General term for series $0,2,12,70, 408...$ Given the curves $y=\sqrt{2x^2+1}$ and $x=\sqrt{2y^2+1}$, the first line (line with smallest y coordinate) which intersects the curves with integer coordinates is $x+y=1.$ This line intersects the curves at $(0,1)$ and $(1,0)$. Along with this it intersects the curves at $(-2,3)$ and $(3,-2)$. If we use the larger integer coordinates and construct another line $x+y = 2+3=5$ and intersect it with the curves we can find the next integer coordinates which are $(2,3)$ and $(3,2)$, but along with this we find another two points of intersection $(-12,17)$ and $(12,-17).$ If we repeat this process and find the intersection of the curves with $x+y=12+17=29$ we can find the next two integer coordinates: $70,99$. If we continue this process we can iteratively find the next integer coordinates.
So the question at hand is to find the $n^{th}$ largest integer intersecting points on these curves or in other words... the $n^{th}$ largest point of the curve $y=\sqrt{2x^2+1}$ whose both coordinates are integers. I believe by reiterating this line process we can find all integer coordinates, i.e. there are no integer coordinates that can't be found by reiterating this process. We won't skip any. (Feel free to disprove this, this is just my intuition.) Anyways, the main question to generate the $n^{th}$ term in this series, i.e. finding a general term, one the doesn't require redrawing the next line to find the next point.
 A: You are looking at a subsequence of the Pell numbers - probably every other one, with some variant on the initial condition:
$$
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, \ldots
$$
https://oeis.org/A000129
You can write a recursive formula but I am pretty sure there is no closed form solution for the $n$th term.
They come from an analysis of the Pell equation $x^2−2y^2=1$.
http://en.wikipedia.org/wiki/Pell%27s_equation
A: As noted on the OEIS page, the Pell number satisfy the recurrence
$$a_{n+1}-2a_n-1=0,\ n>1$$ with initial data $$a_0=0\\a_1=1$$
This is a homogeneous second-order linear difference equation, with constant coefficients, and it can be solved by the standard method.
The characteristic equation is $$r^2-2r-1=0,$$ which has the roots $$r=1\pm\sqrt2$$ so the general solution is $$a_n=a(1+\sqrt2)^n+b(1-\sqrt2)^n$$ for some constants $a$ and $b$.
Substituting the initial conditions into the formula, we find
$$\begin{align}
a_n&=\frac1{2\sqrt2}(1+\sqrt2)^n-\frac1{2\sqrt2}(1-\sqrt2)^n\\
&=\frac1{2\sqrt2}(\sqrt2+1)^n-\frac{(-1)^n}{2\sqrt2}(\sqrt2-1)^n\\
&=\frac1{2\sqrt2}(\sqrt2+1)^n-\frac{(-1)^n}{2\sqrt2}(\sqrt2+1)^{-n}\\
&=\frac1{2\sqrt2}\left(b_n+\frac{(-1)^{n+1}}{b_n}\right),
\end{align}$$ where $$b_n=(\sqrt2+1)^n$$
A: I would select the smaller absolute value with each pair of coordinates, thus $0,2,12,70,408,...$.
Suppose you have a line $x+y=a_n+b_n$ which intersects $y^2-2x^2=1$ at $(a_n,b_n)$ and thus intersects $x^2-2y^2=1$ at $(b_n,a_n)$.   The other intersection with $y^2-2x^2=1$ is determined by
$(a_n+b_n-x)^2-2x^2=1$
$-x^2+2(a_n+b_n)x+(a_n+b_n)^2-1=0$
The roots of this equation add up to $2(a_n+b_n)$ and by hypothesis one of them is $a_n$, so the other root defined as the next $x$ coordinate on the $y^2-2x^2=1$ hyperbola will be
$a_{n+1}=a_n+2b_n$  Eq. 1
Similarly we can seek the next $x$ coordinate on $x^2-2y^2=1$.  That quadratic equation will be
be
$x^2-2(a_n+b_n-x)^2=1$
$-x^2+4(a_n+b_n)x-2(a_n+b_n)^2-1=0$
With one root at $b_n$ and the two roots adding up to $4(a_n+b_n)$ we identify
$b_{n+1}=4a_n+3b_n$  Eq. 2
To get a recursion containing only $a_n$ terms, first increment Eq. 1 by one term to get
$a_{n+2}=a_{n+1}+2b_{n+1}$  Eq. 3
Then solve Eq. 3 for $b_{n+1}$ and Eq. 1 for $b_n$, and substitute these inte Eq. 2.  You end up with
$\color{blue}{a_{n+2}=6a_{n+1}-a_n}$
Using your initial $x$ coordinates on $y^2-2x^2+1$ you can put in the initial conditions $a_0=1,a_1=2$ and get the sequence
$0,2,12,70,408,2378,...$
matching your subsequent $x$ coordinates on $y^2-2x^2=1$.
A: I noticed sequence you got there,
$$ 1, 5, 29 , 169 , 985, 5741, \dots \dots $$
The formula to describe their $n$th term is
$$f[n] = 1+ 2\cdot \sum_{k=0}^{n-1} { \sum_{q=0}^{k+1} { 2^q \binom{2k+2}{2q+1} } }$$
The function grows very fast, $f[0] =1 , f[\infty ] = \infty$
I compared how fast $f[n]$ was rising with $n^n$ and $n!$ though it appears to rise faster, but it soon slow down with large $n$
