All Questions
6 questions
16
votes
2
answers
2k
views
Does this sequence ever end?
This may help: A080670 A195265
Define $f(n)$ as this:
Take a number $n$, and split it into its prime composition using $^$ and $×$. Now remove all $^$ and $×$, you get a new number, this is $f(n)$ (...
1
vote
0
answers
60
views
On the parity of $(2^{\varphi(n)}-1) \bmod{n^2}$
For odd integer $n$ define the function
$$ J(n)=(2^{\varphi(n)}-1) \bmod{n^2}$$
$J(n)$ is integer in $[0,n^2-1]$ and it is divisible by $n$.
Integer $n$ is Wieferich number
iff $J(n)=0$ and if $n$ is ...
1
vote
1
answer
181
views
On the sequence $a(n)=\gcd(2^n-1,\phi(2^n-1))$
For natural $n$, define the sequence
$$
a(n)=\gcd(2^n-1,\phi(2^n-1))
$$
It doesn't appear to be in OEIS and starts
$1,1,1,1,9,1,1,1,3,1,9,1,3,1,1,1,27,1,75,49$
Q1 Can we unconditionally prove $a(n)=1$...
5
votes
0
answers
161
views
Consecutive integers each of which has a large prime factor
There are many results about consecutive integers all having small prime factors. But what about consecutive integers each of which has a large prime factor?
More precisely, let $P(n)$ be the ...
3
votes
3
answers
696
views
For any prime $p$, is there $C$ such that if $x\ge C$, then all but one integer among $x+1, x+2, \dots, x+p$ has Greatest Prime Factor $> p$
I apologize if this is a naive question about greatest prime factors (gpf). I was thinking about the sequence of integers where $\mathrm{gpf}(x) \le p$ where $p$ is any prime.
Clearly, as $x$ ...
5
votes
0
answers
753
views
Least Prime Factor in a sequence of 2n consecutive integers
I was thinking about consecutive integers and I wondered if anyone had done work exploring whether a sequence of $2n$ consecutive integers (i.e. 101,102,103,...,100+2n) always contains at least one ...