Looking for a certain deterministic card shuffle Does there exist such a number $r$, that for some partition of $r$, the least common multiple of elements of the partition is equal to the factorial of $r$? 
How to determine the partition that gives the largest such least common multiple for a certain $r$?
For example if $r=11$, then one possible partition of $r$ is $5+3+2$ the least common multiple $lcm(5,3,2)=30$, but, sadly $30\neq 39916800=11!$
Context: For r cards there are a total of $r!$ different permutations they can take upon in the deck. Assume you are shuffling cards in some regular, deterministic manner (There is a certain one-to-one correspondence of each cards position before the shuffle and after the shuffle). Your shuffle will then contain certain cycles of cards (for example, the second card might go to the position of the fourth, the fourth card to the position of the fifth, and the fifth to the second, a cycle of length 3). After a certain ammount of such shuffles the cards will obviously go back in order. The amount of shuffles is equal to the least common multiple of the cycles lenghts that compromise the shuffle. And the sum such cycle lenghts is precisely the number of cards. Thus if such an $r$ and partition of $r$ that satisfies the conditions above exists, then there is a determenistic shuffle that takes on each possible permutation of the deck.
So an alternative formulation of the question might be 
Does there exist a deterministic shuffle of r cards that 
when applied multiple times, 
goes through all the possible permutations of those r cards?
What shuffle would go through the maximum number of permutations of r cards?

I have a strong intuition that possibly $r!$ grows too fast and the least common multiple of the partitions isnt going to come even close. I am having a hard time to even find the greatest least common multiple of a number though.
 A: Consider the simple partition $x+y=r$. For this we have
$$
\operatorname{lcm}(x,y)\leq xy=x(x-r)
$$
where the latter is an expression in $x$ that has its maximum at $x=r/2$ corresponding to $x=y$. So the least common multiple is bounded by a product that has its maximum when the factors are equal. This principle extends inductively to a product of $n$ factors whose sum is $r$. So any partition
$$
\sum_{i=1}^n a_i=r
$$
has its least common multiple bounded from above
$$
\operatorname{lcm}(a_1,a_2,...,a_n)\leq \prod_{i=1}^n a_i\leq \left(\frac{r}{n}\right)^n=\frac{r^n}{n^n}
$$
Using calculus, one can show that $\ln(r^x/x^x)=x(\ln(r)-\ln(x))$ has its maximum at $x=r/\operatorname e$ in which case we have $x(\ln(r)-\ln(x))=r/\operatorname e$. So $r^x/x^x\leq \operatorname e^{r/\operatorname e}<1.5^r$. For $r\geq 3$ we have $1.5^r<r!$ which is easy to see since the LHS grows by a factor $1.5$ whereas the RHS grows by a factor $(r+1)$ when $r$ is increased by $1$. Thus for $r\geq 3$ we get
$$
\operatorname{lcm}(a_1,a_2,...,a_n)<1.5^r<r!
$$
for any partition $a_1,a_2,...,a_n$ of $r$. This shows that your desired algorithm for shuffling a deck of cards is not possible.

Regarding just obtaining a partition with the largest possible LCM, my analysis shows that if you could have $n\approx r/\operatorname e$ partitions whose average size would then have to be approximately $\operatorname e\approx 2.718$ and you also somehow succeeded in having those to be relatively prime, you would obtain the maximal LCM. So for $r=17$ we should have $2,3,5,7$ which has $\operatorname{lcm}(2,3,5,7)=210<17!=355687428096000$ as the maximal solution. Not very impressive!
