Mirroring and interleaving in the paperfolding sequence

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1 University of Wollongong Research Online Faculty of Informatics - Papers (Archive) Faculty of Engineering and Information Sciences Mirroring and interleaving in the paperfolding sequence Bruce Bates University of Wollongong Martin Bunder University of Wollongong, mbunder@uow.edu.au Keith Tognetti University of Wollongong Publication Details Bates, B. Philip., Bunder, M. W. & Tognetti, K. P. (). Mirroring and interleaving in the paperfolding sequence. Applied Analysis and Discrete Mathematics, 4 (), Research Online is the open access institutional repository for the University of Wollongong. For further information contact the UOW Library: research-pubs@uow.edu.au

2 Mirroring and interleaving in the paperfolding sequence Abstract Three equivalent methods of generating the paperfolding sequence are presented as well as a categorisation of runs of identical terms. We find all repeated subsequences, the largest repeated subsequences and the spacing of singles, doubles and triples throughout the sequence. The paperfolding sequence is shown to have links to the Binary Reflected Gray Code and the Stern-Brocot tree. Disciplines Physical Sciences and Mathematics Publication Details Bates, B. Philip., Bunder, M. W. & Tognetti, K. P. (). Mirroring and interleaving in the paperfolding sequence. Applied Analysis and Discrete Mathematics, 4 (), This journal article is available at Research Online:

3 Applicable Analysis and Discrete Mathematics available online at Appl. Anal. Discrete Math. 4 (), doi:.98/aadm5b MIRRORING AND INTERLEAVING IN THE PAPERFOLDING SEQUENCE Bruce Bates, Martin Bunder, Keith Tognetti Three equivalent methods of generating the paperfolding sequence are presented as well as a categorisation of runs of identical terms. We find all repeated subsequences, the largest repeated subsequences and the spacing of singles, doubles and triples throughout the sequence. The paperfolding sequence is shown to have links to the Binary Reflected Gray Code and the Stern-Brocot tree.. INTRODUCTION Take a sheet of paper and fold it, right over left, n times. When the paper is unfolded we see a sequence of n creases, some downward and some upward. An analysis of this sequence first appeared in Davis and Knuth [], who labeled these creases D and U. We will label them and as do Dekking et al [8]. Prodinger and Urbanek [4] label them and while Allouche and Bousquet-Mélou [4] allow both and and and. This sequence of n s and s we call S n as do Davis and Knuth [] with their sequence of n Ds and Us. The middle element of S n is always a (a D for Davis and Knuth); S n appears to the left of this and S R n (obtained from S n by reversing the order and swapping s and s a notation adopted by Lothaire [, p. 56], Mendès France and Shallit [] and Prodinger and Urbanek [4]) appears to the right. So as in Davis and Knuth []. Theorem. S n = S n S R n. We use S for lim i S i and call this the paperfolding sequence. Mathematics Subject Classification. B37, B83. Keywords and Phrases. Paperfolding, Binary Reflected Gray Code, Stern-Brocot tree. 96

4 Mirroring and interleaving in the paperfolding sequence 97 Mendès France and Shallit [] give four different methods for representing the sequence. One of their representations, called the Dragon Curve in Davis and Knuth [], is a sequence of lattice points obtained by unfolding the paper so that all the folds are 9 and then looking at the edge of the paper. Their third representation has R for and L for. An important concept, similar to perfect shuffling, is useful in understanding paperfolding. It is called interleaving. Definition. (Interleave Operator). The interleave operator # acting on the two sequences U = u, u,..., u k and V = v, v,..., v n where k > n, generates the following interleaved sequence: U#V = u,..., u p, v, u p+,..., u p, v, u p+,..., u np, v n, u np+,..., u k k where p =. n + Example. Let U = u, u,..., u n+ and V = v, v,..., v n. Then U#V = u, v, u, v,..., u n, v n, u n+. Note that Definition requires the two sequences to have different parity. Accordingly, # does not define a perfect shuffle although there are obvious similarities. Davis and Knuth [] and Prodinger and Urbanek [4] have yet another method for constructing S n and S. This can be expressed through interleaving, as follows. Theorem. Let A k = () k that is, with k s, then S n = A n #A n # #A #, = A n #S n. Allouche and Bacher [] use Toeplitz transformations to construct S n in essentially the same way. Dekking et al. [8] have a similar result. The following theorem appears in various forms in the literature, notably Davis and Knuth [] and Dekking et al. [8]. Our variant arises from using s and s in S. In what follows f i is the ith element in S. Theorem 3. For i, i) f i = f i and ii) f i = ( )i. Proof. By Theorem, i) the sequence of successive even entries in S n is S n, establishing i), ii) the sequence of successive odd entries in S n is A n, establishing ii). The following result, though similar to other formulations (for example, Allouche and Bousquet-Mélou [3]), is compact and yields an interesting corollary.

