Q:

Finding the billionth number in the series: $2, 3, 4, 6, 9, 13, 19, 28, 42, \ldots $?

Series is defined as
$$a_{n+1} = \lfloor\frac{3\cdot a_n}{2}\rfloor,\qquad  a_0 = 2$$
It can be viewed as the number of animals starting from a single pair if any pair of animals can produce a single offspring.
The problem is if the most straightforward computer program was implemented (recurrence relation in a loop), it seems it would not finish for the lifetime of its creator...
How to approach this differently?
EDIT: This number is very large. Its approximation in scientific notation (let's say first 10 decimal digits, and decimal exponent) would be sufficient.

A:

The related series $$x_n=\lceil\frac32x_{n-1}\rceil,\ x_0=1$$ is studied at MathWorld. Since for all integers $n$, $\lceil\frac32n\rceil=\lfloor\frac32(n+1)\rfloor-1$, we can rewrite this as $x_n+1=\lfloor\frac32(x_{n-1}+1)\rfloor$, and $x_0+1=2$; thus we have $a_n=x_n+1$ for all $n$. That page (links to a paper that) derives that there exists a real number $K$ such that for all $n$, $$x_n=\lceil K(3/2)^n\rceil,$$ which tells us that the sequence very nearly follows the expected power law, although the accumulated roundoff error $K$ is hard to calculate in advance.
So let's take the direct route. Define the sequences $a_{k+1}^+=\frac32a_k^+$,  $a_{k+1}^-=\frac32a_k^--1$, where the initial values set $a_m^+=a_m^-=a_m$ for some fixed basepoint $m$. Then it is easy to show that
$$a_n^+=\Big(\!\frac32\!\Big)^{n-m}a_m;\qquad a_n^-=\Big(\!\frac32\!\Big)^{n-m}(a_m-2)+2.$$
Now we have the inequality $a_n^-\le a_n\le a_n^+$ because $x-1\le\lfloor x\rfloor\le x$, so this directly gives bounds on $a_n$ for $n=10^9$. Since you want around 10 digits of accuracy, we need to pick an $a_m\ge 2\cdot 10^{10}$: for example, $a_{58}=26510400994$. Then we have:
$$\log_{10}(a_m-2)\le\log_{10}a_n-(n-m)\log_{10}(3/2)\le\log_{10}a_m$$
$$\log_{10}26510400992\le\log_{10}a_{10^9}-(10^9-58)\log_{10}(3/2)\le\log_{10}26510400994$$
$$176091259.2658045137\underline{02}\le\log_{10}a_{10^9}\le176091259.2658045137\underline{34}$$
Thus the exponent of $a_{10^9}$ is $176091259$, and the fractional part is $10^{0.2658045137}\approx1.844185120$ (where only the digits that agree for both bounds are shown). Putting it all together, we have
$$a_{10^9}=1.844185120\dots\times10^{176091259}.$$
By repeating this process with bigger basepoints, you can confirm as many digits as you want. Here's a few more correct digits:
$$a_{10^9}=1.844185120759922192245258053300812265366889206992486395592885\times10^{176091259}.$$

A:

I think some graphs like these ones can be very helpful...where $$A061418(n) = a_{n+1} = \Big\lfloor\frac{3\cdot a_n}{2}\Big\rfloor,\qquad  a_0 = 2$$

You can easily see that the Logarithimc scatterplot seems to grow linearly, so if you're alloking for an approximation of the billionth number of the sequence you can start from the second graph. With a bit of homemade linear interpolation you get that $$\log(a_n) \approx n\cdot 0,6\overline{1} = n \frac{11}{18} \Rightarrow \\ a_n\approx e^{n \frac{11}{18}}$$

