All Questions
Tagged with prime-factorization sequences-and-series
37
questions
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243
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Square-free integers in the sequence $n^{\operatorname{rad}(n)}+\operatorname{rad}(n)+1$
For integers $n\geq 1$ in this post we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an ...
user243301
0
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1
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163
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On prime-perfect numbers and the equation $\frac{\varphi(n)}{n}=\frac{\varphi(\operatorname{rad}(n))}{\operatorname{rad}(\sigma(n))}$
While I was exploring equations involving multiple compositions of number theoretic functions that satisfy the sequence of even perfect numbers, I wondered next question (below in the Appendix I add a ...
user243301
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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
3
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1
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106
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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
5
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1
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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
2
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1
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68
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On miscellaneous questions about perfect numbers II
Let $\varphi(m)$ the Euler's totient function and $\sigma(m)$ the sum of divisors function. We also denote the product of primes dividing an integer $m>1$ as $\operatorname{rad}(m)$, that is the ...
user243301
2
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1
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162
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On even integers $n\geq 2$ satisfying $\varphi(n+1)\leq\frac{\varphi(n)+\varphi(n+2)}{2}$, where $\varphi(m)$ is the Euler's totient
This afternoon I am trying to get variations of sequences inspired from the inequality that defines the so-called strong primes, see the definition of this inequality in number theory from this ...
user243301
2
votes
1
answer
105
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Square-free integers in the sequence $\lambda+\prod_{k=1}^n(\varphi(k)+1)$, where $\lambda\neq 0$ is integer
While I was exploring the squares in the sequence defined for integers $n\geq 1$
$$\prod_{k=1}^n(\varphi(k)+1),\tag{1}$$
where $\varphi(m)$ denotes the Euler's totient function I wondered a different ...
user243301
0
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2
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72
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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
2
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0
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Compare $\sum_{k=1}^n k^{\operatorname{rad}\left(\lfloor\frac{n}{k}\rfloor\right)}$ and $\sum_{k\mid n}k^{\operatorname{rad}\left(\frac{n}{k}\right)}$
I would like to know how do a comparison between the sizes of these functions defined for integers $n\geq 1$, when $n$ is large
$$f(n):=\sum_{k=1}^n k^{\operatorname{rad}\left(\lfloor\frac{n}{k}\...
user243301
2
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0
answers
64
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Find next larger number with same number of prime factors
Is there are way to determine, given a (composite) number $n$ and a list of its prime factors, the next larger (composite) number with the same number of prime factors? Clarification: Not the same ...
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4
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Does a sequence based on hereditary factorisation always terminate?
The well-known Goodstein sequences are based on the hereditary base-$b$ notation, where you don't just present the digits in base $b$, but also the corresponding exponents etc.
That lead me to the ...
- 1,610
31
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1k
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Have I discovered an analytic function allowing quick factorization?
So I have this apparently smooth, parametrized function:
The function has a single parameter $ m $ and approaches infinity at every $x$ that divides $m$.
It is then defined for real $x$ apart from ...
- 1,091
4
votes
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194
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Sequence generated by $2^k-1$ contains new prime factors
I was playing around with the sequence where the $k^{th}$ number is equal to $2^k-1$. It seems that all numbers except $63$ contain at least one new prime in there prime factorization. That is a prime ...
- 2,086
3
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A very nice pattern involving prime factorization
A while ago I was fiddling around with prime numbers and C++. I defined:
$$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