5 98 Bruce Bates, Martin Bunder, Keith Tognetti Theorem 4. For h, k, f i = {, if i = k (4h + ), if i = k (4h + 3). Proof. We prove this by induction on n. We have S =, so f = = k (4h + ) where k = h =. Suppose the theorem is true for S n. By Theorem, S n = f f 4... f n. That is, the i th entry in S n becomes the i th entry in S n. Also A n =, = f f 3 f 5... f n. That is, for h n, f 4h+ = and f 4h+3 =. Corollary. For k, r, f i = + r mod where i = k (r + ). Proof. From Theorem 4, for k, h, i) f i =, if i = k (4h + ) = k (r + ) where r is even. That is, f i = + r mod, for r even. ii) f i =, if i = k (4h + 3) = k ( (h + ) + ) = k (r + ) where r is odd. That is, f i = + r mod, for r odd. Davis and Knuth [] prove the following result to which we provide an interesting corollary. Theorem 5. S n+ = S n S R n and S R n+ = S n S R n where S =. Corollary. Proof. From Theorem 5, S i and Si R terms. Since we have S i+ = S i S R i, { fk, for k < f i +k = i and k i, for k = i. are identical except in their respective ( i ) th = f f... f k... f i (f i = )f i +... f i +k... f i+ f i+ f i +k = f k, for k < i and k i, f i +k = f 3( i ) = for k = i.

6 Mirroring and interleaving in the paperfolding sequence 99 We note that Theorem 5 offers the simplest way to generate S i+. It requires only S i. In this paper we use a generalised version of A k to produce yet another way of generating S n. We find the number and position of single s and s (that is, instances of and ) as well as those of double s and s, and triple s and s, in S. We link the Paperfolding and Stern-Brocot sequences, examine functions related to S and show that one of these has properties similar to those of the Gray code function of Bunder et al [7].. THE GENERAL ALTERNATING PAPERFOLDING SEQUENCE We define a generalisation of A k which utilises S i n Si n R instead of. Definition. For n < i, A i,n = ( ) n S i n Si n R. Note that A i,n #S n = S i n f Si n R f S i n f 3 Si n R S i nf n Si n R. Theorem 6. S i = A i,n # S n and Si R = A i,n # Sn R. Proof. We prove the first result by induction on n. We have S i = A i, #S. Let S i = A i,t #S t, where t < n. Therefore by Theorem, S i = S i t f Si t R f S i t f 3 Si t R S i tf t Si t R ( ) (, ) ( ) = S i (t+) Si (t+) R f S i (t+) Si (t+) R f S i (t+) Si (t+) R f 3 ( ) ( ) S i (t+) Si (t+) R f t S i (t+) Si (t+) R, = A i,t+ #A t#s t, = A i,t+ #S t+. Therefore our result is true for all n < i. A similar argument holds for S R i. Allouche and Bacher [] generate S using a particular Toeplitz transform. They commence with the sequence B = ωωωω..., = (ωω).

7 Bruce Bates, Martin Bunder, Keith Tognetti To obtain B k+ from B k, for i =,,..., the i th ω in B k is replaced by the i th term in B. Thus B = ωωω..., B = ωω...,. Continuing in this way, we have S = lim k B k. This transformation can be restated through interleaving. Let W n = (ω) n. Then B = S ω S R ω S ω S R ( ) ω..., = S ω S R ω, = lim n A n+,n#w n. B is obtained from B by replacing the i th ω in B by the i th term in B. That is, ( ) B = S S R ω S S R ω, ( ) = S ω S R ω, = lim n A n+,n#w n. Continuing in this way, B k = lim A n+k+,n#w n. It follows that, S = lim B k, as n k before. Summary. There are three equivalent representations of the paperfolding sequence, S i and Si R. i) Mirroring: S i = S i Si R and SR i = S i Si R, ii) Interleaving: S i = A i,n #S n and S R i = A i,n #S R n where < n < i. In particular, S i = A i #S i and Si R = A i #Si R ; and iii) Alternation: S i = A i #A i # #A # and Si R = A i #A i # #A #. 3. SOME FUNCTIONS RELATED TO THE PAPERFOLDING SEQUENCE Davis and Knuth [] define the following two functions, which are in our notation: Definition 3. d (n) = { if fn = if f n =, n g (n) = d (i). i=

8 Mirroring and interleaving in the paperfolding sequence It follows that g (n) is the excess of s over s in f... f n. Clearly: Theorem 7. i) d (n) = f n and n ii) g (n) = n + f i. i= Davis and Knuth [] prove the following interesting results, which have obvious f n counterparts. Theorem 8. i) d ( n+ m ) = d (m) if < m < n, ii) g ( n+) =, iii) g ( n+ m + ) = + g (m) if m n and k+ iv) min n (g (n) = k) =. 3 Prodinger and Urbanek [4] have a function n (k), which represents the number of s in the first k places of their S, which has s and s swapped from our S. We will use: Definition 4. N (k) is the number of s in f f... f k. Thus N (k) = k n (k) and Theorem 9. N (k) = k f i. i= Prodinger and Urbanek [4] also introduce v (k), the number of changes of consecutive digits in the binary expansion of k, where the leftmost counts as a change. So, for example, v (7) = as 7 is in binary form and v () = 4 as is in binary form. They adopt the following definition: Definition 5. v () = and v (j + i) = v (j) + δ where i, δ {, } and δ = i + j (mod ). They give the following interesting connection with the function N (k): Theorem. n (k) = ( ) k v (k), that is, N (k) = ( ) k + v (k) and v (k) = N (k) k. Theorem. v (k) = k (f i ) = g (k + ). i= Prodinger and Urbanek [4] also have the interesting result (converted to N (k)):

