Dragons, Folds and Fractions

Exploring the Summed Paperfold Sequence

By now, we should be well familiar with the peaks and troughs of our dragon curve/paperfold sequence:

But are there other patterns hidden within the dips and rises of this sequence?

Perhaps if we simply count the repeated runs of 1s and 0s to find a summed paperfolding sequence (OEIS A088431)?

\[1,1,0,1,1,0,0\] 3 folds (7 bends)
⇒
\[2,1,2,2\]
Can you see how one sequence gives rise to the other?

1 15

When exploring sequences these days, one of my first strategies is to see how the terms express themselves as a continued fraction - does the sequence appear to converge to a particular value and, if so - what value?
\[2+\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{2 }}}\]\[= \frac{19}{7}\]\[\approx 2.714285714285714...\]

In this case, the summed paperfolding sequence appears to converge to a value around \(2.708914768398756726684863375035394176...\).

While I was initially tempted to see the possibilities of convergence to Euler's number \(e\), there appears no evidence that this is where the pattern is headed, and so should be consigned to the basket of wishful thinking.

However, a visit to the amazing Online Encyclopedia of Integer Sequences and delving into some online references (listed below) quickly revealed some different connections.

Perhaps \(\frac{13}{10}\cdot (5^{\frac{3}{4}}-2^{\frac{1}{3}})\) or even \(\sum_{(x=1)}^{\infty}{(\frac{100}{5^n+(\frac{33 \cdot n^2}{2} - \frac{7 \cdot n}{2} - 49})}\)? But probably not.

In fact, if we wished to build the Summed PaperFolding sequence ourselves (with little or no reference to the regular paperfold sequence), several different methods present themselves, three of these described below..

In this way, it is possible to actually derive the regular paperfolding sequence in a most efficient way, to as many terms as desired.

Unfurling the Dragon's Wings

As described in Bunder, Bates and Arnold (2024: Theorem 3), thinking about the way that our dragon curve unfolds, perhaps if we look at our summed paperfold sequence as mirrored and reversed (just as our regular paperfold sequence, not surprisingly)?

2 folds (2²-1: 3 bends): 1,1,0
⇒ 2¹ = 2 terms:
\[ [2,1]\]

3 folds (2³-1: 7 bends): 1,1,0,1,1,0,0
⇒ 2² = 4 terms:
\[[2,1,2,\]\[2]\]

4 folds (2⁴-1: 15 bends): 1,1,0,1,1,0,0,1,1,1,0,0,1,0,0
⇒ 2³ = 8 terms:
\[[2,1,2,\]\[2,3,\]\[2,1,2]\]

5 folds (2⁵-1: 31 bends):
1,1,0,1,1,0,0,1,1,1,0,0,1,0,0,1,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0
⇒ 2⁴ = 16 terms:
\[[2,1,2,2,3,2,1,\]\[2,3,\]\[1,2,3,2,2,1,2]\]
Seeing a pattern?

6 folds (2⁶-1: 63 bends)
1,1,0,1,1,0,0,1,1,1,0,0,1,0,0,1,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0,1,
1,1,0,1,1,0,0,1,1,1,0,0,1,0,0,0,1,1,0,1,1,0,0,0,1,1,0,0,1,0,0
⇒ 2⁵ = 32 terms:
\[[2,1,2,2,3,2,1,2,3,1,2,3,2,2,1,\]\[2,3,\]\[1,2,2,3,2,1,3,2,1,2,3,2,2,1,2]\]

Can you write down the 64 terms of the next expansion?

Using this approach, we could now continue to build our summed paperfold sequence indefinitely - with complete accuracy!

A Recursive Formula

As noted in the OEIS (A088431): To construct the sequence use the rule: a(1)=2, then a(a(1) + a(2) + ... + a(n) + 1) = 2 and fill in any undefined places with the sequence 1,3,1,3,1,3,1,3,1,3,1,3,....

Step by step:

a(1) = 2

a(2) is undefined ⇒ so a(2) = 1

a(1) + 1 = 2 + 1 = 3 ⇒ so a(3) = 2

a(1) + a(2) + 1 =
2 + 1+ 1 = 4 ⇒ so a(4) = 2

a(5) is undefined ⇒ so a(5) = 3

a(1) + a(2) + a(3) + 1 =
2 + 1 + 2+ 1 = 6 ⇒ so a(6) = 2

a(7) is undefined ⇒ so a(7) = 1

a(1) + a(2) + a(3) + a(4) + 1 =
2 + 1 + 2 + 2+ 1 = 8 ⇒ so a(8) = 2

a(9) is undefined ⇒ so a(9) = 3

a(10) is undefined ⇒ so a(10) = 1

a(1) + a(2) + a(3) + a(4) + a(5) + 1 =
2 + 1 + 2 + 2 + 3+ 1 = 11 ⇒ so a(11) = 2

a(12) is undefined ⇒ so a(12) = 3

An Interesting Infinite Sum

As noted in the OEIS (A07400), consider a relatively simple series:

