# Questions tagged [divisors-multiples]

For questions on divisors and multiples, mainly but not exclusively of integers, and related and derived notions such as sums of divisors, perfect numbers and so on.

220 questions
Filter by
Sorted by
Tagged with
167 views

107 views

253 views

### Analogue of the second Hardy-Littlewood conjecture for numbers of divisors?

Let $f(n)$ denote the proposition "There exists some $k>1$ such that $$\sum_{m=k}^{k+n-1}\tau(m) < \sum_{m=1}^n\tau(m)$$ where $\tau(m)$ is the number of the divisors of $m$." (This ...
1 vote
272 views

164 views

### Proportionality constant in Montgomery-Vaughan Theorem 7.20

In Multiplicative Number Theory - Vol. I by Montgomery and Vaughan the following result is proved. Theorem 7.20 Let $A(x,r)$ denote the number of $n\leq x$ such that $\Omega(n)\leq r \log \log x,$ and ...
131 views

### A modern reference for the Piltz divisor problem

apparently, the Dirichlet hyperbola method is no longer up-to-date, and instead Voronoi's identity is used in order to establish good bounds on the Dirichlet divisor problem. The same applies to the ...
1 vote
90 views

### Lower bound on a Truncated Divisor Sum

Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$. I am interested in estimating, the following sum $$A(a,x)=\sum_{n\leq x} \min[ d(n), M]^a$$ ...
296 views

### Factors of polynomials of bounded height

Let $f(x)=a_nx^n+\cdots+a_0 \in \mathbb{Z}[x]$ be an integer polynomial in one variable. Recall that the height $H(f):=\textrm{max}\,|a_n|$ is the largest coefficient. Consider the set of polynomials ...
### On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$
Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$. We could prove that: (1) $2^r t > 2n$. (We can modestly ...
### Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$
Let $n=a_3z^3+a_2z^2+a_1z+1$ where $a_1<z, \ a_2<z, \ 1 \le a_3<z, z>1$ are non negative integers. To obtain proper divisors of $n$ of the form $xz+1$, one may perform trial divisions \$xz+...