9 Bruce Bates, Martin Bunder, Keith Tognetti Theorem. N (k) = k i k + i i+. The obvious counterpart via Theorem for v (k) to Theorem 8 iii) is: Theorem 3. If n k < n+, then v (k) = v ( n+ k ). This relation is very similar to the recurrence relation for the Gray code function b (k) found in Bunder et al [7]. Definition 6. If n k < n+, then b (k) = b ( n+ k ) + n. Not surprisingly, some of the results in [7] for b (k) can also be proved for v (k). Theorem 4. i) v (4k + ) = v (k) +, ii) v ( r (4k + )) = v (k) + if r > and iii) v ( r (4k + 3)) = v (k + ) if r. Proof. i) By Definition 5. ii) v ( r (4k + ) ) = v ( r (4k + ) ) if r >,. = v ( (4k + ) ) if r >, = v (4k + ) +, = v (k) + by i). iii) v ( r (4k + 3) ) = v ( r (4k + 3) ) if r >,. = v ( (4k + 3) ), Theorem 5. i) v (k + ) v (k) = ( ) k+ ii) v ( r (4k + 3)) v ( s (4k + )) = i) By Theorems 3 and, ii) By Theorem 4, = v (4k + 3) if r >, = v (k + ). { v (k + ) v (k) = f k+, = ( ) k+. ( ) k+, if s > ( ) k+, if s = v ( r (4k + 3)) v ( s (4k + )) = v (k + ) v (k) if s >, = v (k + ) v (k) if s =

10 Mirroring and interleaving in the paperfolding sequence 3 and the result follows by i). 4. TRIPLES, DOUBLES AND SINGLES We now consider consecutive runs of identical terms. Definition 7. (n-tuple). An n-tuple is an instance of n, and only n, consecutive identical values in a binary sequence. Example. S 4 = contains: one instance of the triple, ; two instances of the double,, and three instances of the double, ; one instance each of the single and the single. Theorem 6. For n 4, S n contains: i) n 4 instances of the triple,, and n 4 instances of the triple, ; ii) n 3 instances of the double,, and n 3 + instances of the double, ; iii) n 4 instances each of the single,, and the single,. Proof. We prove i) by induction. Similar proofs exist for ii) and iii). The theorem holds for n = 4. Suppose for some k, where k 4, S k contains k 4 instances of the triple,, and k 4 instances of the triple,. Consider S k+ = S k Sk R. Since S k contains k 4 instances of the triple,, Sk R contains k 4 instances of the triple, (they are mirrors of each other). Similarly, since S k contains k 4 instances of the triple,, Sk R contains k 4 instances of the triple,. Now because S k is a mirror sequence that begins with, it concludes with. Therefore Sk R begins with. It follows that the middle term of S k+ = S k Sk R generates a triple of in addition to those already found in S k and Sk R. Thus S k+ contains k 4 + k 4 + = k 3 instances of the triple,. S k+ also contains k 4 + k 4 = k 3 instances of the triple,. It follows that our result holds for all k 4. Corollary 3. The middle element of S n is located at the beginning of the ( n 5 + ) th instance of the triple,. Proof. By Theorem 6, S n contains n 5 instances of the triple,. Since S n = S n Sn R and SR n begins with an instance of the triple,, the result follows. Theorem 7. S n contains only singles, doubles or triples. Proof. Since S i = A i #S i, every odd term in S i is different to adjacent odd terms. Therefore the maximum run of like terms is three before a change occurs. That is, only singles, doubles and triples can exist within S i. We discover that singles are regularly spaced in the paperfolding sequence un-

11 4 Bruce Bates, Martin Bunder, Keith Tognetti like doubles and triples. The following theorem shows that, beginning at f 3, single instances of repeat every 6 spaces in S, and beginning at f 3, single instances of also repeat every 6 spaces in S. Theorem 8. In S: i) The set of all single instances of is given by {f 6h+3 : h =,,,...}. ii) The set of all single instances of is given by {f 6h+3 : h =,,,...}. Proof. From Theorem 6, for i = n + 3 and letting n, we have (4.) S = S 3 S R 3 S 3 S R 3... where the bracketed entries in (4.) are consecutive terms in S. Each triple is formed at these bracketed entries and nowhere else. This is because S 3 and S3 R contain no triples and both begin with and end with. Singles can only then be found within the S 3 and S3 R components of (4.) without the first two and last two terms of each component. Consider the following typical subsequence in (4.) beginning with the component S 3 and deleting the middle term (4.) S 3 ( or ) S R 3 = ( or ). Note that in (4.) the third term is the single and the thirteenth term is the single. There are no other singles in (4.). Since the bracketed term does not affect the incidence of singles (which is the reason for us not including it in (4.)), it follows that single s occur every third term in repeats of S 3 ( or ) S3 R, that is, at a spacing of 6 terms. Similarly, single s occur every thirteenth term in repeats of S 3 ( or ) S3 R. The result follows. The following theorem shows that triples are not regularly spaced, but nonetheless can be identified exactly within the paperfolding sequence. Theorem 9. In S : i) The set of all first terms of all triples is given by { fk (4h+) : k = 3, 4, 5,... ; h =,,,... }. ii) The set of all last terms of all triples is given by { fk (4h+3) : k = 3, 4, 5,... ; h =,,,... }. Proof. From Theorem 6, for i = n + 3 and letting n, we have (4.3) S = S 3 S3 R S 3 S3 R... where the bracketed entries in (4.3) are consecutive terms in S. The bracketed entries in (4.3) are at f 8, f 6, f 4,... and by Theorem 4, bracketed entries with

