All Questions
12
questions
3
votes
1
answer
84
views
Dirichlet series of $\ln(n) \tau(n)$
I was experimenting with a technique I developed for double/multiple summation problems, and thought of this problem:
Find
$$S(p)=\sum_{n=1}^{\infty} \frac{\ln(n) \tau(n)}{n^p}$$
where $\tau(n)=\sum_{...
user1214193
3
votes
2
answers
391
views
A proof of $\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}$
I'm trying to proof the following statement:
Let $n \in \mathbb{Z}$ and the $\sum$ are on the divisors $d$ of $n$. Show that
$$\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}.$$
...
- 500
4
votes
1
answer
637
views
Identity for the divisor function: $\tau(mn)=\sum\limits_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d)$
Let $\tau$ denote the classical divisor function and $\mu$ be the
Möbius function.
Then for each pair of integers $n,m$ we have
$$\tau(mn)=\sum_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d),$$
where the ...
- 1,061
1
vote
2
answers
156
views
Stuck at evaluating $\sum\limits_{d \mid n} \tau(d)$
It seems the number of nonnegative integer solutions to the equation $xyz=n$ is given by
$$\sum\limits_{d \mid n} \tau(d)$$
$\tau$ is the number of divisors function. I'm wondering if there is a way ...
- 12.2k
2
votes
2
answers
151
views
Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
Can somebody prove or disprove? Let $\tau$ be the divisors function, so that $\tau(6) = \#\{ 1,2,3,6\} = 4$
$$ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$$
Here I am using $b \vee d = \mathrm{...
- 24.5k
6
votes
3
answers
2k
views
Prove that $\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)$
How can I prove:
$$\sum \limits_{d|n}(n/d)\sigma(d) = \sum \limits_{d|n}d\tau(d)?$$
Few observations :
Left side is a sum function and the right side is a number of divisors function. Both the sides ...
- 813
7
votes
3
answers
3k
views
For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$
For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$
My try :
Left hand side :
$\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) +...
- 12.2k
2
votes
1
answer
377
views
Evaluating a sum with Mobius function $\sum_{d|n}\tau(d)\mu(d)$
I'm not sure how to evaluate the sum $\sum_{d|n}\tau(d)\mu(d)$, since it is also not a convolution.
- 81
5
votes
2
answers
522
views
Finding the summation of the floor of the series identity [closed]
I would appreciate if somebody could help me with the following problem:
Q: How to proof ?
The number of positive divisors of $n$ is denoted by $d(n)$
$$\sum_{i=1}^n\left\lfloor\frac{n}i\right\...
- 5,506
4
votes
2
answers
628
views
Number theory Exercise: $\sum_{d \mid n} \mu(d) d(d) = (-1)^{\omega(n)}$ and $\sum_{d \mid n} \mu(d) \sigma (d)$
for positive integer $n$, how can we show
$$ \sum_{d | n} \mu(d) d(d) = (-1)^{\omega(n)} $$
where $d(n)$ is number of positive divisors of $n$ and $mu(n)$ is $(-1)^{\omega(n)} $ if $n$ is square ...
- 10.7k
5
votes
2
answers
3k
views
Sum of Positive Divisors: $\sum_{d|n} \mu(n/d)\nu(d)=1$ and $\sum_{d|n} \mu(n/d)\sigma(d)=n$
If $\nu(n)=$ Number of positive divisors of $n,$ $\mu$ is the Möbius function and
$\sigma(n)$ is the sum of positive divisors.
show that;
$\sum\limits_{d|n} \mu(n/d)\nu(d)=1$ for all $n.$
$\sum\...
- 191
4
votes
3
answers
4k
views
How to prove $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$
For every positive integer $d$, we let $\tau\left(d\right)$ be the number of positive divisors of $d$.
Prove that
\begin{align}
\sum_{d|n} \tau^3(d)
= \left(\sum_{d|n} \tau (d)\right)^2
\end{align}
...
- 2,879