Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
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.
1
vote
A truncated divisor function sum
The $n=ab$ trick is very effective. There is a similar trick of writing in a unique way $n=ab$ with $n$ squarefree, $b$ a square and I was wondering whether this could also work here.
1
vote
Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots...
There is a cheap way via Cauchy's inequality in case you do no want to use Erdos--Kac. I will only do it for $a=1$ just to outline the idea.
$$\frac{x}{2} (1+o(1) ) \leq
\sum_{n \leq x } 1_A(x)=
\su …
14
votes
3
answers
1k
views
On the number of consecutive divisors of an integer
Define for $n \in \mathbb{N}$ the function $$\tau_1(n):=\sum_{\substack{d|n, \\ d+1|n}}1,$$ i.e. the number of consecutive divisors of an integer. The average of $\tau_1(n)$ is $1$ since $$\sum_{n\leq …