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.
 A: 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.
