Looking for the best way to find Pythagorean triples where $B-A=\pm1$. Pythagorean triples where $A-B=\pm1$ are some of the rarest; the $19^{th}$ has terms $A,B,C$ in the quadrillions. I found a formula in a book, "Pythagorean Triangles" that generates them in sequence beginning with a seed triple $T_1=(3,4,5)$: $A=3A+2C+1\quad B=3A+2C+2\quad C=4A+3C+2$ will generate $T_2=(20,21,29)\quad T_3=(119,120,169)\quad T_4=(697,696,985)$ and so on. Nineteen iterations gives you the first $19$ triples and that is great but I developed a formula which uses less computation until you get to the $n^{th}$ triple you want to view. It generates the parameters $(m,n)$ to feed Euclid's formula:$$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$

The formula is $\quad m_{x+1}=2m_x+n_x\quad n_{x+1}=m_x\quad $ and generates the following pairs with a seed: $P_0=(1,0)$. $$P_1=(2,1)\quad P_2=(5,2)\quad P_3=(12,5)\quad P_4=(29,12)\quad P_5=(70,29)\quad P_6=(169,70)\quad ...$$

I would like to be able to generate the $6^{th}$ or the $1000^{th}$ pair directly without generating $1$-thru-$5$ or $1$-thru-$999$ to get there but I haven't been able to figure out any way to generate an individual pair directly. I have tried $2=2^1, 5=2^2+2^1, 12=2^3+2^2, hmm, 29=2^4+2^3+2^2+2^1+2^0 $ and other things like factors of $2,5,12,29...$ and I'm out of ideas.

Is it possible to generate an $x^{th}$ member pair using just $x$ as an input number or, by the nature of this sequence, is it required to generate all of them in order until I get to the desired pair?

Someone said my formula does not work but here it is working in a spreadsheet.
The computation of the Pythagorean triples of the form $T_n=(a_n,a_n+1,c_n)$ is equivalent to the computation of some convergents of the continued fraction of $\sqrt{2}$: in particular
$$ [1;\underbrace{2,2,\ldots,2,2}_{2n\text{ times}}]=\frac{2a_n+1}{c_n} $$
where
$$ c_n = \frac{(1+\sqrt{2})^{2n+1}-(1-\sqrt{2})^{2n+1}}{2\sqrt{2}},\qquad 2a_n+1 =\frac{(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}}{2}=d_n $$
both fulfill the recurrence $\ell_{n+2}=6\ell_{n+1}-\ell_n$. They can be expressed in terms of $D_n$ and $D_{n+1}$, where
$$ D_n = (3+2\sqrt{2})^n+(3-2\sqrt{2})^n =\sigma^n+{\bar{\sigma}}^n$$
is the trace of the $n$-th power of a $2\times 2$ matrix. This sequence fulfills

$$ D_{2n} = D_n^2-2,\qquad D_{2n+1}=D_n D_{n+1}-6 \tag{R}$$
so the couple $(D_n,D_{n+1})$ can be computed by a repeat-squaring algorithm. A concrete example will hopefully clarify how. Let us assume to want to compute $D_{23}$ and $D_{24}$. The binary representation of $23$ is $10111_2$, so we compute the couples $(D_m,D_{m+1})$ for $m=1_2,10_2,101_2,1011_2$ and finally $10111_2$ via $(R)$.
$$ (D_1,D_2)=(6,34) $$
$$ (D_2,D_3)= (34,198)$$
$$ (D_5,D_6)=(6726,39202) $$
$$ (D_{11},D_{12})=(263672646,15367968024) $$
$$ (D_{23},D_{24})=(405211279147678086,2361744410637427202)$$
This gives us $c_{23}$ and $d_{23}$, thus $T_{23}$, with no more than $3\log_2(23)$ multiplications.
Edited answer...

which starts with a comment.

The title Looking for the best way to find Pythagorean triples where $B−A=\pm1$ is possibly misleading for the actual question, which after many comments can be extracted only from the fact that the initial answer for the $(A,C)$ recursion (remained in the sequel) is not the wanted answer. Please always make clear in the OP which is the question (without alternatives, best stated as question, not as a wish). 

The OP gives two ways to construct Pythagorean triples, one is "from a book" and constructs triples $T_x=(A_x,B_x,C_x)$ recursively, the other one comes from a formula, it constructs intermediate pairs $P_x=(m_x,n_x)$ given by an other recursion, that may be used to reconstruct $(A_x,B_x,C_x)$. As i extract  the information from the comments, we need only the $P_x$ (despite of the title, and the initial digression / divagation on triples). OK, this is even simpler.



