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Paperfolding and Catalan numbers
Roland Bacher
To cite this version:
Roland Bacher. Paperfolding and Catalan numbers. 2004. �hal-00002104�
ccsd-00002104, version 1 - 17 Jun 2004
Paperfolding and Catalan numbers
Roland Bacher June 17, 2004
Abstract
1: This paper reproves a few results concerning paperfolding sequences using properties of Catalan numbers modulo 2. The guiding prin- ciple can be described by: Paperfolding = Catalan modulo 2 + signs given by 2−automatic sequences.
1 Main results
In this paper, a continued fraction is a formal expression b
0+ a
1b
1+
ba22+...
= b
0+ a
1.
b
1+ a
2.
b
2+ . . .
with b
0, b
1, . . . , a
1, a
2, . . . ∈ C [x] two sequences of complex rational functions.
We denote such a continued fraction by b
0+ a
1|
|b
1+ a
2|
|b
2+ . . . + a
k|
|b
k+ . . .
and call it convergent if the coefficients of the (formal) Laurent-series ex- pansions of its partial convergents
P
0Q
0= b
01 , P
1Q
1= b
0b
1+ a
1b
1, . . . , P
kQ
k= b
0+ a
1|
|b
1+ a
2|
|b
2+ . . . + a
k|
|b
k, . . . are ultimately constant. The limit P
k≥n
l
kx
k∈ C [[x]][x
−1] is then the obvious formal Laurent power series whose first coefficients agree with the Laurent series expansion of P
n/Q
nfor n huge enough.
Let W
1be the word (−x
1) x
1of length 2 in the alphabet {x
1, −x
1}. For k ≥ 2, we define recursively a word W
kof length 2(2
k− 1) over the alphabet {±x
1, ±x
2, . . . , ±x
k} by considering the word
W
k= W
k−1x
k(−x
k) W
k−11Math. Class: 11B65, 11B85, 11C20, 15A23. Keywords: Catalan numbers, paperfold- ing, automatic sequence, continued fraction, Jacobi fraction, binomial coefficient.
constructed by concatenation where W
k−1denotes the reverse of W
k−1ob- tained by reading W
k−1backwards. The first instances of the words W
kare
W
1= (−x
1) x
1W
2= (−x
1) x
1x
2(−x
2) x
1(−x
1)
W
3= (−x
1) x
1x
2(−x
2) x
1(−x
1) x
3(−x
3) (−x
1) x
1(−x
2) x
2x
1(−x
1) and these words are initial subwords of a unique, well-defined infinite word W
∞= w
1w
2w
3, . . . with letters w
i∈ {±x
1, ±x
2, . . .}.
The following result has been known for some time, cf [6] or [13]. Similar or equivalent equivalent statements are for instance given in [9], Theorem 6.5.6 in [2]. It is closely related to paperfolding, see [2], [11], [5]. Related results are contained in [4], [12] and [13].
Theorem 1.1 Consider a sequence x
1, x
2, . . . with values in C . If the cor- responding continued fraction
1|
|1 + w
1|
|1 + w
2|
|1 + . . . + w
k|
|1 + . . .
associated with the sequence W
∞= w
1w
2. . . defined by the letters of the infinite word W
∞converges, then its limit is given by
1 + x
1+ x
21x
2+ x
41x
22x
3+ x
81x
42x
23x
4+ . . . = 1 + X
∞j=1 j
Y
i=1
x
2ij−i.
Remarks. (i) In fact, the proof will follow from the identity P
2(2k−1)Q
2(2k−1)= 1|
|1 + w
1|
|1 + w
2|
|1 + . . . + w
2(2k−1)|
|1 = 1 +
k
X
j=1 j
Y
i=1
x
2ij−i(cf. Lemma 6.5.5 in [2]) showing that the 2(2
k− 1)−th partial convergent of this continued fraction is a polynomial in x
1, . . . , x
k.
(ii) It is of course possible to replace the field C by any other reasonable field or ring.
Examples. (1) Setting x
i= x for all i and multiplying by x we get x|
|1 − x|
|1 + x|
|1 + x|
|1 − x|
|1 + x|
|1 + . . .
= x + x
2+ x
4+ x
8+ . . . = X
∞k=0
x
2k.
