# When is it easy to write down the Bhargava S-factorial?

In Manjul Bhargava's The Factorial Function and Generalizations he motivates a new type of factorial $n!_S$ using by generalizing three theorems:

• For $k, l \in \mathbb{Z}$, we have $k! \times l!$ divides $(k+l)!$.

• For any primitive polynomial $f(x) \in \mathbb{Z}[x]$ with $\deg f = k$ then $\mathrm{gcd}\{ f(a): a \in \mathbb{Z}\}$ divides $k!$

In the process of solving generalizing these two results, he invents a factorial for any set of integers $S \in \mathbb{Z}$. For any prime $p$, order the element of $S$ by:

• choose $a_0 \in S$
• find $a_1$ giving the smallest power of $(a_1 - a_0)$
• find $a_2$ giving the smallest power of $(a_2 - a_0)(a_2 - a_1)$
• ...
• find $a_k$ giving the smallest power of $\prod_{i< k} (a_k - a_i)$

Here is my question:

• Bhargava observes that the integers in order $1,2,3, \dots, n$ form a $p$-ordered sequence for all $p$.
• Later in the same paper he observes the squares do as well $1^2,2^2,3^2, \dots, n^2$
• He also observes the powers of $2$ form a simultaneous $p$-ordering $1,2,4,8,\dots$

Does there exist a characterization of $S$ for when such a simultaneous ordering exists for all $p$? How can we read off the primes for which this isn't the case?

To my knowledge, this question remains open. It is difficult to characterize such sets. I would point out that all Fermat numbers is one of those sets [1]. More generally, if $E = O_f (a)$ is an the orbit of an integer $a$ under the iterated action of a polynomial $f \in \mathbb{Z}[X]$, then the sequence is simultaneous ordering exists for all $p$ (for sequence $E$).

Y fares

• Sorry, I meant: 1) the set of Fermat's numbers 2) The sequence( f ^ n (a)) is a simultaneous ordering of E = O_f (a) Nov 2 '14 at 14:29