\[\sum_{x=0}^{\infty}{(\frac{1}{2})}^{(2^x)}\] \[\frac{1}{2} + \frac{1}{4} + \frac{1}{16} + \frac{1}{256} +... = \sum_{x=0}^{\infty}{(\frac{1}{2})}^{(2^x)} = \sum_{x=0}^{\infty}{2^{(-2^x)}}\] \[\approx 0.816421509021893143708079737...\] \[⇒ [0,1,4,2,4,4,6,4,2,4,6,2,4,6,4,4,2,4,6,2,4,4...]\] \[⇒ 0+\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{2 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{6 + \cfrac{1}{4 + \cfrac{1}{2 + \cfrac{1}{4 + \cfrac{1}{6...}}}}}}}}}}\]
Still not seeing it?

Have a closer look at the continued fraction terms.

Remove the leading 0, 1 - and halve each of the remaining terms.
\[[(0,1,) 4,2,4,4,6,4,2,4,6,2,4,6,4,4,2,4,6,2,4,4...]\] \[ ⇒⇒ [2,1,2,2,3,2,1,2,3,1,2,3,2,2,1,2,3,1,2,2...]\]
So taking increasing terms of the sum above, with minor adjustments, allows us to generate our summed paperfold sequence - and, subsequently, the original paperfold sequence.

References

Martin Bunder, Bruce Bates and Stephen Arnold (2024): The Summed Paperfolding Sequence
(Complete version: Bulletin of the Australian Mathematical Society: Published online by Cambridge University Press: 25 March 2024)

A014577 Online Encyclopedia of Integer Sequences: The regular paper-folding sequence (or dragon curve sequence)

A088431 Online Encyclopedia of Integer Sequences: Half of the (n+1)-st component of the continued fraction expansion of Sum_{k>=0} 1/2^(2^k)

A007400 Online Encyclopedia of Integer Sequences: Continued fraction for Sum_{n>=0} 1/2^(2^n) = 0.8164215090218931...

Or try the Reverse Symbolic Calculator!

Or perhaps try the Ramanujan Machine!

A.J. van der Poorten An Introduction to Continued Fractions (Note: Page 6)

A.J. van der Poorten and J. Shallit Folded Continued Fractions

Index

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Summed Paperfold Sequences and Series

Key properties and proofs associated with the Summed PaperFolding Sequence are developed in detail by Bunder, Bates and Arnold (2024) in their recent publication, and some fundamentals may be explored here.

Theorem 1: Expression for the PaperFolding Sequence f (Test yourself?)

Theorem 2: PaperFolding Runs of f (Test yourself?)

Theorem 3: Interleaved PaperFolding of g

Theorem 4: Length of the Summed PaperFolding Sequence g (Test yourself?)

Theorem 5: Defining an Expression for g

Theorem 6: Relationships in g

Theorem 7: Summing the Summed PaperFolding Sequence: h (Test yourself?)

Theorem 8: Limits on h

Theorem 9: Proof that OEIS A088431 = g

Index

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Dragons, (Continued) Fractions and Folds

Suppose you wanted to make a list of all the rational numbers?

One approach might be to begin with the denominators of the fractions: first, list those with denominator 1 - zero and 1: \(\frac{0}{1}, \frac{1}{1}\).

Next, add those with denominator 2: \(\frac{0}{1}, \frac{1}{2}, \frac{1}{1}\)

Add those with denominator 3 (\(\frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{1}{1}\)) and you are on your way to the Farey Sequence! This last would be referred to as the Farey Sequence of Order 3.

There are many wonderful ways to visualise such arrays, from Ford Circles to Fraction Trees. One such tree is the Stern-Brocot Tree, a binary listing of ALL the rationals - not just those between 0 and 1! It does this by mirroring the reciprocals of Farey Numbers above at each level. So for order 3,
\[(\frac{0}{1}), \frac{1}{3}, \frac{1}{2}, \frac{2}{3} + \frac{1}{1} + \frac{3}{2}, \frac{2}{1}, \frac{3}{1}, (\frac{1}{0})\]

The Stern-Brocot Tree even has its own continued fraction which can generate every convergent at each level! (With thanks to Dr Bruce Bates from the University of Wollongong) AND it can also serve to generate the PaperFold Sequence (and consequently our Dragon Curve)!

To build our paperfold sequence from the extended Stern-Brocot continued fraction, just follow two simple steps:

Delete 0/1 and 1/0 from the Stern-Brocot sequence and represent every other term, except 1, in its short-form continued fraction - 1 is best thought of as the continued fraction \([0,1]\).

Find each continued fraction with an even number of terms and replace each with 1, and those with an odd number of terms with 0 to produce the PaperFold Sequence!