12 Mirroring and interleaving in the paperfolding sequence 5 value are of the form f k (4h+). Thus each triple starts at an f n with n of the form n = k (4h + ) and n 8, establishing i). Similarly, by Theorem 4, bracketed entries in (4.3) with value are of the form f k (4h+3). Accordingly, each triple ends at an f n with n of the form n = k (4h + 3) and n 8, establishing ii). We now give attention to doubles which have a semi-regular pattern of spacings. Theorem. In S: i) The set of all first terms of double instances of is the union of the sets {f }, { f k (4h+3)+ : k = 3, 4, 5,... ; h =,,,... } and {f 6h+4 : h =,,,...}. ii) The set of all first terms of double instances of is the union of the sets { f k (4h+) : k = 3, 4, 5,... ; h =,,,... } and {f 6h+ : h =,,,...}. Proof. i) Since S =..., f begins a double. In (4.3) where a appears, it is always succeeded by a double. From Theorem 9, {f k (4h+3)+ : k = 3, 4, 5,... ; h =,,,...} must therefore be the set of first terms of doubles following triples. From (4.), a double always begins at the fourth term of each S 3 component of (4.3). These are all the possibilities for doubles and so the result follows. ii) In (4.3) where a appears, it is always preceded by a double. From Theorem 9, { f k (4h+3) : k = 3, 4, 5,... ; h =,,,... } must therefore be the set of first terms of doubles preceding triples. From (4.), a double always begins at the third term of each S3 R component of (4.3). These are all the possibilities for doubles and so the result follows. 5. REPEATED SUBSEQUENCS OF THE PAPERFOLDING SEQUENCE We now consider repeated subsequences of S n. Our aim is to discover all the repeated subsequences of S n and their size. By so doing, we discover the largest repeated subsequences of S n. Definition 8. (Repeated Subsequences). Let S n = f f... f n. Let also U k,s = f k f k+... f k+s and V j,s = f j f j+... f j+s be subsequences of S n. If for j k, i) U k,s = V j,s and ii) one of j =, k =, or f j f k is true, and

13 6 Bruce Bates, Martin Bunder, Keith Tognetti iii) one of j + s = n, k + s = n, or f j+s+ f k+s+ is true, then U k,s and V j,s are called repeated subsequences (rsss) of S n. Consider S n for n > 4. We have (5.) S n = S n Sn R, (5.) = S n Sn R S n Sn R, (5.3) = S n 3 Sn 3 R S n 3 Sn 3 R S n 3 Sn 3 R S n 3 Sn 3 R,.. (5.4) = S i+ Si+ R SR i+ S i+ S i+ Si+ R, (5.5) = S i Si R Si R S i S i Si R,.. (5.6) = S 5 S5 R SR 5 S 5 S 5 S5 R, (5.7) = S 4 S4 R SR 4 S 4 S 4 S4 R, = S 3 S3 R S 3 S3 R, S R 3 S 3 S 3 S R 3 S 3 S R 3, (5.8) = S S R S S R, (5.9) S R S S S R S S R. Lemma. If n > i >, there is no occurrence of S i or S R i in S n or S R n other than those shown explicitly in (5.5) or its counterpart for S R n. Proof. (By induction) We prove both the S n and Sn R cases by induction on i. Consider i = 3. If there were an occurrence of S 3 (or S3 R ) other than one explicitly shown in (5.5) or its counterpart for Sn R, S 3 (or S3 R) would have to appear as a part of S 3 S3 R, SR 3 S 3, S 3 S3 R or SR 3 S 3. This can be shown, by inspection, to not be the case. We now assume the result for i. Consider i +. If S i+ (or Si+ R ) appears in (5.4) or its counterpart for SR n, in other than the explicitly shown positions, it must appear in a part of S i+ S R i+, SR i+ S i+, S i+ S R i+ or SR i+ S i+ that includes the or. But then S i (or S R i ) appears in S i Si R, S i Si R, SR i S i or Si R S i other than as an explicit S i (or Si R ). This is impossible by the induction hypothesis. Proof. (By contradiction) Suppose our result is false, that is, there exists an S i

