All Questions
11 questions
5
votes
1
answer
960
views
There at least 4 divisors of $n-1$ which do not divide $\phi(n)$ if $n$ is a composite of the form $6k+1$
If $n$ is composite then $\phi(n) < n-1$ (Euler's totient function) hence there must be one or more divisors of $n-1$ which do not divide $\phi(n)$. For lack of a better terminology, let us call ...
- 5,470
5
votes
2
answers
1k
views
A truncated divisor function sum
Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$. The following estimate is well known
$$
\sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal ...
- 10.4k
2
votes
1
answer
928
views
Is there a formula that can predict the primes in the sequence of ratios of consecutive superior highly composite numbers? : $2, 3, 2, 5, 2, 3, 7,...$
This is the sequence of prime numbers which are the elementary building blocks for the superior highly composite numbers:
$2, 3, 2, 5, 2, 3, 7, 2, 11, 13, 2, 3, 5, 17, 19, 2, 23, ...$
The $n^{th}$ ...
- 341
36
votes
2
answers
7k
views
Why do primes dislike dividing the sum of all the preceding primes?
I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I ...
- 5,470
10
votes
4
answers
4k
views
Sum of the sum-of-divisors function
I was looking at the abstract of a paper 1 which claims that [2] and [3] prove
$$
\sum_{n\le x}\sigma(n)-\frac{\pi^2}{12}x^2=\Omega(x\log\log x).
$$
But I cannot find the above—or indeed, ...
- 9,114
9
votes
1
answer
558
views
Is the divisor counting function equidistributed mod $p$?
Let $\sigma_0(n)$ be the divisor counting function:
$$\sigma_0(n) = \sum_{d \vert n} 1.$$
I ran some numerical experiments that showed when $p$ is prime, the function $\sigma_0(n)$ is equidistributed ...
- 1,949
8
votes
1
answer
1k
views
Sum of divisors below threshold
Let $\sigma(n)$ denote the sum of divisors of $n$, that is,
$$
\sigma(n) = \sum_{d | n} d.
$$
It is known that $\sigma$ can have values as large as order $n \log \log n$. However, obviously the sum is ...
- 2,207
5
votes
2
answers
402
views
Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots,x\}.$
I apologise for the long-windedness of this question.
Let $a$ be a positive real constant and let $d(n)$ denote the number of divisors of $n.$ Define
$$
S_a(x)=\sum_{n\leq x} d(n)^a.
$$
For $a=1,$ ...
- 10.4k
3
votes
2
answers
795
views
Estimate about primes
Can anyone give an estimate (upper bound or lower bound) for the number of divisors $d\mid P_r$ such that $\frac{\sqrt{P_r}}{2}< d < \sqrt{P_r}$, where $P_r$ is the product of the $r$ smallest ...
- 499
1
vote
0
answers
106
views
Lower bound on a Truncated Divisor Sum
Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$.
I am interested in estimating, the following sum
$$
A(a,x)=\sum_{n\leq x} \min[ d(n), M]^a
$$
...
- 10.4k
0
votes
1
answer
146
views
On $\mathsf{LCM}$ of a set of integers
For integers $a,b$ define $$\mathcal R(a,b)=\{q\in\mathbb Z\cap[1,\min(a,b)]: a\equiv b\bmod q\}$$ and $\mathsf{LCM}(\mathcal R(a,b))$ to be $\mathsf{LCM}$ of all entries in $\mathcal R(a,b)$.
How ...
- 1,826