# Doubling dimension of a Euclidean space

The doubling dimension of a metric space $X$ is the smallest positive integer $k$ such that every ball of $X$ can be covered by $2^k$ balls of half the radius.

It is well known that the doubling dimension $d(n)$ of the Euclidean space $\mathbb R^n$ is $O(n)$, which means that there is a constant $C$ such that for large $n$ one has $d(n)\leq Cn$. A posteriori, I can find a new constant $D$ that works for all $n$. I would like to have an explicit description of this new constant. In other words,

Question: What explicit and possibly nice and small constant $D>0$ would guarantee that $d(n)\leq Dn$, for all $n$?

Edit. As observed by Igor Rivin, $D=\log 2$ should be good for $n\geq7$, by a theorem of Verger-Gaugry. Any idea for all $n$? I have to clarify that at the moment I am not interested in the best possible constant, but in some good-looking constant, something to make aesthetically pleasant a certain formula that I found out.

Just as a side remark: afaik $d(1)=2$ and $d(2)=7$, hence $D \geq 3.5$ – cubic lettuce Feb 21 '15 at 23:34
As shown in this paper,Theorem 1.2, $D \leq \log 2.$ I remark that this paper came up in my answer to this question, and there is a bug for small $n$ ($n < 7$), but the author's interest was apparently similar to yours, so the large $n$ results should be correct. (the paper is: "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n," by Jean-Louis Verger-Gaugry) - Thank you very much. What can we say for small$n$? I am really interested in all values of$n$. Maybe also to know that say$D=2$is good enough would be OK. Indeed, for the moment I want to put this constant in a as nice as possible formula for all$n$and then maybe discuss the fact that can be sharpened for$n\geq7$... – Valerio Capraro Jun 27 '12 at 14:23 I am having a look at the paper and maybe I am missing something. He fixes a radius$T>\frac12$and answers the question of how many balls of radius$\frac12$are needed to cover a ball of radius$T$. My case is little different, because the covering balls have radius$T/2$. Well, it is possible that one can go down inductively and apply that formula, but I am little in trouble with that terrible formula. Moreover, that formula holds only for$T\leq\frac{n}{2\log(n)}$... In two words: I'm confused! – Valerio Capraro Jun 27 '12 at 14:40 For your problem$T=1,$and the inequality is vacuously satisfied... – Igor Rivin Jun 27 '12 at 15:15 Yes, indeed! thanks again – Valerio Capraro Jun 27 '12 at 15:33 I am sorry, but I cannot understand how you get$\log 2$. Indeed, let$f(n)$be Verger-Gaugry's estimation in Theorem 1.2 but without$2^n$. It seems to me that$d$should verify the property that$\log_2(f(n))\leq(d-1)n$, for all$n$. One can easily see that every$d>1$is eventually good, but the function$f(n)$diverges and then it seems to me impossible to find a constant$d\leq1\$. Am I missing something? – Valerio Capraro Jun 30 '12 at 16:48