The sequence of signs w
1, w
2, . . . ∈ {±1} given by X
∞k=0
x
2k= x|
|1 + w
1x|
|1 + w
2x|
|1 + w
3x|
|1 + . . .
can be recursively defined by
w
4i+1= −w
4i+2= (−1)
i+1, w
8i+3= −w
8i+4= (−1)
i, w
8i+7= −w
8i+8= w
4i+3and is 2−automatic (cf. [2]).
(2) Setting x
i= x
1+3i−1and multiplying by x we get x|
|1 − x
2|
|1 + x
2|
|1 + x
4|
|1 − x
4|
|1 + x
2|
|1 + . . .
= x + x
3+ x
9+ x
27+ . . . = X
∞k=0
x
3k.
(3) Setting x
i= x
1+(i−1)i!and multiplying by x we get x|
|1 − x|
|1 + x|
|1 + x
3|
|1 − x
3|
|1 + x|
|1 + . . .
= x + x
2+ x
6+ x
24+ . . . = X
∞k=1
x
k!.
Consider the sequence µ = (µ
0, µ
1, . . .) = (1, 1, 0, 1, 0, 0, 0, 1, . . .) defined by x P
∞i=0
µ
ix
i= P
∞k=0
x
2k. We associate to µ = (µ
0, µ
1, . . .) its Hankel matrix H whose (i, j)−th coefficient h
i,j= µ
i+j, 0 ≤ i, j depends only on the sum i+j of its indices and is given by (i+j)−th element of the sequence µ.
Consider the (2−automatic) sequence s = (s
0, s
1, s
2, . . .) = 1 1 (−1) 1 (−1) (−1) . . . defined recursively by s
0= 1 and
s
2i= (−1)
is
i, s
2i+1= s
i.
It is easy to check that s
n= (−1)
B0(n)where B
0(n) counts the number of bounded blocks of consecutive 0
′s of the binary integer n (or, equivalently, B
0(n) equals the number of blocks 10 appearing in the binary expansion of the n). E.g, 720 = 2
9+ 2
7+ 2
6+ 2
4corresponds to the binary integer 1011010000. Hence B
0(720) = 3. This description shows that we have s
2n+1+n= −s
nfor n < 2
n. Since we have B
0(2
n+1+ n) = 1 + B
0(n) for n < 2
nand B
0(2
n+1+ n) = B
0(n) for 2
n≤ n < 2
n+1, the sequence s
0, s
1, . . . can also be constructed by iterating the application
W
αW
ω7−→ W
αW
ω(−W
α)W
ωwhere where W
α, W
ωare the first and the second half of the finite word
s
0. . . s
2n−1.
Denote by D
sthe infinite diagonal matrix with diagonal entries s
0, s
1, s
2, . . . and by D
athe infinite diagonal matrix with diagonal entries given by the alternating, 2−periodic sequence 1, −1, 1, −1, 1, −1.
We introduce the infinite unipotent lower triangular matrix L with co- efficients l
i,j∈ {0, 1} for 0 ≤ i, j given by
l
i,j≡ 2i
i − j
− 2i
i − j − 1
≡ 2i
i − j
+ 2i
i − j − 1
≡
2i + 1 i − j
(mod 2) and denote by L(k) the k × k submatrix with coefficients l
i,j, 0 ≤ i, j < k of L. The matrix L(8) for instance is given by
L(8) =
1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1
.
Theorem 1.2 (i) We have
H = D
sL D
aL
tD
s.
(ii) (BA0BAB−construction for L) We have for n ≥ 0
L(2
n+2) =
A B A
0 B A
B A B A
where A and B are the obvious submatrices of size 2
n× 2
nin L(2
n+1).
Remarks. (i) Assertion (ii) of Theorem 1.2 yields an iterative construc- tion for L by starting with L(2) =
1 1 1
.
(ii) Conjugating H with an appropriate diagonal ±1−matrix, we can get the the LU-decomposition of any Hankel matrix associated with a sequence having ordinary generating function x
−1P
∞k=0
ǫ
kx
2k, ǫ
0= 1, ǫ
1, ǫ
2, . . . ∈ {±1}. More precisely, the appropriate conjugating diagonal matrix has co- efficients d
n,n= (−1)
Pj=0νjǫj+1for 0 ≤ n = P
j=0
ν
j2
ja binary integer.
The case ǫ
0= −1 can be handled similarly by replacing the diagonal matrix D
awith its opposite −D
a.
