# Crazy pattern in the simple continued fraction for ∑∞k=11(2k)!\sum_{k=1}^\infty \frac{1}{(2^k)!}

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

$$∞∑k=11(2k)!=0.5416914682540160487415778421\sum_{k=1}^\infty \frac{1}{(2^k)!}=0.5416914682540160487415778421$$

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:

$$An=(2n2n−1)−1A_n= \left( \begin{array}( 2^n \\ 2^{n-1} \end{array} \right) -1$$

$$A1=1A_1=1$$

$$A2=5A_2=5$$

$$A3=69A_3=69$$

$$A4=12869A_4=12869$$

$$A5=601080389A_5=601080389$$

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 $$∑∞k=11(ak)!\sum_{k=1}^\infty \frac{1}{(a^k)!}$$?

Edit

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

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

As a side note if we could set $$1,1=21,1=2$$ then all these strings will be the same.

Now we rewrite the sequence. I will denote $$AnA_n$$ by just their indices $$nn$$:

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

$$[a3b4c3c5d3a4b3c6d3b4c3d5a3a4b3c7d3b4c3c5d3a4b3d6a3b4c3d5a3a,...][a3b4c3c5d3a4b3c6d3b4c3d5a3a4b3c7d3b4c3c5d3a4b3d6a3b4c3d5a3a,...]$$

Now we have new large numbers $$AnA_n$$ appear at positions $$2n2^n$$. And postitions of the same numbers are in a simple arithmetic progression with a difference $$2n2^n$$ as well.

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

The $$10 00010~000$$ 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.

actually your pattern is true and it is relatively easy to prove. Most of it can be found in an article from Henry Cohn in Acta Arithmetica (1996) (“Symmetry and specializability in continued fractions”) where he finds similar patterns for other kind of continued fractions such as $\sum \frac{1}{10^{n!}}$. Curiously he doesn’t mention your particular series although his method applies directly to it.

Let $[a_0,a_1,a_2,\dots,a_m]$ a continued fraction and as usual let

we use this lemma from the mentioned article (the proof is not difficult and only uses elementary facts about continued fractions):

[Folding Lemma]

This involves negative integers but it can be easiy transformed in a similar expresion involving only posive numbers using the fact that for any continued fraction:

So

With all this consider the write the $m$th partial sum of your series as

Where we can take always $m$ even (ie if $m$ is odd and $a_m>1$ then we can consider instead the continued fraction $[0,a_1,a_2,\dots,a_m-1,1]$ and so on).

Now $q_m = 2^n!$ we see it using induction on $n$: it is obvious for $n=1$ and if $S_{n-1} = P/2^{n-1}!$ then

now any common factor of the numerator and the denominator is has to be a factor of $2^{n-1}!$ dividing also $P$, but this is impossible as both are coprime so $q_m = 2^n!$ and we are done.

Using the “positive” form of the folding lemma with $x = \binom{2^{n+1}}{2^n}$ we get:

And we get the “shape” of the continued fraction and the your $A_m$. Let’s see several steps:

We start with the first term wich is

as $m$ is odd we change and use instead the continuos fraction

and apply the last formula getting

We can dispose of the zeros using the fact that for any continued fraction:

so

this time $m$ is even so we apply again the formula getting

again $m$ is even (and it will always continue even as easy to infer) so we apply again the formula getting as the next term

and we reduce it using the zero trick:

from now it is easy to see that the obtained continued fraction always has an even number of terms and we always have to remove the zero leaving a continued fraction ending again in 1,1. So the rule to obtain the continued fraction from here to an arbitrary number of termes is repeating the follewing steps: let the shape of the last continued fraction be $[0,1,1,b_1,\dots,b_k,1,1]$ then the next continued fraction will be

from this you can easily derive the patterns you have found for the position of apperance of the different integers.