We have the linear recursion formula for the pair written as column vector
$$
\begin{aligned}
\begin{bmatrix}
m_{x+1}\\
n_{x+1}
\end{bmatrix}
&amp;=
\begin{bmatrix}
2&amp;1\\1&amp;0
\end{bmatrix}
\begin{bmatrix}
m_{x}\\
n_{x}
\end{bmatrix}
\\
&amp;=
\underbrace{
\begin{bmatrix}
1+\sqrt 2&amp;1-\sqrt 2\\1&amp;1
\end{bmatrix}}_C
\underbrace{
\begin{bmatrix}
1+\sqrt 2&amp;\\&amp;1-\sqrt 2
\end{bmatrix}}_D
\underbrace{
\frac 1{2\sqrt 2}
\begin{bmatrix}
1 &amp;\sqrt 2-1\\-1&amp;\sqrt 2+1
\end{bmatrix}}_{C^{-1}}
\begin{bmatrix}
m_{x}\\
n_{x}
\end{bmatrix}
\ ,
\end{aligned}
$$
with a diagonal matrix $D$ above. The powers of
$\begin{bmatrix}2&amp;1\\1&amp;0\end{bmatrix}$ are not immediate, but the powers of $D$ are.
We get directly for your pairs formula the explicit version with proof: 
$$
\begin{aligned}
\begin{bmatrix}
m_x\\n_x
\end{bmatrix}
&amp;=
(CDC^{-1})^x\begin{bmatrix}m_0\\n_0\end{bmatrix}
=
CD^xC^{-1}\begin{bmatrix}1\\0\end{bmatrix}
\\[2mm]
&amp;=
\frac 1{2\sqrt 2}
\begin{bmatrix}
1+\sqrt 2&amp;1-\sqrt 2\\1&amp;1
\end{bmatrix}
\begin{bmatrix}
(1+\sqrt 2)^x&amp;\\&amp;(1-\sqrt 2)^x
\end{bmatrix}
\begin{bmatrix}
1 &amp;\sqrt 2-1\\-1&amp;\sqrt 2+1
\end{bmatrix}
\begin{bmatrix}1\\0
\end{bmatrix}
\\[2mm]
&amp;=
\frac 1{2\sqrt 2}
\begin{bmatrix}
(1+\sqrt 2)^{x+1} &amp; (1-\sqrt 2)^{x+1}\\
(1+\sqrt 2)^{x  } &amp; (1-\sqrt 2)^{x  }
\end{bmatrix}
\begin{bmatrix}
1 &amp;\sqrt 2-1\\-1&amp;\sqrt 2+1
\end{bmatrix}
\begin{bmatrix}1\\0
\end{bmatrix}
\\[2mm]
&amp;=
\frac 1{2\sqrt 2}
\begin{bmatrix}
(1+\sqrt 2)^{x+1} - (1-\sqrt 2)^{x+1} &amp; ? \\
(1+\sqrt 2)^{x  } - (1-\sqrt 2)^{x  } &amp; ?
\end{bmatrix}
\begin{bmatrix}1\\0
\end{bmatrix}
\\[2mm]
&amp;=
\frac 1{2\sqrt 2}
\begin{bmatrix}
(1+\sqrt 2)^{x+1} - (1-\sqrt 2)^{x+1}  \\
(1+\sqrt 2)^{x  } - (1-\sqrt 2)^{x  } 
\end{bmatrix}
\ ,
\end{aligned}
$$
which gives $\displaystyle n_x=\frac 1{2\sqrt 2}\Big(\ (1+\sqrt 2)^{x  } - (1-\sqrt 2)^{x  }\ \Big)$. 

$\square$

In particular, for $x&gt;0$ we can compute $n_x$ by rounding to an integer the number $\displaystyle n_x=\frac 1{2\sqrt 2} (1+\sqrt 2)^x$. For instance, for $x=7$ we obtain $168.99926035179\dots$ and after rounding we get $n_7=169$.



Older answer for the explicit form of the $(A,C)$ recursion:

You may try to write explicitly the recursion formula as passing from one pair $(A,C)$ to the next pair $(A',C')$ in the form:
$$
\begin{aligned}
\begin{bmatrix}
A'\\C'\\1
\end{bmatrix}
&amp;=
\underbrace{
\begin{bmatrix}
3&amp;2&amp;1\\4&amp;3&amp;2\\0&amp;0&amp;1
\end{bmatrix}}_{=:T}
\begin{bmatrix}
A\\C\\1
\end{bmatrix}
\ ,
\\
&amp;\qquad\text{ extracted from the mentioned passage}
\\
A' &amp;=3A+2C+1\ ,\\
C' &amp;=4A+3C+2\ ,
\end{aligned}
$$
thus introducing a matrix $T$. Its charactristic polynomial is $(x - 1)  (x^2 - 6x + 1)$, so we have the following diagonalization over $\Bbb Q(a)$ with $a=\sqrt 2$. We expect a diagonal matrix with diagonal entries corresponding to the roots of the above characteristic polynomial, they are $1$ and $3\pm 2\sqrt 2$. Computer aided typing:

