# A recurrence involving the radical of an integer

I've considered the recurrence defined as $a_1=b_1=2$ and for integers $n\geq 1$ $$\left. \begin{array}{l} a_{n+1}=a_1b_1+a_2b_2+\ldots+a_nb_n\\ b_{n+1}=\operatorname{rad}(a_nb_n) \end{array} \right\} \tag{1}$$ where $\operatorname{rad}(m)$ denotes the product of distinct primes dividing $m>1$, see the Wikipedia Radical of an integer if you need it.

I believe that it is an interesting recurrence of positive integers. As a simple motivation of our equation $(1)$ one can to prove next easy claim.

Claim. A) For $n\geq 1$ one has $a_{n+2}-a_{n+1}=a_{n+1}\operatorname{rad}(a_nb_n)$ and thus $$a_{n+2}=a_{n+1}(1+b_{n+1}).$$

B) If there are no mistakes one can also deduce for $N>1$ that $$b_2+b_3+\ldots+b_N=-N+1+\frac{a_3}{a_4}+\frac{a_4}{a_3}+\ldots+\frac{a_{N+2}}{a_{N+1}}.$$

In my exploration of $(1)$ I've calculated the first terms of our sequence: $a_2=2^2$, $a_3=2^2\cdot3$ with $b_2=b_3=2$; $a_4=2^2\cdot3^2$ and $a_5=2^2\cdot3^2\cdot7$ with $b_4=b_5=2\cdot 3$; $a_6=2^2\cdot3^2\cdot7^2$ and $a_7=2^2\cdot3^2\cdot7^2\cdot 43$ with $b_6=b_7=2\cdot 3\cdot 7$; $a_8=2^2\cdot3^2\cdot7^2\cdot 43^2$ and $a_9=2^2\cdot3^2\cdot7^2\cdot 43^2\cdot 13\cdot 139$ with $b_8=b_9=2\cdot 3\cdot 7\cdot 43$. And my last calculation was $a_{10}=2^2\cdot3^2\cdot7^2\cdot 13^2\cdot 43^2\cdot 139^2$ with $b_{10}=2\cdot 3\cdot 7\cdot 13\cdot 43\cdot 139$.

Question. Maybe the interesting thing of this sequence is as a prime generator. It seems obvious that the number of prime factors in the factorization of $b_{n+1}$ grows as $n\to\infty$. Can you be more precise? Prove or calculate an approximation of how many (different) prime factors has $b_{n+1}$ as $n$ grows. Many thanks.

I've added previous claim as motivation, but maybe it is useful for our proof. On the other hand now I don't know if my question is easy to solve.

• Many thanks @Rohan – user243301 Dec 22 '17 at 14:55
• Learn a programming language, plot those things, you'll see the main pattern is randomness. Also : do you expect us to spend several hours investigating this nonsense recurrence, searching for patterns ? There are no miraculous theorems for those kind of things, see the dozens of conjectures about prime numbers you read on wikipedia. – reuns Dec 22 '17 at 14:55
• Has mathematical meaning my Question @reuns ? You can read it as: has $b_{n+1}$ a large number or prime factors as $n\to\infty$ or can we say anything how many primes factors has it? If has mathematical meaning the question, then what's your problem? – user243301 Dec 22 '17 at 14:58