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: 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 $$\log(a_n) \approx n\cdot 0,6\overline{1} = n \frac{11}{18} \Rightarrow \\ a_n\approx e^{n \frac{11}{18}}$$
A: Computing by brute force, we get that $a_{62}=134208905031$. That has $11$ significant figures. Since the error is multiplied by the same factor as the significant part, if we use this number and multiply by $\left(\frac32\right)^{999999937}$ and $\left(\frac32\right)^{999999938}$, we get
$$
a_{999999999}\doteq1.2294567472\times10^{176091259}
$$
and
$$
a_{1000000000}\doteq1.8441851208\times10^{176091259}
$$
to $10$ places, depending on which you mean by "billionth".
A: The OEIS entry for $2,3,4,6,9,13,19,28,42,\ldots$, adjusted to agree with the OP's starting index, says that
$$a_n=\lceil K(3/2)^{n+1}\rceil$$
where $K\approx1.0815136685898448773046339885995$, with more digits available at a linked-to reference.  If all you want is an approximation in scientific notation, then the ceiling function round-up can be ignored.
Let's let $n=10^9$.  Then Wolfram Alpha gives us
$$\left({3\over2}\right)^{n+1}\approx 1.70518891653445589648412113069806868 × 10^{176091259}$$
and thus
$$a_n\approx1.8441851207599221922452580533 × 10^{176091259}$$
with agrees with what robjohn and others found, with a few more digits.  (I truncated a few digits from the WA output, to stay on the safe side.)
