Solving a Recurrence: $a_0 = 2$

[magneto1]

Not sure if Discrete Math is the correct category, but I'm looking for some idea / hint on how to tackle the following recurrence.

$a_0 = 2$, and $a_{n+1} = 2a_{n} + \sqrt{3(a_n)^2 - 12}$ for $n \in \Bbb{N}$.

Some attempts to massage the equation got me: $(a_{n+1}-a_n)^2 = 2a_na_{n+1} - 12$, which is equally messy.

 

[John Harris1]

Not sure if Discrete Math is the correct category, but I'm looking for some idea / hint on how to tackle the following recurrence.

$a_0 = 2$, and $a_{n+1} = 2a_{n} + \sqrt{3(a_n)^2 - 12}$ for $n \in \Bbb{N}$.

Some attempts to massage the equation got me: $(a_{n+1}-a_n)^2 = 2a_na_{n+1} - 12$, which is equally messy.

What do you want to do? If it is to find the limit for a_n, then a nice 'trick' is to just drop all the subscripts. That is, if a_n goes to a as n goes to infinity, so does a_n-1, a_n-2, a_n-3, ..., a_n-k for any fixed k. Of course, that only works if there is a 'nice' limit. However this a_n goes to infinity which is not nice.

EDIT: Read more here
A003500 - OEIS

 

[magneto1]

The original question was to show all members of the sequence are integers. However, I am interested in a "closed form" formula for this recurrence. (Since the sequence is increasing, its limit does not exist.)

 

[John Harris1]

The original question was to show all members of the sequence are integers. However, I am interested in a "closed form" formula for this recurrence. (Since the sequence is increasing, its limit does not exist.)

See EDIT above

 

[Nono713]

Some thoughts: if you rewrite the equation as:
$$a_n = 2a_{n - 1} + \sqrt{3a_{n - 1}^2} - \delta(a_{n - 1})$$
Where $\delta$ is defined for $z \geq 2$ as:
$$\delta(z) = \sqrt{3z^2} - \sqrt{3z^2 - 12}$$
Then we have:
$$a_n = 2a_{n - 1} + \sqrt{3} a_{n - 1} - \delta(a_{n - 1}) = (2 + \sqrt{3}) a_{n - 1} - \delta(a_{n - 1})$$
Expanding out we find:
$$a_n = (2 + \sqrt{3}) \left ( (2 + \sqrt{3}) a_{n - 2} - \delta(a_{n - 2}) \right ) - \delta(a_{n - 1})$$
$$a_n = (2 + \sqrt{3})^2 a_{n - 2} - (2 + \sqrt{3})\delta(a_{n - 2}) - \delta(a_{n - 1})$$
Iterating we (eventually) get:
$$a_n = 2 \cdot (2 + \sqrt{3})^n - \sum_{k = 1}^n (2 + \sqrt{3})^{k - 1} \delta(a_{n - k})$$
So now the problem is to bound the error term in the sum. We see that $\delta(z)$ tends to $2 \sqrt{3} / z$ quite quickly, which combined with the asymptotic growth of the sequence of $a_m \approx (2 + \sqrt{3})^m$ yields the good approximation:
$$\delta(a_m) \approx \frac{2 \sqrt{3}}{(2 + \sqrt{3})^m}$$
Allowing us to approximate our expression as:
$$a_n \approx 2 \cdot (2 + \sqrt{3})^n - \sum_{k = 1}^n (2 + \sqrt{3})^{k - 1} \frac{2 \sqrt{3}}{(2 + \sqrt{3})^{n - k}} = 2 \cdot (2 + \sqrt{3})^n - \frac{2 \sqrt{3}}{(2 + \sqrt{3})^{n + 1}} \sum_{k = 1}^n (2 + \sqrt{3})^{2k}$$
Which is a geometric series, and we get:
$$a_n \approx 2 \cdot (2 + \sqrt{3})^n - \frac{(2 + \sqrt{3})^{2n} - 1}{(2 + \sqrt{3})^{n + 1}} \left [ \frac{12 + 7 \sqrt{3}}{3 + 2 \sqrt{3}} \right ]$$
Which seems to be a good, if not exact approximation, and probably actually reduces to the closed form in John Harris' OEIS link.​