Other "natural" simple continued fractions with closed-forms?

Question

Given the simple continued fraction,

$$x =b_0 + \cfrac{1}{b_1 + \cfrac{1}{b_2 + \cfrac{1}{b_3 + \ddots}}} $$

for some well-defined sequence of positive integers $b_k$ for $k>0$. More compactly, for $b_0 = 0$,

$$x = [0\color{red}{;}b_1,b_2,b_3,\dots]$$

we compile some old results from MSE. The first three have closed-forms, and we inquire about the last three. Let $I_n(x)$ be the "modified Bessel function of the first kind".

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I. The $b_k$ as multiples of all positive integers.

$$D_1 =\cfrac{1}{1m +\cfrac{1}{2m + \cfrac{1}{3m + \ddots}}} = \frac{I_1(2/m)}{I_0(2/m)}\qquad$$

See this 2015 post for $m=1$, or A073821 for $m=2$, or this Mathworld article.

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II. The $b_k$ as multiples of all odd integers.

$$D_2 =\cfrac{1}{1m +\cfrac{1}{3m + \cfrac{1}{5m + \ddots}}} = \frac{I_{1/2}(1/m)}{I_{-1/2}(1/m)} =\frac{e^{2/m}-1}{e^{2/m}+1} = \operatorname{tanh}(1/m) $$

See a 2014 post for $m=1$, this Mathworld article for $m=2$, or a 2019 post for general $m$.

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III. The $b_k$ as multiples of $4n+1$.

$$D_{3a} =\cfrac{1}{1m + \cfrac{1}{5m + \cfrac{1}{9m + \ddots}}} = \frac{I_{1/4}\big(\tfrac{1}{2m}\big)}{I_{-3/4}\big(\tfrac{1}{2m}\big)}\quad$$

For multiples of $4n-1$, then $D_{3b}=\dfrac{I_{3/4}\big(\tfrac{1}{2m}\big)}{I_{-1/4}\big(\tfrac{1}{2m}\big)}.\,$ See A308741 and A308742.

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IV. The $b_k$ as squares. (From a 2020 post.)

$$D_4 =\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{9 + \ddots}}} =\;??$$

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V. The $b_k$ as primes. (From a Sept 2011 post.)

$$D_5 =\cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{5 + \ddots}}} =\;??$$

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VI. The $b_k$ as composites. (From a Nov 2011 post.)

$$D_6 =\cfrac{1}{4 +\cfrac{1}{6 + \cfrac{1}{8 + \cfrac{1}{9 + \ddots}}}} =\;??$$

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Question: In the intervening years, has progress been made for the last three continued fractions, or for others using various sequences like the Fibonacci numbers or factorials?

Note: The first three have $b_k$ in arithmetic progression, hence can be expressed by the modified Bessel function. However, $D_4$ uses a quadratic sequence $b_k$ while the last two use "sieved sequences", thus understandably will be difficult, though one may hope for a closed-form.

Answer 1

There's also $$ \tan(1/k) = \dfrac{1}{k-1+ \dfrac{1}{1+\dfrac{1}{3k-2 + \dfrac{1}{1+\dfrac{1}{5k-2 + \dfrac{1}{1+\ldots}}}}}}$$

Answer 2

comment
On $D_4$. Well, I was too optimistic. The usual method—Kapitel 11 Satz 8 of

Perron, Oskar, Die Lehre von den Kettenbrüchen., Leipzig - Berlin: B. G. Teubner. xiii, 520 S. (8^\circ) (1913). ZBL43.0283.04.

—works, and yields a value. But that value does not involve "well-known" functions.

$$ D_4 = \frac{\varphi(1)}{\psi(1)} \approx 0.804 , $$ where $\varphi$ and $\psi$ are solutions of the differential equation $$ \left( -{x}^{3}+{x}^{2} \right) {\frac {{\rm d}^{4} y}{{\rm d}{x}^{4}}} - \left( 4\,{x}^{2}-4\,x \right) {\frac {{\rm d}^{3 }y}{{\rm d}{x}^{3}}} - \left( -2+2\,x \right) {\frac {{\rm d}^{2}y}{{\rm d}{x}^{2}}} -y =0 $$ with $\varphi(0)=0, \varphi'(0)=1, \psi(0)=1, \psi'(0)=0$ .

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Another. Define sequences $p_n, q_n$ by $$ p_n = p_{n-1}+\frac{p_{n-2}}{n^2(n+1)^2},\qquad p_{-1}=1, p_0=1, \\ q_n = q_{n-1}+\frac{q_{n-2}}{n^2(n+1)^2},\qquad q_{-1}=0,q_0=1 . $$ Then $$ \frac{1}{D_4} = \lim_{n\to\infty}\frac{p_n}{q_n} \approx 1.243 . $$ But here the sequences converge, $p = \lim_{n\to\infty}p_n \approx 1.293, q = \lim_{n\to\infty}q_n \approx 1.040$. The recurrences yield absolutely convergent expressions, $$ p = 1+{\frac {1}{ 1 ^2 2 ^2 }}+{\frac {1}{ 2 ^2 3 ^2 }}+{\frac {1}{ 3 ^2 4 ^2 }}+{\frac {1}{ 3 ^2 4 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 4 ^2 5 ^2 }}+{\frac {1}{ 4 ^2 5 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 4 ^2 5 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 }}+{ \frac {1}{ 5 ^2 6 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 3 ^2 4 ^2 }}+{ \frac {1}{ 5 ^2 6 ^2 3 ^2 4 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 1 ^2 2 ^2 }}+{ \frac {1}{ 6 ^2 7 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 3 ^2 4 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 3 ^2 4 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 4 ^2 5 ^2 }}+{\frac {1 }{ 6 ^2 7 ^2 4 ^2 5 ^2 1 ^2 2 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 4 ^2 5 ^2 2 ^2 3 ^2 }} +\cdots , \\ q = 1+{\frac {1}{ 2 ^2 3 ^2 }}+{\frac {1}{ 3 ^2 4 ^2 }}+{\frac {1}{ 4 ^2 5 ^2 }}+{\frac {1}{ 4 ^2 5 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 5 ^2 6 ^2 3 ^2 4 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 }}+{ \frac {1}{ 6 ^2 7 ^2 2 ^2 3 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 3 ^2 4 ^2 }}+{\frac {1}{ 6 ^2 7 ^2 4 ^2 5 ^2 }}+{ \frac {1}{ 6 ^2 7 ^2 4 ^2 5 ^2 2 ^2 3 ^2 }} +\cdots . $$

Is there a possibility of a closed form for these?