All Questions
Tagged with prime-factorization sequences-and-series
37
questions
31
votes
0
answers
1k
views
Have I discovered an analytic function allowing quick factorization?
So I have this apparently smooth, parametrized function:
The function has a single parameter $ m $ and approaches infinity at every $x$ that divides $m$.
It is then defined for real $x$ apart from ...
- 1,091
13
votes
1
answer
344
views
Prime divisors of the sequence terms $a_n=a\cdot 2017^n+b\cdot 2016^n$
I am dealing with the test of the OBM (Brasilian Math Olimpyad), University level, 2017, phase 2.
As I've said at another topic (question 1), I hope someone can help me to discuss this test.
The ...
- 1,525
9
votes
2
answers
840
views
Is 641 the Smallest Factor of any Composite Fermat Number?
Consider the sequence $a_n = 2^{2^n}+1$ of so-called Fermat numbers. It's well known that $a_5$ isn't prime ($a_5 = 641 \cdot 6700417$, this is due to Euler). What I want to know about this sequence ...
- 2,195
5
votes
2
answers
194
views
Periodic sequences of integers generated by $a_{n+1}=\operatorname{rad}(a_{n})+\operatorname{rad}(a_{n-1})$
Let's define the radical of the positive integer $n$ as
$$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p$$
and consider the following Fibonacci-like sequence
$$a_{n+1}=\...
- 1,828
5
votes
1
answer
135
views
What about sequences $\{\sum_{k=1}^n (\operatorname{rad}(k))^p\}_{n\geq 1}$ containing an infinitude of prime numbers, where $p\geq 1$ is integer?
We denote the radical of the integer $n> 1$ as
$$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p,$$
taking $\operatorname{rad}(1)=1$ that is this definition from Wikipedia.
In ...
user243301
4
votes
2
answers
199
views
Lower bound related to the number of distinct prime numbers
Let $\omega(n)$ be the number of distinct prime factors of $n$ (without multiplicity, of course). I know some results about average of $\omega(n)$. But I didn't find any result about the following: ...
- 629
4
votes
1
answer
194
views
Sequence generated by $2^k-1$ contains new prime factors
I was playing around with the sequence where the $k^{th}$ number is equal to $2^k-1$. It seems that all numbers except $63$ contain at least one new prime in there prime factorization. That is a prime ...
- 2,086
4
votes
1
answer
168
views
Does a sequence based on hereditary factorisation always terminate?
The well-known Goodstein sequences are based on the hereditary base-$b$ notation, where you don't just present the digits in base $b$, but also the corresponding exponents etc.
That lead me to the ...
- 1,610
3
votes
1
answer
114
views
What comes next in this sequence $1,4,8,16,17,44,58,76,\dots$? [closed]
My friend John asked me what is the next number in the sequence
$$1,4,8,16,17,44,58,76,\dots ?$$
I told him it could be anything, and I asked for a clue. He told me it involves adding numbers that ...
user913631
3
votes
1
answer
402
views
Consider $A$ the set of natural numbers with exactly 2019 divisors
Consider $A$ the set of natural numbers with exactly 2019 divisors, and for each $n \in A$ denote
$$
S_n = \frac{1}{d_1+\sqrt{n}} + \frac{1}{d_2+\sqrt{n}} + ... + \frac{1}{d_{2019}+\sqrt{n}},
$$
...
- 61
3
votes
1
answer
106
views
A problem similar than that of the amicable pairs using the function $\operatorname{rad}(k)$: a first statement or conjecture
In this post we denote the product of distinct primes dividing an integer $k> 1$ as $\operatorname{rad}(k)$, with the definition $\operatorname{rad}(1)=1$, that is the so-called radical of an ...
user243301
3
votes
0
answers
88
views
Do the sequences defined by $a_n=a_{n-1}+(\text{the least prime factor of }a_{n-1})+1$ starting with $2,6,14,\ldots$ merge?
Let $S_k$ be the sequence defined by $a_k(1)=k,\ a_k(n)=a_k(n-1)+(\text{the least prime factor of }a_k(n-1))+1$.
A diagram of these sequences for around $k<100$ is shown below. As you can see, $S_3$...
- 1,451
3
votes
0
answers
950
views
A very nice pattern involving prime factorization
A while ago I was fiddling around with prime numbers and C++. I defined:
$$f_a(b)= \text{ the amount of numbers } 2^a\leq n<2^{a+1}\text{ with } b \text{ prime factors}$$
I calculated $f_a(b)$ for ...
- 8,421
3
votes
0
answers
107
views
Special $\omega(n)$-sequence
Let $k$ be a natural number, $\omega(n)$ the number of distinct prime factors of $n$.
The object is to find a number $n$ with $\omega(n+j)=j+1$ for each $j$ with
$0\le j\le k-1$. In other words, a ...
- 85.1k
3
votes
0
answers
105
views
Do there exist any cycles for these number sequences?
We define, for $k\in\mathbb{N}$, the sequence $\left(S_{k,n}\right)_{n\in\mathbb{N}}$:
$$S_{k,1}=k,\;\;\; S_{k,n+1}=p_1q_1\cdots p_mq_m \text{ (written out in decimal)}$$
Where $p_1^{q_1}*\cdots *p_m^{...
- 3,540