $N$ is said to be perfect if $\sigma(N)=...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
40 views

Fractions with $a < \sigma(b)$ are dense

Prove that any open interval of real numbers in $[1, \infty)$ contains a rational number $\frac{a}{b}$, $(a,b)=1$, with $b\leq a < \sigma(b)$. (Here $\sigma$ denotes sum of all positive divisors.) ...
DesmondMiles's user avatar
  • 2,803
1 vote
1 answer
42 views

For what $p$ and $k$ (where $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$) is this algebraic expression divisible by $3$?

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$. (...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
118 views

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

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$. I discovered an interesting identity involving ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
1 answer
57 views

If $k + 1$ is prime and $(k + 1) \mid (q - 1)$, then $\sigma(q^k)$ is divisible by $k + 1$, but not by $(k + 1)^2$ (unless $k+1=2$).

I tried Googling for the keywords "the theory of odd perfect numbers" and one of the search results that came up was this document, titled ON THE DIVISORS OF THE SUM OF A GEOMETRICAL SERIES ...
Jose Arnaldo Bebita Dris's user avatar

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