For clarification: as used throughout this question, the $n$th primorial is "the product of the first $n$ primes" (counting $2$ as the first prime, not the zeroth, such that the primorial of $0$ is $1$.).
Define this integer sequence as follows: begin with the number $1$ as $a_0$. Multiply by $1$, then divide by the largest primorial dividing the result. This gives you $1$ again; therefore, $a_1 = 1$. Now repeat, but multiply by $2$ instead of $1$. Since you get $2$, which can be divided by $2$ (since $2$ is a primorial), we find again $a_2 = 1$. The sequence gets more interesting as you keep going; for example, $a_3 = 3, a_4 = 2, a_5 = 5, $ and $a_6 = 1$. After $a_6$, I have found no more $1$'s. The first few terms of the sequence are
$$1,1,1,3,2,5,1,7,28,42,2,11,22,143,1001,15015, 8, 68, 204, 646, 6460, 646...$$
The sequence is characterized by stretches of subfactorial growth separated by large falls. For example, $a_{50} = 14347428260398132900$, but $a_{51} = 1190$, since $a_{50} \cdot 51$ is divisible by $$614889782588491410 = 2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47.$$
We can also make a sequence documenting the ordinal of the primorial that we divide by for each $a_n$; the following sequence is defined as $b_n = $ the number of prime factors of $\dfrac{a_{n-1} \cdot n}{a_n}$:
$$0, 1, 0, 2, 1, 3, 0, 1, 2, 4, 1, 2, 1, 1, 2, 1, 1, 0, 6...$$
It becomes immediately apparent that $b_n = 0$ if and only if $n = 2^{k} - 1$. Besides that, it's hard to identify other patterns in this sequence.
My primary question is:
Is there some $n > 6$ such that $a_n = 1$?
Other ramblings: the only numbers besides $1$ that I've seen repeated have been $a_{19} = a_{21} = 646$ and $a_{29} = a_{30} = 29$. Are there any other numbers that repeat?
Also, I cannot ascertain whether there is some $f(n)$ such that $\displaystyle\lim_{n \to \infty}f(n) = \infty$ and $f(n) < a_n$ for all $n \in \mathbb{N}$, that is, whether there is some function that approaches infinity that is a lower bound for the $a_n$ sequence. Obviously, if there were, it would prove that the set of all $n$'s for which $a_n = 1$ is finite, if not limited to the aforementioned examples.
Can someone help me answer my question? Any help predicting the behavior of the sequence would be appreciated. Sorry if my explanation was confusing... I couldn't figure out how to write it out in mathematical notation.
Thanks for your help!
EDIT: With the use of a computer program, I've identified that there are no $1$'s up to $300,000$. Since the numbers grow so quickly I can't calculate actual values, but I know they've gone past a googol for sure. Still, this doesn't answer the question of repeated values, or prove the $1$ question...