The matrix P = (D
sL D
s)
−1= D
sL
−1D
swith coefficients p
i,j, 0 ≤ i, j encodes the formal orthogonal polynomials Q
0, Q
1, . . . , Q
n= P
nk=0
p
n,kx
kwith moments µ
0, µ
1, . . . given by P
∞i=0
µ
ix
i+1= P
∞k=0
x
2k.
In order to describe the matrix P we introduce a unipotent lower trian- gular matrix M with coefficients m
i,j∈ {0, 1}, 0 ≤ i, j by setting
m
i,j= i + j
2j
(mod 2) .
As above (with the matrix L) we denote by M (k) the k × k submatrix with coefficients m
i,j, 0 ≤ i, j < k of M.
Theorem 1.3 (i) We have
L D
aM = D
a(where D
adenotes the diagonal matrix with 2−periodic diagonal coefficients 1, −1, 1, −1, . . .).
(ii) (BA0BAB−construction for M ) We have have for n ≥ 0
M(2
n+2) =
A
′B
′A
′A
′B
′A
′B
′0 B
′A
′
where A
′and B
′are the obvious submatrices of size 2
n× 2
nin M(2
n+1).
The matrix P = D
sD
aM D
aD
s= (D
sL D
s)
−1encoding the coeffi- cients of the orthogonal polynomials with moments µ
0, µ
1, . . . has thus all its coefficients in {0, ±1}, see also [1] and [8].
The (formal) orthogonal polynomials Q
0, Q
1, . . . satisfy a classical three- term recursion relation with coefficients given by the continued Jacobi frac- tion expansion of P
µ
ix
i=
x1P
∞k=0
x
2k. This expansion is described by the following result.
Theorem 1.4 We have
∞
X
k=0
x
2k=
|1−xd|1x+
|1−x2d|2x+
|1−x2d|3x+ . . .
=
x||1−x
+
x2||1+2x
+
x2||1
+
x2||1
+
x2||1−2x
+ . . . where d
n=
sns−sn−2n−1
with s
−1= 0, s
0= 1 and s
2i+1= (−1)
is
2i= s
iis the recursively defined sequence introduced previously (extended by s
−1= 0).
The formal orthogonal polynomials with moments µ
0, µ
1, . . . are thus recursively defined by Q
0= 1, Q
1= x + 1 and
Q
n+1= (x − d
n)Q
n+ Q
n−1for n ≥ 2 .
The rest of the paper is organised as follows.
Section 2 contains a proof of Theorem 1.1.
Section 3 recalls a few definitions and facts concerning Hankel matrices, formal orthogonal polynomials etc.
Section 4 is a little digression about Catalan numbers. All objects ap- pearing in this paper are essentially obtained by considering the correspond- ing objects associated with Catalan numbers and by reducing them modulo 2 using sign conventions prescribed by recursively defined sequences (which are 2−automatic).
Section 5 recalls two classical and useful results on the reduction modulo 2 (or modulo a prime number) of binomial coefficients.
Section 6 is devoted to the proof of Theorem 1.2.
Section 7 contains a proof of Theorem 1.3 and a related amusing result.
Section 8 contains formulae for the matrix products M L and L M . Section 9 contains a proof of Theorem 1.4.
Section 10 displays the LU −decomposition of the Hankel matrix ˜ H as- sociated with the sequences 1, 0, 1, 0, 0, 0, 1, 0, . . . obtained by shifting the sequence 1, 1, 0, 1, 0, 0, 0, 1, . . . associated with powers of 2 by one (or equiv- alently, by removing the first term).
Section 11 contains a uniqueness result.
2 Proof of Theorem 1.1
More or less equivalent reformulations of Theorem 1.1 can be found in several places. A few are [9], [11] and [2]. We give her a computational proof: Symmetry-arguments of continued fractions are replaced by polyno- mial identities.
Given a continued fraction b
0+ a
1|
|b
1+ a
2|
|b
2+ . . .
it is classical (and easy, cf. for instance pages 4,5 of [10]) to show that the product
1 b
00 1
0 a
11 b
10 a
21 b
2· · ·
0 a
k1 b
kequals
P
k−1P
kQ
k−1Q
kwhere
P
kQ
k= b
0+ a
1|
|b
1+ . . . + a
k|
|b
kdenotes the k−th partial convergent.
Given a finite word U = u
1u
2. . . u
lwe consider the product M (U ) =
0 u
11 1
0 u
21 1
· · ·
0 u
l1 1
.
