# On a conjecture about cutting hypercubes

Consider the equation (1)

where *n* and all *c _{k}* are integers with

*n*≥ 0,

*c*≥ 1 and without loss of generality

_{k}*c*

_{k+1}≥

*c*. We can ask for the number of solutions of this equation for a given

_{k}*n*and call this number

*s*, thereby creating a sequence. This is sequence A263207 in the OEIS.

_{n}Geometrically interpreted, the sequence (*s _{n}*) provides the number of distinct ways to cut an

*n*-dimensional cube orthogonally into equally many outer parts, i.e. those which can be seen from the outside, and inner parts. All cuts must go through the whole body.

*c*is the number of cuts for the

_{k}*k*-th dimension.

Joerg Arndt asked whether there is some bound for the *c _{k}* when

*n*is fixed. Due to

*c*

_{k+1}≥

*c*, we can ask for a bound for

_{k}*c*instead.

_{n}I conjectured that the supremum is given by A204321(*n*) − 1 for *n* ≥ 1, where A204321 is the sequence defined by *a*_{1} = 4 and *a _{m}* =

*a*

_{m−1}(

*a*

_{m−1}− 1). The following is an attempt to prove this conjecture. At the end I will describe why I think that the proof attempt is not complete. I would appreciate any help in making it complete.

## Proof attempt

We attempt to construct a solution with the largest possible *c _{n}*.

We can safely assume *c _{k}* ≠ 1; otherwise the left side of equation (1) would be zero while the right side never is. Now transform equation (1) by dividing both sides by $(c_n-1)\prod_{k=1}^{n-1}(c_k+1)$ into equation (2):

The fraction on the right side of this equation is greater than 1 and converges to 1 as *c _{n}* approaches infinity. We attempt to make

*c*as large as possible, so we have to get the right side – and hence the left side – of the equation as close above 1 as possible.

_{n}On the left side we deal with a product which consists of the constant 2 multiplied by fractions which are all between 0 and 1. These fractions increase as *c _{k}* increases. So we will now choose values for

*c*

_{1}to

*c*

_{n−1}that are just big enough to raise the whole product above 1, but not greater than that.

### Step 1

At the start the left side of equation (2) is simply 2. With the definition *a*_{1} := 4 we can write the number two like this:

### Step 2

At *k* = 1 we will have to multiply the number two from the previous step by $\frac{c_1-1}{c_1+1}$ and choose *c*_{1} so that this product reaches just above 1:

To find *c*_{1}, multiply both sides of this inequality by the denominators to get *a*_{1}(*c*_{1} − 1) > (*a*_{1} − 2)(*c*_{1} + 1). Expanding yields *a*_{1}*c*_{1} − *a*_{1} > *a*_{1}*c*_{1} − 2*c*_{1} + *a*_{1} − 2. Add 2*c*_{1} + *a*_{1} and subtract *a*_{1}*c*_{1} to get 2*c*_{1} > 2*a*_{1} − 2. Dividing by two yields: *c*_{1} > *a*_{1} − 1. We need *c*_{1} to be just big enough to satisfy the inequality, so we choose *c*_{1} := *a*_{1}. Therefore

and when we define *a*_{2} := *a*_{1}(*a*_{1} − 1):

### Step *m*

In general, at *k* = *m* − 1 for any *m* > 1 we will take the result from the previous step $\frac{a_{m-1}}{a_{m-1}-2}$ and multiply that by $\frac{c_{m-1}-1}{c_{m-1}+1}$, again keeping the product just above 1:

Following the same path as described above we choose *c*_{m−1} := *a*_{m−1} because that is the smallest number which satisfies the inequality. We define *a _{m}* :=

*a*

_{m−1}(

*a*

_{m−1}− 1) and get:

### Step *n*

So, in step *n* we choose *c*_{n−1} := *a*_{n−1}, define *a _{n}* :=

*a*

_{n−1}(

*a*

_{n−1}− 1) and get:

Using the definitions made in the process we can rewrite this equation as:

We see that – by having chosen values for *c*_{1} to *c*_{n−1} so that the left side of equation (2) always reaches as close above 1 as possible – we get *c _{n}* =

*a*− 1, which is what the conjecture says.

_{n}## What the proof attempt lacks

In every step we bring the product as close above 1 as possible – always based on the previous step where we did the same. But could there be a better strategy? Could we get even closer to 1 in a certain step, if we deliberately did not approach 1 as close as possible in one of the earlier steps?

So far I do not have a compelling argument to rule that hypothetical possibility out. Can you help? That would be great. You can send an e-mail to the address provided at the bottom of this page.