Continued fraction for $c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $ - is there a systematic expression?

Question

I want to use the convergents of the continued fraction for $$c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $$ - but of course a numeric software is very limited here, so I hope there exists a systematic expression for the entries of the continued fraction itself which I can use and evaluate directly.
So

Q: is there a functional description of the terms of the continued fraction for $c$

(I have a vague memory that I had once a copy of an article which dealt with systematic, but a-periodic continued fractions possibly related to constants like $c$ here, but I cannot remember more details, so a useful reference would help as well. At least I just found an old article of J. Shallit "Simple continued fractions for some irrational numbers" which discusses this but only for series with bases $3$ instead of $2$ in the denominators $c_3= \sum_{k=0}^\infty \frac 1{3^{2^k}} $ , and also I do not get the recipe to determine the entries of the continued fractions correctly)

Using Pari/GP with internal decimal precision of \p 200 I get the following:

 c = sum(k=0,24,(2.0)^-(2^k))*1.0   \\ generate an approximation
 %575 = 0.816421509022              \\ result: first digits of the constant
 contfrac(c)
 %574 = [0, 1, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2,
         4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4,
         6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 
         2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2,
         4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4,
         4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4,
         6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6,
         4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4,
         2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2,
         6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4]

Answer 1

Shallit's paper, which you've cited, gives a simple algorithm for generating these coefficients. It works for $\sum_{k=0}^{\infty}u^{-2^k}$ for integer $u\ge 3$; but he notes that it also works for $u=2$ (your case) with a slight modification.

Start with $$B_1=[1,3].$$ Then repeatedly apply the following rule: $B_{n+1}$ is generated from $B_n$ by appending the reverse of $B_n$ to $B_n$ and then adding $1$ and $-1$ to the two central terms. That is, $$ B_2=[1,4,2,1] \\ B_3=[1,4,2,2,0,2,4,1] \\ B_4=[1,4,2,2,0,2,4,2,0,4,2,0,2,2,4,1] \\ ... $$ This is correct as it stands, but generally one excludes zeroes from the entries of a continued fraction. To rectify that (this is the slight modification), generate $C_n$ from $B_n$ by contracting any subsequence $[a,0,b]$ down to $[a+b]$. That is, $$ C_2=[1,4,2,1] \\ C_3=[1,4,2,{\mathbf{4}},4,1] \\ C_4=[1,4,2,{\mathbf{4}},4,{\mathbf{6}},{\mathbf{4}},2,4,1]\\ ... $$ Now the elements of each $C_n$ (except the final $1$) are the entries in the continued fraction for $\sum_{k=0}^{\infty}2^{-2^{k}}$. (You'll note that they agree with the ones you've already found.) Just choose a large enough $n$ to get to the entry you need.

Answer 2

Ah, at least one empirical pattern which can be built recursively (if that heuristic holds...) Let $$a(n) = \sum_{k=0}^n \frac 1{2^{2^k}} $$ and let $ c(n) $ be the list of partial denominators of the continued fraction (="entries" of the cf) of $a(n)$ - written as a string because we have only few numbers and can nicely compact the notation to focus on the pattern in the entries. Also let the last entry be unnormalized, so split the last entry $e$ into two entries $e-1,1$ then we get the patterns:

c(2)=[0;1 4 2 1]
c(3)=[0;1 4 2 4 4 1]
c(4)=[0;1 4 2 4 4 6 4 2 4 1]
c(5)=[0;1 4 2 4 4 6 4 2 4 6 2 4 6 4 4 2 4 1]
c(6)=[0;1 4 2 4 4 6 4 2 4 6 2 4 6 4 4 2 4 6 2 4 4 6 4 2 6 4 2 4 6 4 4 2 4 1]

From $c(3)$ to $c(4)$ and then for $c(4)$ to $c(5)$ we find a simple recursive pattern:
"To create $c(m+1)$ from $c(m)$ replace the trailing $1$ by a $6$ and concat the string of entries (except the second-last) in reverse" such that we have

c(3)=[0;1424  41 ]
c(4)=[0;1424  46  4241 ]

and from $c(4)$ to $c(5)$ the pattern of generating is the same:

c(4)=[0;14244642   41]
c(5)=[0;14244642   46   24644241]

So it seems, the whole infinite string of entries can be defined by a recursive "string"-operation.
I've seen one time an arcticle, which considers more of such recursive patterns in the entries of continued fractions and especially that mirroring to the reverse order when concatenating - but I don't remember the author.

Answer 3

Since the ratio test can be applied, we know there exists the limit L.

Now we can say $a_n=a_{n+1}$ for $n\to\infty$.

If you are just looking for a close approximation, take wolfram alpha https://www.wolframalpha.com/input/?i=sum+%281%2F2^2^k%29

Note that after k=3 nothing dramatic happens anymore, since the fraction gets so small very quickly

I don't know exactly if it can be applied here, but maybe adic expansion might help here too (preferably, like Helms said base 2)