Continued fraction (2020)

The Continued fraction for $e^x$ , found in The Application of Continued Fractions and Their Generalizations to Problems in Approximation Theory by A. N. Khovanskii (1963), pg 114, is:

$\large \displaystyle e^x = 1 + \frac{2x}{2-x + \frac{x^2}{6 + \frac{x^2}{10 + \frac{x^2}{14 + \ddots}}}}$

So:

$\large \displaystyle e = 1 + \frac{2}{1 + \frac{1}{6 + \frac{1}{10 + \frac{1}{14 + \ddots}}}}$

Let us call:

$\large \displaystyle B \doteq \frac{1}{6 + \frac{1}{10 + \frac{1}{14 + \ddots}}}$

So:

$\large \displaystyle e = 1 +\frac{2}{1+B}$

$\color{#20A900} \boxed{ \large \displaystyle B =\frac{3-e}{e-1} }$

But, also:

$\large \displaystyle A = 2 + B$

So:

$\color{#20A900} \boxed{ \large \displaystyle A =\frac{e+1}{e-1} }$

Thus:

$\color{#3D99F6} \large \displaystyle a=b=c=d=1 \rightarrow \boxed{\large \displaystyle 4a+3b+2c+d = 10}$