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 recursion in the literature.
Definition. I define the sequence with $a_1=1$, and for integers $k\geq 2$ as $$a_k=a_{k-1}+\operatorname{rad}(k+a_{k-1}),\tag{1}$$ where $\operatorname{rad}(n)$ is the radical of the integer $n$, see its definition from this Wikipedia.
I've calculated the first terms ($a_k$, with $2\leq k\leq 10$) as $4, 11, 26, 57, 78, 163, 220, 449, 500$.
Computational fact. Thus for $2\leq k\leq 10$, I've observed that the absolute value of the differences $a_k-a_{k-1}$ is always a squarefree integer.
I would like to know when fails this conjecture and, if it is possible to set a new conjecture using a new definition.
Question.
A) If you know that this definition was in the literature, please refer it. Find the first $K$ such that there exists a prime $p$ with $p^2$ dividing $|a_{K}-a_{K-1}|$ (that is, $|a_{K}-a_{K-1}|$ isn't a squarefree integer).
B) Can you set an improvement of my idea definining a Rowland's sequence using the radical of an integer $\operatorname{rad}(n)$ with different $a_1$ and/or with a different formula $(1)$? Thus I am asking to you for a different definition, always using the arithmetic function $\operatorname{rad}(n)$, that provide us a nice statemement or a conjecture for $$|a_{k}-a_{k-1}|\in\text{ a sequence of positive integers},$$ for $k\geq 2$. Thanks in advance.
I believe this should answer most of your question:
By definition, $a_k-a_{k-1} = \mathrm{rad}(k+a_{k-1})$. Since $\mathrm{rad}(n)$ is a product of distinct primes for all $n$, it is necessarily squarefree.