I have performed a heuristic calculation of the density of the Klarner-Rado sequence. My calculation suggests that the Klarner's conjecture is false. Instead, I expect the number of terms in the Klarner-Rado sequence between $1$ and $N$ approaches $c N / \log N$ as $N$ approaches infinity, for some positive constant $c$.

The Klarner-Rado sequence is the range of the function $f : \{0, 1, 2\}^* \to \mathbb {N}$ whose domain is the set of sequence of the letters $\{0, 1, 2\}$, defined by

\begin{align*}
f (0x) &= 2 f (x) \\
f (1x) &= 3 f (x) + 2 \\
f (2x) &= 6 f (x) + 3 \\
f (e) &= 1
\end{align*}

where $e$ is the empty sequence. I will start with the easier question of counting Klarner-Rado numbers less than $N$ with multiplicity: that is, how many $x$ are there with $f (x) < N$? To simplify this further, I will replace $f$ with a purely multiplicative version of itself which differs from $f$ only by a constant factor:

\begin{align*}
\tilde {f} (0x) &= 2 \tilde {f} (x) \\
\tilde {f} (1x) &= 3 \tilde {f} (x) \\
\tilde {f} (2x) &= 6 \tilde {f} (x) \\
\tilde {f} (e) &= 1
\end{align*}

Then $g (x) = \log \tilde {f} (x)$ has a simple expression $g (c_0 c_1 \dots c_{r-1}) = a_{c_0} + a_{c_1} + \dots + a_{c_{r-1}}$ where $a_0 = \log 2$, $a_1 = \log 3$, $a_2 = \log 6$. To understand the distribution of $g$, and hence $\tilde {f}$, I will consider the function

$$ G (s) = \sum _{x \in \{0, 1, 2\}^*} e^{- s g (x)} = \sum _{x \in \{0, 1, 2\}^*} Z^{g (x)} $$

where $Z = e^{-s}$. The second expression displays how $G (s)$ is a sort of generating function, except that arbitrary real-valued exponents are allowed. (In a similar manner the Riemann zeta function can be thought of as the generating function $\zeta (s) = \sum_{n=1}^\infty Z^{\log n}$ where $Z = e^{-s}$.) $G (s)$ can be evaluated as

$$ G (s) = \sum_{n=0}^\infty (e^{-s \log 2} + e^{-s \log 3} + e^{-s \log 6})^n = \frac {1} {1 - (2^{-s} + 3^{-s} + 6^{-s})} $$

$2^{-s} + 3^{-s} + 6^{-s}$ is decreasing in $s$, and $2^{-1} + 3^{-1} + 6^{-1} = 1$. It follows that $G (s)$ converges for $s > 1$ and approaches infinity as $s \to 1$. In fact, $G (s) \sim c / (s-1)$ for $s \sim 1$ and some $c$.

This suggests that the number of $x$ with $g (x) < k$ is approximately $e^k$, with an averaged multiplicative error that is $O (e^{\epsilon k})$ for all $\epsilon > 0$. Moreover, the similar series

$$ \sum_{n=0}^{\infty} e^{-n s} = \frac {1} {1 - e^{-s}} $$

has a similar behavior near the divergence point, it is $\sim \frac {1} {s}$ for $s \sim 0$, and it is distributed uniformly. This suggests the multiplicative error for my estimate is even smaller than subexponential. On the other hand, I believe the analytic continuation of $G (s)$ has infinitely many poles with imaginary parts arbitrarily close to $1$, which suggests a significant oscillatory component to the density.

Ignoring these difficulties, I will assume that the number of $x$ with $g (x) < k$ is around $c e^k$ for some $c$. This is the same as the number of $x$ with $\tilde {f} (x) < N = e^k$, and this number is $c N$. Since $f$ and $\tilde {f}$ only differ by a bounded multiplicative factor, I expect that density of the Klarner-Rado sequence with multiplicity is bounded below by a constant.

What about the density of the Klarner-Rado sequence without multiplicity? With the above, this can be reformulated as the question: If $x$ is a sequence with $f (x) < N$, what is the probability that $x$ is first in dictionary order with this value for $f (x)$? If the density of the Klarner-Rado sequence is bounded below, so should this probability. On the other hand, this occurs for $x$ if and only if $x$ doesn't have any suffix $1 y$ with $f (1y) = f (0z)$ for some $z$. Heuristically the probability that such a $z$ exists for a given $y$ should be the same as the density of $\mathrm {ran} f$, times a constant to account for the nonuniformity of the sequence at different residues. So this probability is bounded below. But as $N$ goes to infinity, so does the number of suffixes of $x$ of the form $1 y$, and so the probability the $x$ is least in dictionary order goes to zero! We get a (heuristic) contradiction. Therefore I don't expect Klarner's conjecture about the density to hold.

Note that these suffixes are distributed in logarithmically uniform distribution on $[1, N]$. Let $h (k)$ be the density of the Klarner-Rado sequence on $[1, e^k]$. Then the probabilistic argument above suggests that

$$ \log h (k) \sim - c^{-1} \int_0^k h (t) dt $$

and so

$$ \frac {h' (k)} {h (k)} \sim - c^{-1} h (k) $$
$$ h' (k) \sim - c^{-1} h (k)^2 $$

Since the integral actually takes the form of a discrete sum close to $k$, this "differential equation" is more properly a complicated difference equation with the same general behavior as the differential equation I wrote. Ignoring this complication, we get a solution $h (k) \sim c/(k-k_0) \sim c/k$ ($k_0$ is irrelevant for the lowest-order asymptotic estimate). Therefore I expect the density on $[1, N]$ is $h (\log N) = c/\log N$, and there are $c N / \log N$ terms of the sequence on this interval.