14 Mirroring and interleaving in the paperfolding sequence 7 or S R i not explicitly identified in (5.5) or its counterpart for S R n. Since (5.) S i or S R i = S 3 S R 3 ( or ) S 3 S R 3 and (5.8) contains every instance of S 3 or S3 R in S n and Sn R, the right hand side of (5.) must be explicitly identified in (5.8) or its Sn R counterpart. But each S 3 and S3 R entry in (5.8) is used to form only the S i and Si R entries of (5.5), a contradiction. Lemma. Every repeated subsequence (rss) of S n (n > 5) of length i) greater than 7, must start with S 3 or S R 3. ii) 7, is a) S 3 or S R 3, or b) SR ( or ) S. iii) less than 7, does not contain triples and can be discovered within the following two subsequences of S n : S 3 and S R 3. Proof. There are two cases to consider:.) Repeated subsequences containing triples. Triples only form at or in (5.8) since each or is preceded by and is followed by. Consider the first triple in the rss. From (5.8) any or is preceded by a from S 3 or S3 R. Accordingly, the smallest possible rsss containing a triple is S R ( or ) S, establishing ii)b). If the rss is any longer and contains a preceding from (5.8) it must contain a whole S 3 and, as it doesn t contain an earlier triple, must start with S 3. Alternatively, if it contains a preceding from (5.8), it must start with S3 R. Thus i) is established..) Repeated subsequences that do not contain triples. The largest sequences that do not contain triples in S n, both of length, are ( ) (5.) S 3 or S3 R ( ) where the subsequences S 3 or S3 R are entries in (5.8). Any rss of length 7 or greater, without containing a triple, must contain an S 3 or S3 R from (5.), or the initial or final (6/7)th or (5/7) th of S 3 or S3 R from (5.). Now S 3 and S3 R are rsss, but any other rss in (5.) of length 7 or greater will include an explicit or from (5.8). Any such will be preceded by and any such will be followed by, which have to be part of the rss. This is impossible, as then a triple is included. So there are no rsss of length 7 to, without triples, other than S 3 or S3 R, establishing ii)a). From.), the smallest possible rss containing a triple has length 7. Accordingly, all rsss of length less than 7 do not contain triples and are discoverable, by observation, from S 3 and S3 R, establishing iii). Lemma 3. If i > and Si R i) n = i +, the rss is Si R. starts a repeated subsequence (rss) of S n then for

15 8 Bruce Bates, Martin Bunder, Keith Tognetti ii) n = i + 3, the rsss are S R i and S R i S i. iii) n > i + 3, the rsss are S R i, SR i S i and S R i S i. Proof. i) By Lemma and (5.), S R i is the only rss of S i+ starting with S R i. ii) By Lemma and (5.3), S R i, SR i S i are the only such rsss of S i+3. iii) By Lemma and (5.5), if n > i + 3, S R i, SR i S i and S R i S i are such rsss of S n. If there are other such rsss starting with a particular instance of S R i, by Lemma and (5.5), these start with a) S R i S i S R i = S R i S R i+ b) S R i S i S R i = S R i S i+ c) Si R S i Si R = Si R Si+ R d) Si R S i Si R = Si R S i+. In (5.4) it is clear that each Si R is part of an Si+ R or S i+, and that this is the case for all occurrences of the rss in S n. Thus the rss has Si R as part of S i+ or Si+ R and so does not start with Si R. This is a contradiction, so the given rsss starting with Si R are the only ones. Theorem. The repeated subsequences (rsss) of S n, by length, are: Length Repeated Subsequences of: S S 3 S 4 S 5 S n, n > 5,,,,,,,,,,,, 3 S, S R, S, S R,, 4, S, S R,,,,, 5,, n 4 S 3, S3 R, S 3, SR S R S R S S, S R 3, S S 4, S4 R, S3 R S3, S R 3 S3. S n 4, S R n 4, S R n 5 Sn 5, S R n 5 Sn 5

16 Mirroring and interleaving in the paperfolding sequence 9 n 3 S n 3, S R n 3, S R n 4 Sn 4, S R n 4 Sn 4 n S n, Sn R, Sn 3 R Sn 3 where the S i s are those shown explicitly in (5.) to (5.9). Proof. There are two cases: i) Lengths to 7: For n 5, by observation. For n > 5, use Lemma ii) and iii). ii) Lengths greater than 7: By Lemma part i), any rss of S n must start with an S i or Si R for i 3. Lemma 3 provides all the rsss starting with Si R. If an rss starts with S i, it can be just S i for 3 i n, or it can, by (5.5), continue as a) S i Si R = Si+ R, or b) S i S R i = S i+. In case a), by Lemma 3, we have one of the above rsss. In case b), the rss is S i+ or, applying the above again, starts with Si+ R or S i+, so the same rsss are obtained. Corollary 4. The longest rsss of S n for n > 4, are S n, S R n and SR n 3 S n 3. Proof. The case for n = 5 is proven by observation. The case for n > 5 follows from Theorem. Corollary 5. Let L be the length of the largest repeated subsequence of S n. Then n for n < 4 L = 5 for n = 4 n for n > 4. Proof. The case for n 4 is proven by observation. The case for n > 4 follows from Corollary 4. [8]. 6. POWER SERIES AND THE PAPERFOLDING SEQUENCE We conclude with the generalisation of a result mentioned in Dekking et al Theorem. Let S = f f f 3... and F (x) = m =,,..., F (x) F f n x n where x <. Then for n= ( x m) m x i =. x i+ i=

