Recurrence relation on a set of integers. Do we have $v_{n+q} = \sum_{j=0}^{q-1}\lambda_{j}v_{n+j}$? Order the set of positive integers of the form $2^{a}3^{b}$ $(a,b \in \mathbb{N}\cup\{0\})$ as $v_{1}<v_{2}<...$.
I.e $v_{1} = 1, v_{2} = 2^{1}3^{0} = 2, v_{3} = 2^{0}3^{1} = 3, v_{4} = 4, v_{5} = 6$ and so on.
Does there exist $q \in \mathbb{N}$ and constants $\lambda_{0},...,\lambda_{q-1} \in \mathbb{R}$ so that
$$v_{n+q} = \sum_{j=0}^{q-1}\lambda_{j}v_{n+j}$$
whenever $n \geq 1$?
I asked this question because I wanted to know whether we can get additive relations on certain sequences where the prime factorization properties of the studied sequence are somewhat well known (thereby somewhat linking addition and prime factorization).
Edit:
We know that $\lim_{n\rightarrow \infty}\frac{v_{n+1}}{v_{n}} = 1$ (as mentioned by Apass.Jack) and prove this result down here;
Since $\log_{2}(3)$ is irrational, the numbers $\log_{2}(3), 2\log_{2}(3),...$ are equidistributed modulo $1$.
Hence for each $k \in \mathbb{N}$ and $u \in \{0,1,...,k-1\}$ there exists $j_{u,k} \in \mathbb{N}$ for which
$$j_{u,k}\log_{2}(3) \equiv \eta_{u,k} \mod 1$$
where $\eta_{u,k} \in [\frac{u}{k},\frac{u+1}{k}]$. Next choose $r_{u,k} \in \mathbb{Z}$ so that $r_{u,k}+j_{u,k}\log_{2}(3) = \eta_{u,k}$.
Hence for all $n$ large enough for which $v_{n} \geq \max_{u \in \{0,...,k-1\}}\{3^{j_{u,k}}\}$ there exists $\tau(n,k)\in \mathbb{N}$ for which $\log_{2}(v_{n+\tau(n,k)}) - \log_{2}(v_{n}) \leq \frac{2}{k}$
Hence
$$\limsup_{n\rightarrow \infty} \frac{v_{n+1}}{v_{n}} \leq 2^{\frac{2}{k}}$$
By taking $k$ to infinity we know that the limit is 1.
 A: This sequence does not satisfy a linear recurrence. Any sequence of non-negative integers which does satisfy a linear recurrence must grow like $\sup_{1 \le i \le n} a_i \sim  Cn^k r^n$ for some non-negative integer $k$ and some real $r$ and we can show that this sequence does not have a growth rate of this form.
I will be a little sloppy in what follows but this argument could be tightened if necessary using explicit upper and lower bounds. Let's count the number of terms of this sequence which are less than or equal to some $N$. This is the number of non-negative integer solutions to $2^a 3^b \le N$, or $a \log 2 + b \log 3 \le \log N$. This is in turn asymptotically the area of a triangle with side lengths $\frac{\log N}{\log 2}, \frac{\log N}{\log 3}$, so it is asymptotically $\frac{(\log N)^2}{2 \log 2 \log 3} + O(\log N)$. Inverting this gives
$$v_n = \exp \left( \sqrt{(2 \log 2 \log 3)n} + O(1) \right)$$
which is faster than any polynomial but slower than any exponential and so cannot be the growth rate of a sequence satisfying a linear recurrence.
Some more comments: these are the $3$-smooth numbers, A003586 on the OEIS. You can look at the graphs on the OEIS, both the ordinary graph and the log graph, to get a hint that this sequence grows faster than polynomial but slower than exponential:

Edit: I've been asked to justify the claim about linear recurrences. The details of the precise claim are slightly tedious to get right (in particular, strictly speaking the original statement above was wrong and has been amended); for precise statements and proofs see Theorems IV.6, IV.7, IV.9, and Lemma IV.1 in Flajolet and Sedgewick's Analytic Combinatorics. The basic point is that the generating function of a sequence satisfying a linear recurrence is rational and so the asymptotic behavior of such a sequence is controlled by the dominant poles of this rational function, but there may be more than one, which leads to quasi-periodicities. However, that's not an issue in this case because 1) the quasi-periodicities don't affect the overall growth rate (defined e.g. as above via $\sup$), and 2) the sequence $v_n$ is increasing so exhibits no such quasi-periodicities anyway.
In any case, we can bypass writing down a precise general statement and argue as follows. If $a_n$ is a sequence satisfying a linear recurrence then $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = r$ exists and is equal to the absolute value of the largest root of the characteristic polynomial of the recurrence. The above argument (suitably tightened up) shows that $\lim_{n \to \infty} \sqrt[n]{v_n} = 1$, so the roots of the characteristic polynomial have absolute value $\le 1$. It follows (using the fact that a sequence $a_n$ satisfying a linear recurrence is a linear combination of sequences of the form $n^k \lambda^n$) that if $v_n$ satisfied a linear recurrence it would grow at most as fast as a polynomial, which the above argument (again, suitably tightened up) shows is impossible. More generally we have the following:

Proposition: If a sequence $b_n$ of positive integers has intermediate growth between polynomial and exponential, in the sense that $\liminf_{n \to \infty} \frac{\log b_n}{\log n} = \infty$ ($b_n$ grows faster than any polynomial) and $\limsup_{n \to \infty} \sqrt[n]{b_n} = 1$ ($b_n$ grows slower than any exponential), then $b_n$ cannot satisfy a linear recurrence.

