A friend of mine asked me the following question: Imagine three circles on a plane, each having radius $1$ and all touching each other. If two of the circles lie on a table, what is the height of the third circle from the table? He also asked me the three dimensional version, we image four spheres, each having radius $1$ and they all touch each other. If three of them lie on a table, what is the height of the fourth one from the table?
I am sure there are many clever, geometric ways to calculate these heights, but I am not that clever so I wanted to go with what I know. I know Descartes' theorem and I wanted to use it. Recall that the Descartes' theorem states that if there are $n+2$ $n$-spheres, all touching each other in the $n$-dimensional space, than the radii are subject to the formula $$
(\sum_{i=1}^{n+2} 1/r_i)^2 = n(\sum_{i=1}^{n+2}1/r_i^2).
$$ Or setting $x_i=1/r_i$ to be the curvatures of the spheres, we have $$
(\sum_{i=1}^{n+2} x_i)^2 = n(\sum_{i=1}^{n+2}x_i^2).
$$ I thought I can use this formula by exhausting the space between the three spheres 
I can step-by-step calculate the radii of the smaller circles, then I can add them all up to find the answer.
Note that if the $n$-th circle has curvature $x_n=1/r_n$, then the $n+1$-th circle satisfies the formula $$
(1+1+x_{n}+x_{n+1})^2 = 2(1+1+x_n^2+x_{n+1}^2),
$$
so
$$
x_{n+1}^2-(2x_n+4)x_{n+1}+x_{n}^2-4x_n=0.
$$ Of course, $x_{n-1}$ is necessarily a solution of this equation (since $n-1$th, $n$th and the original two circles already touch each other), hence we obtain $$
x_{n+1} = 2 x_n+4-x_{n-1}.
$$ Using $x_1=1, x_2=3+2\sqrt{3}$, I found a closed formula for the sequence, and to the height $$
h = 1+\sum_{i=2}^{\infty} 2/x_i.
$$ Indeed, this gives me the correct number, that is, $\sqrt{3}$ (which, in hindsight, was actually easy to see). You might check WolframAlpha.
Now I wanted to use the same idea for $3$-dimensional case. Then, Descartes' Theorem gives $$
x_{n+1}-(x_n+3)x_{n+1}+x_n^2-3x_n=0,
$$ hence the relation is $$
x_{n+1} = x_n+3-x_{n-1}.
$$ Using $x_1 = 1, x_2 = 2+\sqrt{6}$, I get the sequence $1,2+\sqrt{6}, 4+\sqrt{6}, 5, 4-\sqrt{6}…$ However there is something wrong, this should have been an increasing sequence but $5<4+\sqrt{6}$ and also it becomes negative at some point.
What went wrong in dimension $3$?
Edit: It is also interesting to note that $1+2/(2+\sqrt{6})+2/(4+\sqrt{6})>\sqrt{3}$, so the height of the fourth sphere is more than $\sqrt{3}$. Of course, this is nonsense since in the $2$-dimensional case the answer is $\sqrt{3}$ and in the $3$-dimensional case the answer must be less than the $2$-dimensional case. But I don't understand what's wrong.
Best Answer
Your results are correct. It simply happens that spheres "go down through the hole", see figure below.