sage: K.&lt;a&gt; = QuadraticField(2)
sage: a^2
2
sage: T = matrix( K, 3,3, [3,2,1, 4,3,2, 0,0,1] )
sage: T
[3 2 1]
[4 3 2]
[0 0 1]
sage: D, change = T.jordan_form(transformation=True)
sage: D
[ 2*a + 3|       0|       0]
[--------+--------+--------]
[       0|       1|       0]
[--------+--------+--------]
[       0|       0|-2*a + 3]
sage: change
[ 1  1  1]
[ a  0 -a]
[ 0 -2  0]
sage: change * D * change^-1 == T
True


Later edit: Let us use, and type explicitly the obtained formula for $T$, and thus also for the powers of $T$. Below, $S$ is the base change matrix.
$$
\begin{aligned}
T
&amp;=
\underbrace{
\begin{bmatrix}
1&amp;1&amp;1\\
a&amp;0&amp;-a\\
0&amp;-2&amp;0
\end{bmatrix}}_S
\begin{bmatrix}
3+2a&amp;&amp;\\
&amp;1&amp;\\
&amp;&amp;3-2a
\end{bmatrix}
\underbrace{
\frac 14
\begin{bmatrix}
2 &amp; a &amp; 1\\
0&amp;0&amp;-2\\
2&amp;-a&amp;1
\end{bmatrix}}_{S^{-1}}
\\
T^n
&amp;=
\frac 14
\begin{bmatrix}
1&amp;1&amp;1\\
a&amp;0&amp;-a\\
0&amp;-2&amp;0
\end{bmatrix}
\begin{bmatrix}
(3+2a)^n&amp;&amp;\\
&amp;1&amp;\\
&amp;&amp;(3-2a)^n
\end{bmatrix}
\begin{bmatrix}
2 &amp; a &amp; 1\\
0&amp;0&amp;-2\\
2&amp;-a&amp;1
\end{bmatrix}
\\[3mm]
&amp;\qquad\text{ Above }a=\sqrt 2\ .
\\
&amp;\qquad\text{ This gives an }\color{blue}{\text{explicit formula}:}
\\[3mm]
\color{blue}{
\begin{bmatrix}
A_n\\ C_n\\ 1
\end{bmatrix}}
&amp;=
T^n
\begin{bmatrix}
A_0\\ C_0\\ 1
\end{bmatrix}
=
T^n
\begin{bmatrix}
0\\ 1\\ 1
\end{bmatrix}
\\
&amp;=
\color{blue}{
\frac 14
\begin{bmatrix}
1&amp;1&amp;1\\
a&amp;0&amp;-a\\
0&amp;-2&amp;0
\end{bmatrix}
\begin{bmatrix}
(3+2a)^n&amp;&amp;\\
&amp;1&amp;\\
&amp;&amp;(3-2a)^n
\end{bmatrix}
\begin{bmatrix}
2 &amp; a &amp; 1\\
0&amp;0&amp;-2\\
2&amp;-a&amp;1
\end{bmatrix}
\begin{bmatrix}
0\\ 1\\ 1
\end{bmatrix}}\ .
\end{aligned}
$$

$\square$



Please stop reading here if computer experiments are not welcome. (I posted this since the shape of the OP also has similar collection of data. It is often for me a check, and in this case it may be helpful for some readers, too.)



Example of an explicit computation:

To compute -say- the $15$.th term (or so) we already have $D^{15}$, so compute with sage the product:
$$
\frac 14
\begin{bmatrix}
1&amp;1&amp;1\\
a&amp;0&amp;-a\\
0&amp;-2&amp;0
\end{bmatrix}
\begin{bmatrix}
(3+2a)^{15}&amp;&amp;\\
&amp;1&amp;\\
&amp;&amp;(3-2a)^{15}
\end{bmatrix}
\begin{bmatrix}
2 &amp; a &amp; 1\\
0&amp;0&amp;-2\\
2&amp;-a&amp;1
\end{bmatrix}
\begin{bmatrix}
0\\1\\1
\end{bmatrix}
\ ,
$$
and we get:

sage: Z = matrix(3, 1, [0,1,1])    # initial solution, A=0, C=1
sage: change * D^15 * change^-1 * Z
[183648021599]
[259717522849]
[           1]
sage: A15, C15 = 183648021599, 259717522849
sage: B15 = A15 + 1
sage: A15^2 + B15^2 == C15^2
True


This is the kind of direct computation needed. (The diagonal matrix $D$ has diagonal entries $1$, and $3\pm 2\sqrt 2$, so the only involved involved computation is the one of the powers of these diagonal entries.)