The continued fraction of this series exhibits a truly crazy pattern and I found no reference for it so far. We have:

$$\sum_{k=1}^\infty \frac{1}{(2^k)!}=0.5416914682540160487415778421$$

But the continued fraction is just beautiful:

`[1, 1, 5, 2, 69, 1, 1, 5, 1, 1, 12869, 2, 5, 1, 1, 69, 2, 5, 1, 1, 601080389, 2, 5, 2, 69, 1, 1, 5, 2, 12869, 1, 1, 5, 1, 1, 69, 2, 5, 1, 1, 1832624140942590533, 2, 5, 2, 69, 1, 1, 5, 1, 1, 12869, 2, 5, 1, 1, 69, 2, 5, 2, 601080389, 1, 1, 5, 2, 69, 1, 1, 5, 2, 12869, 1, 1, 5, 1, 1, 69, 2, 5, 1, 1, 23951146041928082866135587776380551749, 2, 5, 2, 69, 1, 1, 5, 1, 1, 12869, 2, 5, 1, 1, 69, 2, 5, 1, 1, 601080389, 2, 5, 2, 69, 1, 1, 5, 2, 12869, 1, 1, 5, 1, 1, 69, 2, 5, 2, 1832624140942590533, 1, 1, 5, 2, 69, 1, 1, 5, 1, 1, 12869, 2, 5, 1, 1, 69, 2, 5, 2, 601080389, 1, 1, 5, 2, 69, 1, 1, 5, 2,...]`

All of these large numbers are not just random - they have a simple closed form:

$$A_n= \left( \begin{array}( 2^n \\ 2^{n-1} \end{array} \right) -1$$

$$A_1=1$$

$$A_2=5$$

$$A_3=69$$

$$A_4=12869$$

$$A_5=601080389$$

And so on. This sequence is not in OEIS, only the larger sequence is, which contains this one as a subsequence https://oeis.org/A014495

What is the explanation for this?

Is there a regular pattern in this continued fraction (in the positions of numbers)?

Is there generalizations for other sums of the form $\sum_{k=1}^\infty \frac{1}{(a^k)!}$?

**Edit**

I think a good move will be to rename the strings of small numbers:

$$a=1, 1, 5, 2,\qquad b=1, 1, 5, 1, 1,\qquad c=2,5,1,1,\qquad d=2, 5, 2$$

As a side note if we could set $1,1=2$ then all these strings will be the same.

Now we rewrite the sequence. I will denote $A_n$ by just their indices $n$:

$$[a, 3, b, 4, c, 3, c, 5, d, 3, a, 4, b, 3, c, 6, d, 3, b, 4, c, 3, d, 5, a, 3, a, 4, b, 3, c, 7, \\ d, 3, b, 4, c, 3, c, 5, d, 3, a, 4, b, 3, d, 6, a, 3, b, 4, c, 3, d, 5, a, 3, a,...]$$

$$[a3b4c3c5d3a4b3c6d3b4c3d5a3a4b3c7d3b4c3c5d3a4b3d6a3b4c3d5a3a,...]$$

Now we have new large numbers $A_n$ appear at positions $2^n$. And postitions of the same numbers are in a simple arithmetic progression with a difference $2^n$ as well.

Now we only have to figure out the pattern (if any exists) for $a,b,c,d$.

The $10~000$ terms of the continued fraction are uploaded at github here.

I also link my related question, from the iformation there we can conclude that the series above provide a greedy algorithm Egyptian fraction expansion of the number, and the number is irrational by the theorem stated in this paper.