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

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=(2n2n1)1

A1=1

A2=5

A3=69

A4=12869

A5=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)!?


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=2,5,2

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 An 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 An appear at positions 2n. And postitions of the same numbers are in a simple arithmetic progression with a difference 2n 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.

Answer

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 110n!. Curiously he doesn’t mention your particular series although his method applies directly to it.

Let [a0,a1,a2,,am] a continued fraction and as usual let
pmqm=[a0,a1,a2,,am]
we use this lemma from the mentioned article (the proof is not difficult and only uses elementary facts about continued fractions):

[Folding Lemma]
pmqm+(1)mxq2m=[a0,a1,a2,,am,x,am,am1,,a2,a1]

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:
[,a,β]=[,a1,1,β1]
So
pmqm+(1)mxq2m=[a0,a1,a2,,am,x1,1,am1,am1,,a2,a1]

With all this consider the write the mth partial sum of your series as
Sn=nk=112k!=[0,a1,a2,,am]=pmqm

Where we can take always m even (ie if m is odd and am>1 then we can consider instead the continued fraction [0,a1,a2,,am1,1] and so on).

Now qm=2n! we see it using induction on n: it is obvious for n=1 and if Sn1=P/2n1! then
Sn=P2n1!+12n!=P(2n!/2n1!)+12n!
now any common factor of the numerator and the denominator is has to be a factor of 2n1! dividing also P, but this is impossible as both are coprime so qm=2n! and we are done.

Using the “positive” form of the folding lemma with x=(2n+12n) we get:

\frac{p_m}{q_m} + \frac{(-1)^m}{\binom{2^{n+1}}{2^n}(2^n!)^2} =
\frac{p_m}{q_m} + \frac{1}{2^n!} =
[0,a_1,a_2,\dots,a_m,\binom{2^{n+1}}{2^n}-1,1,a_m-1,a_{m-1},\dots,a_1]

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
\frac{1}{2} = [0,2]
as m is odd we change and use instead the continuos fraction
\frac{1}{2} = [0,1,1]
and apply the last formula getting
\frac{1}{2}+\frac{1}{2^2!} = [0,1,1,5,1,0,1]
We can dispose of the zeros using the fact that for any continued fraction:
[\dots, a, 0, b, \dots] = [\dots, a+b, \dots ]
so
\frac{1}{2}+\frac{1}{2^2!} = [0,1,1,5,2]
this time m is even so we apply again the formula getting
\frac{1}{2}+\frac{1}{2^2!}+\frac{1}{2^3!} = [0,1,1,5,2,69,1,1,5,1,1]
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
\frac{1}{2}+\frac{1}{2^2!}+\frac{1}{2^3!}+\frac{1}{2^4!} = [0,1,1,5,2,69,1,1,5,1,1,12869,1,0,1,5,1,1,69,2,5,1,1]
and we reduce it using the zero trick:
\frac{1}{2}+\frac{1}{2^2!}+\frac{1}{2^3!}+\frac{1}{2^4!} = [0,1,1,5,2,69,1,1,5,1,1,12869,2,5,1,1,69,2,5,1,1]

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
[0,1,1,b_1,\dots,b_k,1,1,A_n,2,b_k,\dots,b_1,1,1 ]
from this you can easily derive the patterns you have found for the position of apperance of the different integers.

Attribution
Source : Link , Question Author : Yuriy S , Answer Author : Esteban Crespi

Leave a Comment