All Questions
11
questions
2
votes
1
answer
86
views
Show that for an odd integer $n ≥ 5$, $5^{n-1}\binom{n}{0}-5^{n-2}\binom{n}{1}+…+\binom{n}{n-1}$ is not a prime number.
I would prefer no total solutions and just a hint as to whether or not I’m at a dead end with my solution method.
So far this is my work:
From the binomial expansion,
$$\sum_{j=0}^n 5^{n-j}(-1)^j\...
- 183
5
votes
2
answers
194
views
Periodic sequences of integers generated by $a_{n+1}=\operatorname{rad}(a_{n})+\operatorname{rad}(a_{n-1})$
Let's define the radical of the positive integer $n$ as
$$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p$$
and consider the following Fibonacci-like sequence
$$a_{n+1}=\...
- 1,828
1
vote
1
answer
279
views
What's Special about Rowland’s Prime-Generating Sequence?
Recently I asked this question and quickly got back some excellent responses.
I asked the question because I came across a paper by Eric Rowland called "A Natural Prime-Generating Recurrence"...
- 692
0
votes
1
answer
67
views
Is the sequence infinite? Finite? Is there a general formula to determine n th term?
Sequence of numbers whose factorial on prime factorisation contains prime powers of prime numbers, whose power is greater than $1$ or contains multiplicity of one for all prime numbers less than equal ...
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
0
votes
1
answer
166
views
Natural density of integers such that $\forall 1\leq k\leq n-1$ satisfy $\operatorname{rad}(k)+\operatorname{rad}(n-k)\geq \operatorname{rad}(n)$
For an integer $n>1$ we defined its square-free kernel $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p \text{ prime}}}p$$
as the product of distinct prime factors dividing it, with the ...
user243301
0
votes
2
answers
72
views
On variations of Rowland's sequence using the radical of an integer $\prod_{p\mid n}p$
This afternoon I tried to create a Rowland's sequence using the radical of an integer in my formula. I don't know if it was in the literature, but I know that also there were variations on Rowland's ...
user243301
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
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
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