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Let $f(d)$ be the least universal number of smaller-diameter parts in $\mathbb R^d$. For $m=4k$ with $k$ a prime power, set

$$d_m=\binom m2-1, \qquad Q_m=\frac{\binom m{m/2}}{\binom m{m/4}}.$$

The equal-cut construction proves $f(d_m)\ge Q_m$. This page completes the two asymptotic steps in the last sentence of Section 2 on physical PDF p. 2 (journal p. 61).

Exponential rate on the construction dimensions

For $0<\alpha<1$, put

$$H(\alpha)=-\alpha\log\alpha-(1-\alpha)\log(1-\alpha).$$

The fixed-density consequence of Stirling's formula recorded in external_inputs gives

$$\begin{aligned} \log Q_m &=m\bigl(H(1/2)-H(1/4)\bigr)+O(\log m)\\ &=m\left(\frac34\log3-\log2\right)+O(\log m). \end{aligned}$$

Since $d_m=m(m-1)/2-1$, we have $\sqrt{d_m}=m/\sqrt2+O(1)$, and hence

$$\lim_{\substack{m\to\infty\\4\mid m}} \frac{\log Q_m}{\sqrt{d_m}} =\sqrt2\left(\frac34\log3-\log2\right).$$

The limiting base is therefore

$$B=\exp\!\left(\sqrt2\left(\frac34\log3-\log2\right)\right) =1.2032138141823219\ldots>1.203.$$

For a finite check of the strict decimal comparison, use

$$\log x=2\sum_{j=0}^{N}\frac{z^{2j+1}}{2j+1}+R_N, \quad z=\frac{x-1}{x+1}, \quad 0<R_N<\frac{2z^{2N+3}}{(2N+3)(1-z^2)}.$$

Taking $N=25,40,10$ for $x=2,3,1203/1000$, respectively, gives the exact rational enclosures

$$\begin{aligned} 0.69314718055994530941&<\log2 <0.69314718055994530943,\\ 1.09861228866810969138&<\log3 <1.09861228866810969141,\\ \log(1.203)&<0.18481843699254182520. \end{aligned}$$

Squaring the rational endpoints gives

$$1.41421356237309504880<\sqrt2 <1.41421356237309504881.$$

These inequalities imply

$$\sqrt2\left(\frac34\log3-\log2\right) >0.18499615534959264419 >\log(1.203).$$

Thus, for every sufficiently large eligible $m=4k$,

$$f(d_m)\ge Q_m>(1.203)^{\sqrt{d_m}}.$$

Transfer to arbitrary sufficiently large dimensions

For a real $x\ge1$, write

$$d(x)=\binom{4x}{2}-1=8x^2-2x-1.$$

Given a large integer $D$, let

$$x_D=\frac{1+\sqrt{8D+9}}8,$$

so that $d(x_D)=D$, and let $p_D$ be the largest prime at most $x_D$. The prime number theorem gives $p_D/x_D\to1$. Set

$$e_D=\binom{4p_D}{2}-1=8p_D^2-2p_D-1.$$

Then

$$e_D\le D, \qquad \frac{e_D}D\longrightarrow1.$$

The prime $p_D$ is an allowed prime power in the construction. Isometrically embedding $\mathbb R^{e_D}$ into $\mathbb R^D$ shows that $f$ is nondecreasing, so

$$f(D)\ge f(e_D)>(1.203)^{\sqrt{e_D}}$$

for all sufficiently large $D$. Since

$$\sqrt{e_D/D}\to1 \quad\text{and}\quad \frac{\log1.2}{\log1.203}<1,$$

the last quantity is at least $(1.2)^{\sqrt D}$ once $D$ is sufficiently large. This proves the claimed transfer.