Gerhard "Now Show Me Your Moves" Paseman, 2019.06.10.

Update 2019.06.11:

Here is an attempt at justifying an approximation to the conjecture made in the post. While not strictly following the conjecture, we show that a good attempt at an optimum starts with n large enough, about m/2 fixed points, about m/4 cycles of length 2, and so on. I do not claim the result is an absolute optimum. I do suggest that the optimum is not far from this choice for n about 2m. For larger n there is more room to play.

I follow the original post, assume n is at least 2m, and look at the permutation of m cycles with one cycle of length n-m+1. The denominator of (n-m+1)((m-1!)) is easily calculated; what changes in cycle structure can make it smaller?

To start, we borrow from the large cycle to make at least (m-1)/2 cycles of length 2, giving possibly more cycles of length 2 than of length 1. (Don't worry, we will recycle some of these 2-cycles. Hah, I'm so funny.) Indeed, when we borrow about (m-1)/2 elements from the large cycle and convert that many 1-cycles to 2-cycles, we replace some terms in the denominator ((m-1)(m-2)(m-3)...) by terms that resemble (m-1)(m-3)(m-5)... (Or possibly (m)(m-2)(m-4)...), and n-m+1 gets replaced by something like n-3(m-1)/2. I am not bothering to use integer arithmetic, as the messy justification that the denominator is made smaller I leave to you. In any case, I assert we can reduce the denominator by replacing half or more of the 1-cycles by 2-cycles, especially when m is at least 6.

Now we can repeat this: replace slightly more than half the 2-cycles with 3-cycles, borrowing about m/4 from the large cycle. We get another reduction, but with a factor of about (3/2)^(m/4) involved as opposed to (2/1)^(m/2). We can iterate this up to log m times about, taking half or more k cycles and making (k+1) cycles from them. With care we borrow about m elements from the large cycle, and make cycle lengths up to log m.

Do we stop there? Not if n is big enough. We can now use a 1-cycle (we have about m/2 of them) and borrow from the large cycle to create a single cycle of length slightly larger than log m. We can repeat this as long as the cycle created has length less than the number of remaining 1-cycles. Similarly, we can reuse a few of the m/4 2-cycles this way.

We do not get quite up to cycles of length sqrt(m) this way because of the repetitions of small cycles, but for large n we have made a series of reductions in the denominator. Using this as a threshold, one can try perturbing this structure slightly to improve the denominator, without having to search the whole space.

End Update 2019.06.11.

Gerhard "Now Show Me Your Moves" Paseman, 2019.06.10.

Let's consider having c cycles of size k, d > c cycles of size k+1, and some number of largest cycles of size p >k+1. Let's shift an element out of d-c of the k+1 cycles, and put them on one p cycle. Just considering cycle lengths, in the product this represents a differential of 1 + (d-c)/p against (1+1/k)^(d-c). If we started with b many p cycles, we now have b-1, and so the product is reduced by an additional factor of b. So "shifting the excess" to a large enough cycle results in a reduction in denominator. Thus almost any cycle that optimizes the product will have more smaller cycles than larger cycles, with all lengths between 1 and l represented, followed by a solitary cycle of length p or more possibly larger than l+1.

(In case we have p=k+1, a similar argument with d-c>1 also works, as does the case when there is more than one cycle of maximal length. So, with the exception of a gap between l and p, we have cycles represented in decreasing number as length grows.)

So with care, the above can be turned into a proof of Aaron's unsurprising result. Now let us see if we can predict how big is l, the length of the largest (but one) cycle .

Suppose the two largest cycle lengths are l and p, with l less than p, and assume they occur uniquely. We split the l cycle into a 1 cycle and increment the p cycle by l-1. If we started with d many 1 cycles, we win (by shrinking the denominator) if (d+1)(p+l-1) is less than or equal to pl. If we have c many cycles of length l, replace pl by cpl in the previous inequality. So we win for sure if (d+1) is less than or equal to l/2.

If we are giving too much attention to one cycles, we can consider having c many k cycles and increasing their count by 1. The inequality now becomes a win if (c+1)k(p+l-k) is at most lp. In particular, if kc is at most l/2, consider shortening the l cycle to k and adding the excess up.

By considering a number of moves of this type, the search space for a cycle that optimizes the associated product should be readily obtained, even by hand, for large enough n.

Gerhard "Now Show Me Your Moves" Paseman, 2019.06.10.

Let's consider having c cycles of size k, d > c cycles of size k+1, and some number of largest cycles of size p >k+1. Let's shift an element out of d-c of the k+1 cycles, and put them on one p cycle. Just considering cycle lengths, in the product this represents a differential of 1 + (d-c)/p against (1+1/k)^(d-c). If we started with b many p cycles, we now have b-1, and so the product is reduced by an additional factor of b. So "shifting the excess" to a large enough cycle results in a reduction in denominator. Thus almost any cycle that optimizes the product will have more smaller cycles than larger cycles, with all lengths between 1 and l represented, followed by a solitary cycle of length p or more possibly larger than l+1.

(In case we have p=k+1, a similar argument with d-c>1 also works, as does the case when there is more than one cycle of maximal length. So, with the exception of a gap between l and p, we have cycles represented in decreasing number as length grows.)

So with care, the above can be turned into a proof of Aaron's unsurprising result. Now let us see if we can predict how big is l, the length of the largest (but one) cycle .

Suppose the two largest cycle lengths are l and p, with l less than p, and assume they occur uniquely. We split the l cycle into a 1 cycle and increment the p cycle by l-1. If we started with d many 1 cycles, we win (by shrinking the denominator) if (d+1)(p+l-1) is less than or equal to pl. If we have c many cycles of length l, replace pl by cpl in the previous inequality. So we win for sure if (d+1) is less than or equal to cl/2.

If we are giving too much attention to one cycles, we can consider having c many k cycles and increasing their count by 1. The inequality now becomes a win if (c+1)k(p+l-k) is at most lp. In particular, if kc is at most l/2, consider shortening the l cycle to k and adding the excess up. If you lose, it won't be by much.

By considering a number of moves of this type, the search space for a cycle that optimizes the associated product should be readily obtained, even by hand, for large enough n.

Gerhard "Now Show Me Your Moves" Paseman, 2019.06.10.