# On a conjecture about cutting hypercubes

Consider the equation (1)

$$2\prod_{k=1}^n(c_k-1)=\prod_{k=1}^n(c_k+1)$$

where n and all ck are integers with n ≥ 0, ck ≥ 1 and without loss of generality ck+1 ≥ ck. We can ask for the number of solutions of this equation for a given n and call this number sn, thereby creating a sequence. This is sequence A263207 in the OEIS.

Geometrically interpreted, the sequence (sn) 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. ck is the number of cuts for the k-th dimension.

Joerg Arndt asked whether there is some bound for the ck when n is fixed. Due to ck+1 ≥ ck, we can ask for a bound for cn instead.

I conjectured that the supremum is given by A204321(n) − 1 for n ≥ 1, where A204321 is the sequence defined by a1 = 4 and am = am−1(am−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 cn.

We can safely assume ck ≠ 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):

$$2\prod_{k=1}^{n-1}\frac{c_k-1}{c_k+1}=\frac{c_n+1}{c_n-1}$$

The fraction on the right side of this equation is greater than 1 and converges to 1 as cn approaches infinity. We attempt to make cn 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.

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 ck increases. So we will now choose values for c1 to cn−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 a1 := 4 we can write the number two like this:

$$\frac{a_1}{a_1-2}$$

### 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 c1 so that this product reaches just above 1:

$$\frac{a_1}{a_1-2}\cdot\frac{c_1-1}{c_1+1}>1$$

To find c1, multiply both sides of this inequality by the denominators to get a1(c1 − 1) > (a1 − 2)(c1 + 1). Expanding yields a1c1 − a1 > a1c1 − 2c1 + a1 − 2. Add 2c1 + a1 and subtract a1c1 to get 2c1 > 2a1 − 2. Dividing by two yields: c1 > a1 − 1. We need c1 to be just big enough to satisfy the inequality, so we choose c1 := a1. Therefore

$$\frac{a_1}{a_1-2}\cdot\frac{c_1-1}{c_1+1}=\frac{a_1}{a_1-2}\cdot\frac{a_1-1}{a_1+1}=\frac{a_1(a_1-1)}{a_1^2-2a_1+a_1-2}=\frac{a_1(a_1-1)}{a_1(a_1-1)-2}$$

and when we define a2 := a1(a1 − 1):

$$\frac{a_1}{a_1-2}\cdot\frac{a_1-1}{a_1+1}=\frac{a_2}{a_2-2}$$

### 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:

$$\frac{a_{m-1}}{a_{m-1}-2}\cdot\frac{c_{m-1}-1}{c_{m-1}+1}>1$$

Following the same path as described above we choose cm−1 := am−1 because that is the smallest number which satisfies the inequality. We define am := am−1(am−1 − 1) and get:

$$\frac{a_{m-1}}{a_{m-1}-2}\cdot\frac{a_{m-1}-1}{a_{m-1}+1}=\frac{a_m}{a_m-2}$$

### Step n

So, in step n we choose cn−1 := an−1, define an := an−1(an−1 − 1) and get:

$$\frac{a_{n-1}}{a_{n-1}-2}\cdot\frac{a_{n-1}-1}{a_{n-1}+1}=\frac{a_n}{a_n-2}$$

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

$$2\prod_{k=1}^{n-1}\frac{a_k-1}{a_k+1}=\frac{(a_n-1)+1}{(a_n-1)-1}$$

We see that – by having chosen values for c1 to cn−1 so that the left side of equation (2) always reaches as close above 1 as possible – we get cn = an − 1, which is what the conjecture says.

## 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.