17 Bruce Bates, Martin Bunder, Keith Tognetti Proof. By Theorem 4, for x <, F (x) F ( x m) = = x i (4h+) x i+m (4h+), i= h= m i= h= m = i= x i (4h+), i= h= x i x i+. The result in Dekking et al [8] can be represented as a corollary to Theorem. Corollary 6. Let S = f f f 3... and F (x) = F (x) F ( x ) = x x 4. f n x n where x <. Then Proof. This is the case m = in Theorem. 7. LINKING PAPERFOLDING AND THE STERN-BROCOT TREE An excellent presentation of the Stern-Brocot tree can be found in Graham et al [9]. Links between the Paperfolding and Stern-Brocot sequences using continued fractions have been identified previously (See for example, Mendès France et al. [3]). We now continue this linkage. Firstly, we give some background definitions. Definition 9. (Mediants) We define the mediant of m n and r s as m n r s = m + r n + s. The Stern-Brocot tree is made up of successive levels of mediants with the th level consisting of and. The mediants in any one level are generated by forming mediants from terms in previous levels. Thus the mediant of (found in level ) and + (found in level 3) is = 3 which is found in level 4. The Stern-Brocot 5 sequence is related to these mediants according to the following definition: Definition. (Stern-Brocot Sequence) Let H =, and for k, n= H k = H k # med H k where med H k denotes the increasing sequence of mediants that are generated from consecutive terms in H k. That is, if then H k = h k,, h k,,..., h k,k +, med H k = (h k, h k, ), (h k, h k,3 ),, ( h k, k h k, k +).

18 Mirroring and interleaving in the paperfolding sequence H k represents the increasing sequence containing both the first k generations of mediants based on H, and the terms of H itself. It is called the Stern-Brocot sequence. We now explore the presence of a paperfolding pattern in the Stern-Brocot tree. Definition. (Left and Right Mediants) A left mediant is the mediant formed in level k + that is smaller than its parent found in level k. A right mediant is the mediant formed in level k + that is greater than its parent found in level k. From Definition, if m is a term in the i th level of the Stern-Brocot tree, the left mediant of m is the mediant that has m as its parent and appears in the (i + ) th level and to the left of m. Similarly the right mediant of m is the mediant that has m as its parent and appears in the (i + ) th level and to the right of m. Example 3. The left mediant of 3 8 is 4. The right mediant of 3 8 is 5. As expected 3 these mediants appear in the 6 th level because 3 8 is in the 5th level. Since every term possesses a left and right mediant (except and ) and from level onwards there are exactly twice the number of terms in any level to that level immediately above it, all terms in any level are simply a succession of left and right mediants of terms in the preceding level. Moreover since in any one level (except the th and st ), the left mediant of a term is immediately followed by the right mediant of the same term, each level after the first consists of an alternating sequence of left and right mediants. Bates et al. [5] have shown that when expressed in the shortest form of their simple continued fractions: i) Left mediants possess continued fractions of the form [a, a,..., a k ] where k is odd. ii) Right mediants possess continued fractions of the form [a, a,..., a k ] where k is even. Accordingly, if we designate any entry in the Stern-Brocot tree with short form continued fraction [a, a,..., a k ] as for k odd and for k even, then every level of the Stern-Brocot tree represents an alternating sequence. Under this mapping, if we displace vertically downwards levels to j onto level j, (that is, if we generate the Stern-Brocot sequence H j and delete the terms and and then redesignate all terms as either or according to this mapping) we have, by Theorem, the paperfolding sequence of size j. That is, for j unbounded, we have established the following result: Theorem 3. Delete and from the Stern-Brocot sequence and represent every other term, except, in its short form continued fraction [a, a,..., a k ]. If k is odd, replace the continued fraction with ; if k is even, replace the continued fraction with. The resulting sequence is the paperfolding sequence.

19 Bruce Bates, Martin Bunder, Keith Tognetti Theorem 3 tells us that the parity of the length of the short form continued fraction for a (n), where a (n) is Sloane s sequence A735 and b (n) is Sloane s b (n) sequence A47679, is the paperfolding sequence. (See Sloane [5, A735, A47679]). Equivalently, Theorem 3 tells us that if we designate all left mediants as and all right mediants as in the Stern-Brocot tree, and set the term in level as, then for n >, level n converts to the alternating sequence. By then moving every term vertically downward from level to n onto level n we obtain S n. 8. LINKING PAPERFOLDING AND THE BINARY REFLECTED GRAY CODE Our attention now turns to the Binary Reflected Gray Code (BRGC) as yet another example of the way in which the paperfolding sequence, and interleave operators in general, are embedded in many constructions. Gray Codes owe their name to Frank Gray, a research physicist at the Bell Telephone Laboratory. Though they were first used in telegraphy by Emile Baudot (845 93), Gray used these codes to minimise errors that arose in signals transmitted by pulse code modulation. The codes are still used for checking errors in communications systems. There are several types of code with that property. We confine ourselves to the Binary Reflected Gray Code (BRGC). Definition. (Binary Reflected Gray Code, BRGC) Let n = k + j, where n, j and k are non-negative integers such that k n < k+, k and so j < k. Then the Binary Reflected Gray Code of n is defined by where G () =. G (n) = n j + G (n (j + )), = k + G ( k (j + ) ) We designate G(n) to base as G(n) and define the First Forward Difference Function of G, G, as follows: Definition 3. (First Forward Difference of G) The First Forward Difference Function of G, G, is defined as G (n) = G (n + ) G (n). Also we define the Binary First Forward Difference of G sequence, P i, as follows: Definition 4. (Binary First Forward Difference of G Sequence) The Binary First Forward Difference of G Sequence, P i, is defined as P i = p (), p (),..., p ( i ), where for n =,,,... { if G (n) > p (n) = if G (n) <.

