Exact value for the continued fraction of $\tiny 1+\cfrac{1}{3+\cfrac{3}{5+\cfrac{5}{7+\cfrac{7}{9...}}}}$? Does anyone know the exact value for the continued fraction of $$1+\cfrac{1}{3+\cfrac{3}{5+\cfrac{5}{7+\cfrac{7}{9+\ddots}}}}?$$
I already know that $$1+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\cfrac{1}{9\ddots}}}}=\frac{e^2+1}{e^2-1},$$ but I only figured that out by typing the decimal approximation into google of the first few terms of the continued fraction (before I knew the exact value) which took me to a math paper saying that $\frac{e^2+1}{e^2-1}$ roughly equals the decimal approximation I typed in. I then typed in the continued fraction of $\frac{e^2+1}{e^2-1}$ into wolfram alpha and it spat out $$1+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\cfrac{1}{9\ddots}}}}.$$ I have no idea how to solve these so please don't downvote, I'm just doing this in case it's useful to someone one day, and out of curiosity of course.

added
Comments point out that this post did the wrong continued fraction.
For the correct one, use $a=-1$ not $1$.  Then follow the same Satz $2$.
The result is
$$
\frac{2\;{}_2F_1(-\frac12;1;\frac12)}{{}_1F_1(\frac12;2;\frac12)}
=\frac{I_0(\frac14)+I_1(\frac14)}{I_0(\frac14)-I_1(\frac14)}  
\approx 1.2831923 .
$$
original post
Here is the reference for everything on continued fractions (as of 1913):
Perron, Oskar, Die Lehre von den Kettenbrüchen., Leipzig - Berlin: B. G. Teubner. xiii, 520 S. $8^\circ$ (1913). ZBL43.0283.04.
Section 81, Satz 2  evaluates
$$
c + \frac{a+b}{\displaystyle c+d + \frac{a+2b}{\displaystyle c+2d+\frac{a+3b}{\displaystyle c+3d+\ddots}}}
$$
So we need $a=1,b=2,c=1,d=2$.
Value of the continued fraction is
$$
\frac{2\;{}_1F_1(\frac12, 1, \frac12)}{{}_1F_1(\frac32, 2, \frac12)}
\approx 1.779306397
$$
This can be written
$$
\frac{2e^{1/4} I_0(\frac14)}
{e^{1/4} I_0(\frac14)+e^{1/4} I_1(\frac14)}
=
\frac{2}{\displaystyle 1+\frac{I_1(\frac14)}{I_0(\frac14)}}
$$
in terms of Bessel functions.
