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Questions tagged [divisor-counting-function]

For questions that involve the divisor counting function, also known as $\sigma_0$, $\tau$, or $d$.

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6 votes
2 answers
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For positive integers $a$ and $n$, when does $a^n+1$ have $n$ divisors? This was a natural question that popped up when I was investigating numbers of the form $a^n+1$, but surprisingly I can't seem ...
CosmicOscillator's user avatar
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2 answers
106 views

Let $\tau(n)$ be the number of positive divisors of $n$, and let $\sigma(n)$ be the sum of its positive divisors. I was playing around with these functions for small values of $n$ and noticed ...
CEOofCompany's user avatar
1 vote
3 answers
163 views

I think that the fraction of numbers with a number of divisors divisible by $3$ is $1-\frac{6}{\pi^2}\zeta(3)$. To formally define what I mean by the fraction of numbers with a certain property, if $f(...
Neil Peter's user avatar
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2 answers
153 views

This is from Hardy and Wright’s An Introduction to the Theory of Numbers, Section 18.2. "The average order of $d(n)$". Here, $d(n)$ denotes the number of divisors of $n$. The section states ...
John L's user avatar
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2 votes
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267 views

I was given the following question :- Let $\tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. For $k\ge 2$, prove that $$D_k(x) := \sum_{n \leq x} \tau_k(n)...
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I was given the following question :- Let $\tau_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. For $k\ge 2$, prove that $$D_k(x) := \sum_{n \leq x} \tau_k(n)...
user avatar
0 votes
2 answers
140 views

Reference Question: The number of and sum of all divisor which are odd and are of the form $3k+2$ for the number $n=2^{7}\times 3^{2}\times 5^{1}\times 7^{1}$ My approach: Since number is odd and is ...
mathophile's user avatar
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2 answers
112 views

Find number of divisor of $n=3^{5}\cdot 5^{7}\cdot 7^{9}$ which are of the form $4k+1$, $k\geq 0$ My solution: Total number of divisor of $n$ are $(5+1)(7+1)(9+1)=480$ All the divisor can be of the ...
mathophile's user avatar
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3 votes
2 answers
177 views

Notation: Here I take $\sigma_x(n)$ to mean $\displaystyle{\sum_{d|n}} d^{x}$. Is there a good lower bound known for $\dfrac{\sigma_1(n)}{\sigma_0(n)}$? I've seen this post on the average order of ...
BBadman's user avatar
  • 199
3 votes
0 answers
48 views

Please guide me as to how to obtain the below bound and whether it is optimal. Let a squarefree integer $N=\prod_{1 \leq i \leq m} p_i$ be a product of $m$ primes ($p_1 < p_2 < \dots < p_m$) ...
user166305's user avatar
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From the first $10\ 000$ highly composite numbers listed in OEIS , the following $20$ are the middle of a twin-prime that is we have a highly composite number $N$ such that both $N-1$ and $N+1$ are ...
Peter's user avatar
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2 votes
1 answer
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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
  • 155
2 votes
1 answer
290 views

Prove that φ(n) + d(n) ≤ n + 1. d(n) is the number of positive divisors of n. φ(n) is the Euler's Totient Function. Attempt: For a prime number n, φ(n) = n - 1 (all numbers less than n are relatively ...
comp.course.master's user avatar
2 votes
0 answers
81 views

Recently I encountered a problem: If n is a positive integer, then is the number of divisors of $2^n - 1$ less or greater than the number of divisors of n? I tried factoring and taking modulo n but ...
notabot's user avatar
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1 vote
2 answers
206 views

I recently stumbled across a problem about numbers' divisor count (more specifically, how many positive integers are equal to the square of their divisor count - answer was 2: they are 1 and 9). But I ...
Francisco Sierra's user avatar
7 votes
0 answers
136 views

Let $d(n)$ be the divisor function, that is, $$d(n)=\sum_{1\leq k\leq n, \,k|n} 1.$$ Or equivalently, $d(n)$ can be defined as the coefficients of $\zeta^2(s)$: $$\zeta^2(s)=\sum_{n\geq1} d(n)n^{-s}.$$...
Landau's user avatar
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6 votes
1 answer
332 views

Let $d(n)$ denote the number of divisors of a positive integer $n$. It is pretty obvious that $d(n) \ge 2$ for any given number $n \ge 2$, since every number is divisible by $1$ and itself. $2$ is ...
Kinheadpump's user avatar
  • 1,391
2 votes
1 answer
142 views

For $q,a\in \mathbb N$ write $d=(q,a)$. Why might be $$\sum _{h|q}\frac {c_h(a)\log (d/h)}{h}=-\frac {q'}{\phi (q')}\sum _{h|q}\frac {c_h(a)}{h}\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}+\sum _{h|d}\...
tomos's user avatar
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4 votes
2 answers
980 views

The following is a picture of equation from Ramanujan's lost notebook. In this page, Ramanujan gives a closed form for, $$\sum_{n\geq 1}\sigma_{s}(n)x^{n}$$ In an attempt initially it's claimed that, $...
RAHUL 's user avatar
  • 1,620
0 votes
1 answer
130 views

If you have two identity functions, then $f(d) * g(n/d)$ would be just $dn/d = n$. Since we have an $n$ added for each divisor of $n$, would the resulting function just be $n$ times the number of ...
Isaac Wachsman's user avatar
1 vote
1 answer
479 views

I am currently in the middle of the following exercise: Exercise. Compute $$ \sum_{d|n} \left(\sigma(d)\mu\left(\frac{n}{d}\right)+\tau(d)\right),$$ where $\sigma$ is the function that corresponds to ...
xyz's user avatar
  • 1,289
1 vote
1 answer
138 views

Maybe this question is stupid, but there was a problem in a math competition (not even in the highest stage) in my county which asked to Find $ \sum_{n\leq390} d(n)$, where $d(n)$ is the number of ...
Fede1618's user avatar
1 vote
0 answers
64 views

Let us define the following recursive function involving the sum of divisors function $\sigma(n)$: \begin{array}{ l } r(n,1)=\sigma(n) \\ r(n,2)=\sum_{d|n}r(d,1) \\ r(n,3)=\sum_{d|n}r(d,2) \\ \...
Eldar Sultanow's user avatar
5 votes
1 answer
142 views

The sum of divisors function is commonly denoted by $\sigma(n)$. Now let us introduce a recursive definition of divisor functions: $r_{n,1}=\sigma(n)$ $r_{n,2}=\sum_{d|n}r_{d,1}$ $r_{n,3}=\sum_{d|n}...
Eldar Sultanow's user avatar
0 votes
1 answer
116 views

We shall prove that $\sum_{n=1}^{+\infty} \frac{d(n)}{2^n}=\sum_{n=1}^{+\infty} \frac{1}{\phi(2^{n+1}-1)}$, where d(n) the divisor function. I was thinking of making use of the fact that d(n) is ...
MIkeTheSci's user avatar

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