I don’t really know how to approach this kind of problems, is there any trick or formula for this?

I write $\psi(s) = s(2s+1)$ and let $d$ be the divisor function. If $s$ is prime then 4 divides $d(\psi(s))$. For example if $s=37$ then $d(\psi(s)) = d(2775) = 12$ and $4|12$. Is this trivial? I am not sure how to attack and prove this. Here is my approach: Now I am thinking $d$ is […]

This question already has an answer here: Prove that $\tau(2^n-1) \geq \tau(n)$ for all positive integers $n$. 1 answer

Need some help on this question from Victor Shoup Let $\tau(n)$ be the number of positive divisors of $n$. Show that: $\sum_{d\mid n} \mu(d)\tau(n/d)=1$; $\sum_{d\mid n} \mu(d)\tau(d)=(-1)^r$, where $n=p_1^{e_1}\cdots p_r^{e_r}$ is the prime factorization of $n$. I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove […]

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 $n.$ $\sum\limits_{d|n} \mu(n/d)\sigma(d)=n$ for all $n.$

I managed to find an interesting inequality containing the sum of the divisors of a number and the number of them, using AM-GM between each 2 of the divisors of it: $\sigma(n)\geq\sqrt n.d(n)+(\sqrt n-1)^2$. I tried to use this to find a bound for the average of divisors of $n$, and I got: $\frac{\sigma(n)}{d(n)}\geq\sqrt n+\frac{n-2\sqrt […]

This question already has an answer here: Prove that $d(n)\leq 2\sqrt{n}$ 2 answers

Let $n\in\mathbf{N}$. I write $a_n=n^2+1$ and let $d(a_n)$ count the number of divisors of $a_n$. Set $$\Phi_n=\gcd\left(a_n,2^{d\left(a_n\right)}\right)$$ I would like to show and I believe it to be true that $$\Phi_n = \begin{cases} 1, & \text{if $n$ is even} \\[2ex] 2, & \text{if $n$ is odd} \end{cases}$$ My gut instinct is two beak it down […]

For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$ My try : Left hand side : $\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\ &= \dfrac{p^{0+1}-1}{p-1} + \dfrac{p^{1+1}-1}{p-1} + \cdots + \dfrac{p^{k+1}-1}{p-1} \\&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) – (k+1)\right) \\&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\ \end{align}$ […]

A highly composite number is a natural number $\ n\ge 1\ $ that has more divisors than any smaller natural number $\ m\ge 1$. Checking the $\ 1000\ $ entries in the OEIS-sequence, I noticed that only the following factorials are highly composite : $$[1, 2, 6, 24, 120, 720, 5040]$$ Is it known whether […]

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