Calculate the continued fraction

Question

Find the limit

$$ \frac{\color{red}{1+\cfrac{3}{4+\cfrac{7}{8+\cfrac{11}{12+\dots}}}}}{\color{blue}{2+\cfrac{5}{6+\cfrac{9}{10+\cfrac{13}{14+\dots}}}}}. $$ By direct calculation I have got that it is about 0.59*** but I hope there exists an exact expression.

EDIT1: $\color{red}{1},\color{blue}{2},\color{red}{3,4},\color{blue}{5,6},\color{red}{7,8},\color{blue}{9,10},\color{red}{11,12},\color{blue}{13,14},\ldots$

EDIT2: By some calculation I conjecture that

$$ \frac{\color{red}{1+\cfrac{3}{4+\cfrac{7}{8+\cfrac{11}{12+\dots}}}}}{\color{blue}{2+\cfrac{5}{6+\cfrac{9}{10+\cfrac{13}{14+\dots}}}}} \to \frac{\varphi}{e}, $$ where $\varphi$ is the golden ratio and $e$ is as usual $2.71...$

@marty cohen seems numerator is very close to golden ratio and denominator is close to $e$ — Leox, Mar 15 '18 at 20:29
Since $\phi$ is a quadratic irrational, its continued fraction is periodic, so it can't be the numerator. The denominator looks like one of Euler's continued fractions for $e$. — marty cohen, Mar 16 '18 at 03:07

score 1 · Accepted Answer · answered Jul 10 '20 at 19:01

The "exact expression" (as follows from this result) uses the confluent hypergeometric function: $$L=\frac{a_3}{3a_1}\left(1+\frac{3a_4}{5a_2}\right),\qquad a_n={}_1F_1\left(\frac{2n-1}{4};\frac{2n+1}{4};\frac14\right).$$ I don't think it can be simplified. Note that numerically $L\approx0.59518$... while $\varphi/e\approx0.59524$...

Calculate the continued fraction

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