We denote for integers $m>1$ the product of the distinct prime numbers dividing $m$ as $$\operatorname{rad}(m)=\prod\_{\substack{p\mid m\\p\text{ prime}}}p,$$
with the definition $\operatorname{rad}(1)=1$ (see it you want the Wikipedia [*Radical of an integer*](https://en.wikipedia.org/wiki/Radical_of_an_integer)), that is the famous multiplicative function in the statement of the abc conjecture. Also we denote the $k$th primorial as $$N\_k=\prod\_{t=1}^k p\_t$$ that is the the product of the first $k$ primes.


I was inspired in the first paragraph of section **B46** of [1] to propose a variant of the type of problems that was shows the book. I consider solutions $(p,k)$, where $p$ denotes always a prime number for which the identity 
$$\frac{p-1}{\operatorname{rad}(p-1)}=N\_k\tag{1}$$
holds for some primorial $N\_k$, with $k\geq 1$.


I've used a Pari/GP program to write the first primes satisfying $(1)$, those first few primes $p$'s are (this is a sample of my computations) $5,13,29,37,53,61,149,157,173,181,229, 269,277,$ $293,317,\ldots$ that correspond to these indexes $k$'s 
$1,1,1,2,1,1,1,1,1,2,1,1,1,1,1\ldots$ of our primorials $N\_k$ satisfying the equation $(1)$.


The curiosity that I've thus is what about the size or cardinality of the set $$\mathcal{K}=\{k\geq 1:(p,k)\text{ is a solution of the equation }(1)\}.$$


**Conjecture.** *The set* $\mathcal{K}$ *is finite*. 


What I am saying is that thus the set $\mathcal{K}$ remains as a bounded/finite set of positive integers when the other variable $p$ runs over the set of all prime numbers.



> 
> **Question.** Is it possible to refute previous conjecture? Can you provide some reasoning or calculations to 
>  elucidate the veracity of the conjecture? **Many thanks.**
> 
> 
> 


Maybe it can be useful state bounds or an approximation for the cardinality $|\mathcal{K}|$ for increasing segments of primes $2\leq p\leq X$ and primorials with indexes $1\leq k\leq Y$


In this ocassions I have no a great intuition about if previous conjecture holds, but I've stated it, I just to find solutions forthe integers $k=1,2$ and $3$ (I don't know if my post about this sequence has a good mathematical content, the comments are welcome).


References:
-----------


[1] Richard K. Guy, Unsolved Problems in Number Theory, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).



On a problem that equates $\frac{\text{prime}-1}{\operatorname{rad}(\text{prime}-1)}$ with the sequence of primorials
The conjecture is false, and in fact every positive integer $k\geq 1$ lies in $\mathcal{K}$. Here is a proof.


Let us fix $k\geq 1$. Let $\pi(x;N\_k^3,N\_k^2+1)$ be the number of primes $p\leq x$ such that $p\equiv N\_k^2+1\pmod{N\_k^3}$. Let $\pi\_1(x;N\_k^3,N\_k^2+1)$ be the same with the restriction that $q^2\nmid p-1$ for any prime $q>p\_k$. Following the proof of Proposition 15.12 in the book Koninck-Luca: Analytic number theory - Exploring the anatomy of integers (AMS, 2012), we can prove that
$$\frac{\pi\_1(x;N\_k^3,N\_k^2+1)}{\pi(x;N\_k^3,N\_k^2+1)}\to\prod\_{p>p\_k}\left(1-\frac{1}{p(p-1)}\right)\qquad\text{as}\qquad x\to\infty.$$
Indeed, we only need to modify this proof at a few places: in the definition of $\mathcal{P}\_2(x)$ we restrict the prime $q$ by $p\_k<q\leq z$, while in (15.18) we restrict $d>1$ to integers composed of prime factors $p\_k<q\leq z$ and replace $\pi(x;d^2,1)$ by $\pi(x;N\_k^3d^2,a\_d)$, where $a\_d\equiv N\_k^2+1\pmod{N\_k^3}$ and $a\_d\equiv 1\pmod{d^2}$. Noting that, by Dirichlet's prime number theorem,
$$\frac{\pi(x;N\_k^3d^2,a\_d)}{\pi(x;N\_k^3,N\_k^2+1)}\to\frac{\phi(N\_k^3)}{\phi(N\_k^3d^2)}=\frac{1}{\phi(d^2)}\qquad\text{as}\qquad x\to\infty,$$
the rest of the argument goes almost verbatim the same as in the referenced proof. From this density calculation, and Dirichlet's prime number theorem, it follows that there exists a prime $p$ such that $p\equiv N\_k^2+1\pmod{N\_k^3}$ and $q^2\nmid p-1$ for any prime $q>p\_k$. That is, $p=N\_k^2M+1$, where $M\equiv 1\pmod{N\_k}$ and $M$ is square-free. However, this implies that
$$\frac{p-1}{\operatorname{rad}(p-1)}=\frac{N\_k^2M}{N\_kM}=N\_k,$$
whence $k\in\mathcal{K}$. The proof is complete.


