Explain this convergence among Pythagorean triplets Why do the ratios of successive values of integers $a$ and $c$, where $a^{2}+(a+1)^{2}=c^{2}$, appear to converge to
$$\frac{a_{n+1}}{a_{n}},\frac{c_{n+1}}{c_{n}}\rightarrow3+2\sqrt{2}$$
I rigorously generated all the {a,c} pairs where $a^{2}+(a+1)^{2}=c^{2}$ for $a<10^6$:
{3, 5}, {20, 29}, {119, 169}, {696, 985}, {4059, 5741}, {23660, 33461}, {137903, 195025}, {803760, 1136689}...
We can see that the fit to the ratio just gets better and better...
$$\frac{c_{8}}{c_{7}}=\frac{1136689}{195025}\approx5.82842712473...$$
$$3+2\sqrt{2}\approx5.82842712475... $$
Why does this ratio emerge? And with no other pairs between that satisfy the condition?
 A: $$\begin{align}
a^{2}+(a+1)^{2}&=c^{2} \\
2a^2+2a+1&=c^2 \\
2\left(a+\frac{1}{2}\right)^2 +\frac{1}{2}&=c^2 \\
(2a+1)^2+1&=2c^2 \\
2c^2-(2a+1)^2&=1 \\
2c^2-d^2&=1 \quad | \quad d=2a+1
\end{align}$$
The above pell equation:  $2c^2-d^2=1$ factors into $(c\sqrt 2-d)(c\sqrt 2 +d)=1$.  With the initial solution being $(c_0,d_0)=(1,1)$, we have that $(\sqrt 2-1)(\sqrt 2 +1)=1$.  Since we are permitted to multiply any equation by a constant we choose $(\sqrt 2-1)^2(\sqrt 2 +1)^2=(3-2\sqrt2)(3+2\sqrt 2)=1^2=1$.  Doing this results in
$$\begin{align}
(c\sqrt 2-d)(3-2\sqrt 2)(c\sqrt 2 +d)(3+2\sqrt 2)&=1 \\
&\implies \\
2(3c+2d)^2-(4c+3d)^2&=1
\end{align}$$ 
This allows you to show iterative solutions:
$$\begin{cases}
c_{k+1}=3c_k+2d_k \\
d_{k+1}=4c_k+3d_k
\end{cases}$$ 
Which can be solved by the method I outlined here to produce:
$$c_k=\frac{2+\sqrt 2}{4} \left(3+2\sqrt 2\right)^k+\frac{2-\sqrt 2}{4} \left(3-2\sqrt 2\right)^k$$
note that the effect of $(3-2\sqrt 2)^k$ diminishes with increasing k, explaining your observation
A: The sequence $a_n$ is https://oeis.org/A001652.
This sequence is defined by $$\cases{a_n=6a_{n-1}+a_{n-2}+2\\ a_0=0\\a_1=3}$$
Therefore the ratio $r_n={a_n\over a_{n-1}}$ has limit $3+2\sqrt 2$ as observed by Klaus Brockhaus. I'll add a proof later.
A: Supplemental, from the OP:
I generalized AmateurMathPirate's answer to arbitrary $a,a+n,c$:
$$a=\frac{n}{4}\left(\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)^{k}+\left(1-\sqrt{2}\right)\left(3-2\sqrt{2}\right)^{k}\right)-\frac{n}{2}$$
$$b=a+n $$
$$c=\frac{n}{4}\left(\left(2+\sqrt{2}\right)\left(3+2\sqrt{2}\right)^{k}+\left(2-\sqrt{2}\right)\left(3-2\sqrt{2}\right)^{k}\right)$$
It's easy to prove that $a,b,c$ are a triple for any choice of positive $n,k$
This form doesn't cover all triplets, for example {5,12,13}.
And for the record, I only figured out this relation after I read AmateurMathPirate's answer above.