20 Mirroring and interleaving in the paperfolding sequence 3 Table shows G(n) for n = to 3 with corresponding values for G(n), G(n), G(n) and p(n). Notice that each successive value of G (n) differs from its previous value in one bit change. For example, for n = 9,,, G (n) =,,. It is this feature that makes Gray Codes attractive as a tool for checking errors in data transmission. Notice also in Table that we have grouped successive integers into blocks. That is, the k th block, designated as B k, consists of all integers k n < k+. n G(n) G (n) G(n) p (n) Table : The Binary Reflected Gray Code Lemma 4. In the column G (n) in Table, for k and < t < k+, i) G ( k ) = k, and ii) G ( k+ + t ) = G ( k+ t ).

21 4 Bruce Bates, Martin Bunder, Keith Tognetti Proof. i) From Definition 3, G ( k ) = G ( k) G ( k ) and from Definition, G ( k) = k + G ( k ). Thus G ( k ) = k. ii) By Definitions 3 and, G ( k+ + t ) = G ( k+ + (t ) ), = G ( k+ + t ) G ( k+ + (t ) ), = [ k+ + G ( k+ t )] [ k+ + G ( k+ t )], = G ( k+ t ) G ( k+ t ), = G ( k+ t ). Lemma 4 ii) tells us that in the column G (n) in Table, for k >, entries that are equally spaced either side of k have the same magnitude but are opposite in sign. Theorem 4. P i = S i, the paperfolding sequence of length i. Proof. We prove this by induction. Since G () =, P = = S. Suppose P k = S k. From Lemma 4, i) G ( k ) = k, and so p ( k ) =. ii) All entries G ( k + t ) = G ( k t ) for t =,,,..., k. But this means that P k+ = S k S R k = S k+. Note that Theorem 4 can also be proven from Theorem 4 and Lemma in Bunder et al. [7]. We conclude with an interesting aspect of the code whereby a variant of the paperfolding sequence appears through examining the inverse of the code. Definition 5. (Inverse Binary First Forward Difference of G Sequence). Let G (n) = G (n + ) G (n) whereby { if G q (n) = (n) > if G (n) <. The Inverse Binary First Forward Difference of G Sequence, Q i, is given by: Q i = q (), q (),..., q ( i ). Table shows G (n) and q (n) for n = to 3. As with the paperfolding sequence, Q i, can be expressed in three forms, namely, i) Mirroring: Q i+ = Q i Q i, where Q = and Q i is defined as the sequence formed when each in Q i is replaced by and each in Q i is replaced by. ii) Interleaving: Q i+ = D i#q i, where D i = D i D i for which D = A =, Q =. iii) Alternation: Q i+ = D i#d i # #D #.

22 Mirroring and interleaving in the paperfolding sequence 5 n G (n) G (n) G (n) q (n) Table : The Inverse Binary Reflected Gray Code 9. NUMBERED PAPERFOLDING SEQUENCE We introduce a numbering of the entries in the paperfolding sequence to generate the numbered paperfolding sequence. Definition 6. (Numbered Paperfolding sequence) Let S n represent the numbered paperfolding sequence after n folds commencing with S =. That is, S n is formed by the interleaving of the numbered creases n, n +,, n with S n. Note that the numbered paperfolding sequence is simply the stickbreaking sequence without its first term. Details on the stickbreaking sequence can be

23 6 Bruce Bates, Martin Bunder, Keith Tognetti found at [6]. Example 4. S 5 = 6, 8, 7, 4, 8, 9, 9,,,,, 5,,, 3,, 4,, 5, 6, 6, 3, 7, 3, 8, 4, 9, 7, 3, 5, 3 Lemma 5. Every even integer can be expressed in the form k + a k where k =, 3,... and a =,,,... Proof. For p >, every even integer u can be expressed in one of two ways. That is, u = p (4h + ) or u = p (4h + 3), h =,,,... Consider each case: i) u = p (4h + ) : (9.) u = p (4h + ), = p ( h + ), = p+ h + p. (9.) ii) u = p (4h + 3) : u = p (4h + 3), Theorem 5. Let f t,n denote the t th entry in S n. Then, = p (4h + + ), = p+ (h + ) + p. i) for odd entries in S n, f k,n = n + k, where k =,,, n. ii) for even entries in S n, f k +a k,n = n k + a, where k =, 3,, n and a =,,,, n k. Proof. We prove by induction on n. S = f, = + =. S = f,, f,, f 3, = +,, + =,, 3. Suppose our inductive hypothesis is true for all n up to some value m. Since S m+ is formed by interleaving m, m +,, m+ with S m, we have i) For odd entries in S m+, f k,m+ = f k,m + m, = m + k + m, = m + k. The first part of our result follows. ii) For even entries in S m+, consecutive even entries in S m+ are consecutive entries in S m. There are two subcases: a) If f t,m+ t odd, then these are the odd entries in S m which are the entries spaced 4-apart, commencing with f,m+, in S m+. That is, they are all

