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 […]

Prove that $\tau(2^n-1) \geq \tau(n)$ for all positive integers $n$. Note that $\tau(2^n-1)=\sum_{d|2^n-1}{1}$. I try to prove by induction. Base case: When $n=1$, we have $\tau(2-1)=\tau(1)$. Hence, the inequality is true. Fix an $n$. Assume that the inequality $\tau(2^n-1) \geq \tau(n)$, which is $\sum_{d|2^n-1}{1} \geq \sum_{d|n}{1}$ is true for $n$. Note that $\sum_{d|2^{n+1}-1}{1}=\sum_{d|2^{n}-1}{1}+1 \geq \sum_{d|n}{1}+1=\sum_{d|n+1}{1}$ […]

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 $d(n)$ $$\sum_{i=1}^n\left\lfloor\frac{n}i\right\rfloor=\sum_{k=1}^nd(k)$$

I believe I have the solution to this problem but post it anyway to get feedback and alternate solutions/angles for it. For all $n \in \mathrm {Z_+}$ prove $n$ is a perfect square if and only if $n$ has odd # of positive divisors. Thought to use induction but the perfect squares don’t increase by […]

how to prove: $$ \prod_{d|n} d= n^{\frac{\tau (n)}{2}}$$ $\prod_{d|n} d$ is product of all of distinct positive divisor of $n$, $\tau (n)$ is number (count)of all of positive divisor of $n$

In Waclaw Sierpinski’s book Elementary Theory of Numbers on page 168 there is the following exercise: “Exercises. 1. Prove that for natural numbers $n$ we have $d(n) \leq 2\sqrt{n}$,” where $d(n)$ is the number of divisors of n. As a hint right below is given: “The proof follows from the fact that of two complementary […]

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