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For each fixed positive rational $x$, the number of subsets $A\subseteq[n]$ with $\sum_{a\in A}1/a=x$ satisfies

$$N_n(x)=2^{c_xn+o_x(n)},\qquad \lim_{n\to\infty}\frac{\log_2 N_n(x)}n=c_x. \tag{1}$$

Here

$$c_x=\int_0^1h\!\left(\frac1{1+e^{\lambda_x/y}}\right)\,dy, \qquad \int_0^1\frac{dy}{y(1+e^{\lambda_x/y})}=x,\quad\lambda_x>0.$$

The exponent is continuous and strictly increasing, tends to 0 at $x\downarrow0$, and tends to 1 at infinity. In particular $0<c_x<1$. The source-reported decimal $c_1\approx0.91117$ has not been numerically certified in this compilation. Equation (1) is an exponential-rate statement: it does not assert $N_n(x)/2^{c_xn}\to1$.

Source: published PDF, Theorem 1, p. 2, with proof completed on p. 11. The ordinary proof is complete relative to the explicitly listed external estimates. This source unit does not compile the materially distinct Liu–Sawhney counting proof.

Bears on. Problem 297.

Proof

The characterization and properties of $c_x$ were proved in entropy_exponent. Since $N_n(x)\le R_n(x)$, the upper bound in lemma_1 and the fixed-$x$ limit in lemma_2 give

$$\limsup_{n\to\infty}\frac{\log_2 N_n(x)}n\le c_x$$

whenever the logarithm is defined, with the same upper bound trivially true if a count is zero.

Fix any sufficiently small $\varepsilon$ with $0<\varepsilon<x$. The denominator of this fixed rational $x$ is $n^{1-\varepsilon}/2$-powersmooth for large $n$. Also $x\le\xi(\varepsilon)\log n$ eventually. theorem_4 therefore gives a positive count and

$$\liminf_{n\to\infty}\frac{\log_2 N_n(x)}n\ge c_x-8\varepsilon.$$

This argument holds for every sufficiently small fixed $\varepsilon>0$; first take the limit in $n$ with $\varepsilon$ fixed, then let $\varepsilon\downarrow0$. The upper and lower bounds prove (1). For $x=1$, $c_1<1$ answers in the negative Erdős and Graham's question whether $N_n(1)=2^{n-o(n)}$, while determining the exact exponential rate. No assertion about a multiplicative error or a numerical evaluation of the defining integrals is needed.