24 Mirroring and interleaving in the paperfolding sequence 7 terms f k +a k,m+ a =,,,, m+ k and k =. Hence from our induction hypothesis f t,m+ = f +a,m+ a =,,,, m, = m + r r =,,, m, = m + a a =,,,, m, = m+ + a a =,,,, m+. And so this subcase proves the second part of our result for k =. b) If f t,m+ t even, then these are the even entries in S m which are the entries spaced 4-apart, commencing with f 4,m+, in S m+. From Lemma, after excluding the case k = in a), they are all terms f k +a k,m+ a =,,,, m+ k and k = 3, 4,, m +. Hence from our induction hypothesis, f t,m+ = f k +a k,m+ a =,,,, m+ k and k = 3, 4,, m +, = f j +a j,m a =,,,, m j and j =, 3, 4,, m, = m j + a a =,,,, m j and j =, 3, 4,, m, = m+ k + a a =,,,, m+ k and k = 3, 4,, m +. And so, this subcase proves the second part of our result for k = 3, 4,, m +. Combining a) and b) the second part of our result follows. Combining i) and ii) our result follows. The following corollary restates Theorem 5 in a longer form that is easier to work with for even entries of S n. Corollary 7. Let f t,n denote the t th entry in S n. Then, i) for odd entries in S n, f k,n = n + k, where k =,,, n. ii) for even entries in S n, a) f k (4a+),n = n k + a and b) f k (4a+3),n = n k + a + where k =,,, n and a =,,,, n k. Proof. i) is directly taken from Theorem 5 i). ii) a) follows from (9.) and Theorem 5 ii). ii) b) follows from (9.) and Theorem 5 ii). REFERENCES. C. Davis, D. E. Knuth: Number Representations and Dragon Curves. Journal of Recreational Mathematics, 3 (97), 66 8.

25 8 Bruce Bates, Martin Bunder, Keith Tognetti. J. P. Allouche, R. Bacher: Toeplitz Sequences, Paperfolding, Towers of Hanoi and Progression Free Sequences of Integers. L Enseignement Mathématique 38 (99), J. P. Allouche, M. Bousquet-Mélou: Canonical Positions for the Factors in the Paperfolding Sequence. Journal of Theoretical Computer Science, 9 () (994), J. P. Allouche, M. Bousquet-Mélou: Facteurs des Suites de Rudin-Shapiro Généralisées. Bulletin of the Belgian Mathematical Society, (994), B. P. Bates, M. W. Bunder, K. P. Tognetti: Linkages between the Gauss Map and the Stern-Brocot Tree. Acta Mathematica Academiae Paedagogicae Nyíregyháziensis, (3) (6). 6. A. Blanchard, M. Mendès France: Symétrie et Transcendance. Bulletin des Sciences Mathématiques, 6 (98), M. W. Bunder, K. P. Tognetti, G. E. Wheeler: On Binary Reflected Gray Codes and Functions. Discrete Mathematics, 38 (8), F. M. Dekking, M. Mendes France, A. J. van der Poorten: Folds!. The Mathematical Intelligencer 4 (98), 3 38, II: ibid. 73 8, III: ibid R. L. Graham, D. E. Knuth, O. Patashnik: Concrete Mathematics, nd Edition., Addison-Wesley, Reading, Massachusetts, M. Lothaire: Applied Combinatorics on Words. Cambridge University Press, New York, 5.. M. Mendès France: Principe de la Symétrie Perturbée. In Séminaire de Théorie des Nombres, Paris 979 8, Séminaire Delange-Pisot-Poitou, Birkhäuser, (98), M. Mendès France, J. O. Shallit: Wire Bending. Journal of Combinatorial Theory, Series A, 5 (989), M. Mendès France, J. Shallit, A. J. van der Poorten: On Lacunary Formal Power Series and their Continued Fraction Expansion. In Number Theory in Progress, Proceedings of the International Conference in Honour of A. Schinzel, Vol., (997) Zakopane (edited by Györy Kálmán et al.), de Gruyter, (999), H. Prodinger, F. J. Urbanek: Infinite --Sequences without Long Adjacent Identical Blocks. Discrete Mathematics, 8 (979), N. J. A. Sloane: On-Line Encyclopedia of Integer Sequences. njas/sequences 6. L. Ramshaw: On the Gap Structure of Sequences of Points on a Circle. Indagationes mathematicae, 4 (978), School of Mathematics and Applied Statistics, (Received June 6, 9) University of Wollongong, (Revised October, 9) Wollongong, NSW, Australia 5 s: bbates@uow.edu.au mbunder@uow.edu.au tognetti@uow.edu.au