The (extended) Stern Brocot
Continued Fraction
of order 3 is
\[[0,3,1,3,1,5,1,3]_{sb}\] \[<=> [0,3,-1,3,-1,5,-1,3]\] \[0 + \cfrac{1}{3 + \cfrac{1}{-1 + \cfrac{1}{3 + \cfrac{1}{-1 + \cfrac{1}{5 + \cfrac{1}{-1 + \cfrac{1}{3}}}}}}}\]
Convergents (excluding 0) ⇒
\[\frac{1}{3} = [0,3] ⇒ (even) ⇒ 1\] \[\frac{1}{2} = [0,2] ⇒ (even) ⇒ 1\] \[\frac{2}{3} = [0,1,2] ⇒ (odd) ⇒ 0\] \[\frac{1}{1} = [0,1] ⇒ (even) ⇒ 1\] \[\frac{3}{2} = [1,2] ⇒ (even) ⇒ 1\] \[\frac{2}{1} = [2] ⇒ (odd) ⇒ 0\] \[\frac{3}{1} = [3] ⇒ (odd) ⇒ 0\]

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By the Way...

A standard piece of paper is around 0.1 mm in thickness.

It is generally accepted that it is impossible to fold a sheet of paper in half more than 7 times, at which point it will have 127 bends and be over 1 centimetre in thickness.

In theory, after only 15 folds (and 32 767 bends) the paper would be over 3 metres in height - and doubling with each subsequent fold!

Like to dig a little deeper?

The PaperFold Sequence has applications to fields such as error-checking in communication systems - fundamental in cyber-security.

In their exploration of the Summed PaperFolding Sequence (2024: Theorem 7), Bunder, Bates and Arnold define an interesting new sequence. In addition to the (regular) paperfolding sequence f(n) and its related summed paperfolding sequence g(n), they define a third sequence, h(n), by adding the terms of the summed paperfolding sequence:
\[h(n) = \sum_{n=1}^{n} g(n)\]
It is well established that, for n folds, the Paperfolding Sequence has \(2^{n}-1\) terms, and the Summed PaperFolding Sequence, \(2^{n-1}\) terms.

If we ADD the terms of each sequence, then the SUM of the Paperfolding Sequence is \(2^{n-1}\), and the SUM of the Summed PaperFolding Sequence, \(2^{n}-1\)!

What seems curious is that, for the related sequence h(n), for \(n>1\) each term of the form \[h(2^{n-1}) = 2^{n}-1\]

After 4 folds, for example, the sum of the paperfold sequence f (with 15 terms) is 8

f = [1,1,0,1,1,0,0,1,1,1,0,0,1,0,0]
\[f(15) = \sum_{n=1}^{15} f(n) = 8\]
The SUM of the Summed Paperfolding Sequence g (with 8 terms) is 15.

g = [2,1,2,2,3,2,1,2]
\[g(8) = \sum_{n=1}^{8} g(n) = 15\]
Then the Summed PaperFolding SERIES h for 4 folds

h = [2,3,5,7,10,12,13,15] with
\[h(1) = 2, h(2) = 3, h(4) = 7, h(8) = 15...!\]

Just for interest... Treating the PaperFold Sequence as a binary number produces yet another sequence of numbers - although one which grows unmanageably fast!

1 Fold: 1₂ ⇒ 1

2 Folds: 110₂ ⇒ 6 ⇒ \(2 \cdot 3\)

3 Folds: 1101100₂ ⇒ 108 ⇒ \(2^2 \cdot 3^3\)

4 Folds: 110110011100100₂
⇒ 27 876 ⇒ \(2^2 \cdot 3 \cdot 23 \cdot 101\)

5 Folds:
1101100111001001110110001100100₂
⇒ 1 826 942 052 ⇒ \(2^2 \cdot 3 \cdot 13^2 \cdot 139 \cdot 6 481\)

6 Folds:
11011001110010011101100011001001
1101100111001000110110001100100₂
⇒ 7 846 656 369 001 524 000
⇒ \(2^5 \cdot 3^3 \cdot 5^3 \cdot 17 \cdot 4 273 777 978 759\)

7 folds:
1101100111001001110110001100100
11101100111001000110110001100100
11101100111001001110110001100100
01101100111001000110110001100100₂
= 144 745 261 873 314 177 475 604 083 946 266 324 068
⇒ \(2^2 \cdot 3 \cdot 7 \cdot 1 723 157 879 444 216 398 519 096 237 455 551 477\)

Index

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GXWeb Dragons, Folds and Fractions

Introduction: Here There Be Dragons

Twisting the Dragon's Tail

Finding the Fractions Hidden in the Folds

GXWeb Dragon Curve Explorer

Exploring the Summed Paperfold Sequence

Unfurling the Dragon's Wings

A Recursive Formula

An Interesting Infinite Sum

References

Summed Paperfold Sequences and Series

Dragons, (Continued) Fractions and Folds

Mathematical ToolKit

By the Way...

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