This is a really beautiful problem!

As there is some discussion in the question/comments/one of the answers about the history of this problem, let me narrate here what I found. I’m making this community-wiki as it is not exactly an answer to the question.

In *Mathematics Magazine*, which is published five times a year by the Mathematical Association of America, in the November 1982 issue (Vol. 55, No. 5), in the problems section (p. 300), the following (much easier) problem was posed anonymously (or rather, by “Anon, Erewhon-upon-Spanish River”, who seems a prolific contributor) as problem **1158** (I’ve changed the notation slightly):

Set $a_0 = 1$ and for $n \ge 1$, $a_n = a_{\lfloor n/2 \rfloor} + a_{\lfloor n/3 \rfloor}$. Find $\lim_{n\to\infty} a_n/n$.

This is a much easier problem, as $\frac12 + \frac13 \neq 1$. (Hint: try polynomial growth.)

Solutions to this problem 1158 were given in the January 1984 issue (Vol. 57, No. 1)’s Problems section (pp. 49–50), under the title *A Pseudo-Fibonacci Limit,* where it was solved by a host of people.

One of them was **Daniel A. Rawsthorne**, Wheaton, Maryland, who in the same section of the same issue (p. 42), proposed the harder problem ***1185**. The asterisk means that he proposed the problem without supplying a solution himself.

Set $a_0 = 1$ and for $n \ge 1$, $a_n = a_{\lfloor n/2 \rfloor} + a_{\lfloor n/3 \rfloor} + a_{\lfloor n/6 \rfloor}$. Find $\lim_{n\to\infty} a_n/n$.

This is a much harder problem, as we need to determine not just the rate of growth ("linear"), but also the constant proportionality factor.

Solutions were given in the January 1985 issue (Vol. 58, No. 1), in the Problems section (pp. 51–52), under the title *A Very Slowly Converging Sequence*, by (together) **P. Erdős, A. Hildebrand, A. Odlyzko, P. Pudaite, and B. Reznick**.

Note that the same page also says:

Also solved by **Noam Elkies** (student), who used Dirichlet series and the residue theorem; and partially (under the assumption that the limit exists) by Don Coppersmith, who gave the explicit formula
$$ a_n = 1 + 2 \sum \frac{(r+s+t)!}{r!s!t!} $$
where the sum is extended over all triples $(r, s, t)$ of nonnegative integers such that $2^r3^s6^t \le n$.

Anyway, in their solution, the authors EHOPR also say

We are writing a paper inspired by this problem and its generalizations.

and give general results, such as the following (I've simplified the numerator a bit):

Suppose $a_0 = 1$, and $a_n = \sum_{i=1}^{s} \lambda_i a_{\lfloor n/m_i \rfloor}$ for $n \ge 1$. Suppose also that not all $m_i$s are (integer) powers of some common integer. Then
$$ \lim_{n\to\infty} \frac{a_n}{n^{\tau}} =
\frac{ \sum_{i=1}^{s} \lambda_i - 1}{\tau \sum_{i=1}^{s} p_i \log m_i} $$
where $\tau$ is the unique real number satisfying $\sum_{i=1}^{s} \lambda_i / m_i^\tau = 1$, and $p_i = \lambda_i / m_i^\tau$ (assuming $\tau \neq 0$).

So in this case, we have
$$
\lim_{n\to\infty} \frac{a_n}{n}
= \frac{3 - 1}{\frac12\log2 + \frac13\log 3 + \frac16\log 6}
= \frac{12}{\log 432} \approx 1.977
$$
The sequence is OEIS A007731.

For the earlier problem (1158), we have, with $\tau \approx 0.78788$ the solution to $(1/2)^x + (1/3)^x = 1$, $p_1 = \frac1{2^\tau}$, and $p_2 = \frac1{3^\tau} = 1 - p_1$, the ratio $\frac{1}{\tau 2^{-\tau}\log 2 + 3^{-\tau}\log 3} \approx 1.469$, so $$a_n \sim 1.469 n^{0.78788}.$$

The paper they wrote was published as:

P. Erdős, A. Hildebrand, A. Odlyzko, P. Pudaite, and B. Reznick,

*The asymptotic behavior of a family of sequences*,

Pacific Journal of Mathematics, Vol. 126, No. 2 (1987), pp. 227–241: Link 1 (PDF), Link 2