Defining
X
k= 1 + P
k j=1Q
ji=1
x
2ij−iX ˜
k= 1 + P
k−1j=1
Q
ji=1
x
2ij−i− Q
ki=1
x
2ik−i(example: X
3= 1 + x
1+ x
21x
2+ x
41x
22x
3and ˜ X
3= 1 + x
1+ x
21x
2− x
41x
22x
3) one checks easily the identities
X
n+ x
n+1(X
n2−1− X
nX ˜
n) = X
n+1and
X
n− x
n+1(X
n2−1− X
nX ˜
n) = ˜ X
n+1. Lemma 2.1 We have for all n ≥ 1
M(W
n) =
X ˜
n− X
n2−11 − X
nX
n2−1X
nand
M (W
n) =
X
n− X
n2−11 − X ˜
nX
n2−1X ˜
n.
Proof. The computations M(W
1) = M((−x
1) x
1) =
0 −x
11 1
0 x
11 1
=
−x
1−x
11 1 + x
1=
(1 − x
1) − 1
21 − (1 + x
1)
1
21 + x
1and
M(W
1) = M (x
1(−x
1)) =
0 x
11 1
0 −x
11 1
=
(1 + x
1) − 1
21 − (1 − x
1)
1
21 − x
1prove Lemma 2.1 for n = 1.
For n ≥ 1 we have
M(W
n+1) = M (W
nx
n+1(−x
n+1) W
n)
=
X ˜
n− X
n2−11 − X
nX
n2−1X
nx
n+1x
n+11 1 − x
n+1X
n− X
n2−11 − X ˜
nX
n2−1X ˜
n=
(X
n− x
n+1(X
n2−1− X
nX ˜
n)) − X
n21 − (X
n+ x
n+1(X
n2−1− X
nX ˜
n)) X
n2X
n+ x
n+1(X
n2−1− X
nX ˜
n)
=
X ˜
n+1− X
n21 − X
n+1X
n2X
n+1and
M(W
n+1) = M (W
n(−x
n+1) x
n+1W
n)
=
X ˜
n− X
n2−11 − X
nX
n2−1X
n−x
n+1−x
n+11 1 + x
n+1X
n− X
n2−11 − X ˜
nX
n2−1X ˜
n=
(X
n+ x
n+1(X
n2−1− X
nX ˜
n)) − X
n21 − (X
n− x
n+1(X
n2−1− X
nX ˜
n)) X
n2X
n− x
n+1(X
n2−1− X
nX ˜
n)
=
X
n+1− X
n21 − X ˜
n+1X
n2X ˜
n+1which proves Lemma 2.1 by induction on n. 2
Proof of Theorem 1.1. Lemma 2.1 shows that M(1 W
n) =
0 1 1 1
X ˜
n− X
n2−11 − X
nX
n2−1X
n=
X
n2−1X
nX ˜
n1
.
The 2(2
n− 1)−th partial convergent of the continued fraction 1|
|1 + w
1|
|1 + w
2|
|1 + . . . defined by the word W
∞equals thus
X
n= 1 +
n
X
j=1 j
Y
k=1
x
2kj−kand the sequence of these partial convergents has limit X
∞= 1 +
X
∞j=1 j
Y
k=1
x
2kj−kwhich is the statement of Theorem 1.1. 2
3 Formal orthogonal polynomials
The content of this section is classical and can for instance be found in [3]
or [15].
We consider a formal power series g(x) =
X
∞k=0
µ
kx
k∈ C [[x]]
as a linear form on the complex vector space C [x]
∗of polynomials by setting l
g(
n
X
i=0
α
ix
i) =
n
X
i=0
α
iµ
i.
We suppose that the n−th Hankel matrix H(n) having coefficients h
i,j= µ
i+j, 0 ≤ i, j < n associated with µ = (µ
0, . . .) is invertible for all n.
Replacing such a sequence µ = (µ
0, µ
1, . . .) by
µ10
µ = (1,
µµ10
, . . .) we may also suppose that µ
0= 1. We get a non-degenerate scalar product on C [x]
by setting
hP, Qi = l
g(P Q) .
Applying the familiar Gram-Schmitt orthogonalisation we obtain a sequence Q
0= 1, Q
1= x − a
0,
Q
n+1= (x − a
n)Q
n− b
nQ
n−1, n ≥ 1
of mutually orthogonal monic polynomials, called formal orthogonal polyno- mials with moments µ
0, µ
1, . . ..
We encode the above recursion relation for the polynomials Q
0, Q
1, . . . by the Stieltjes matrix
S =
a
01 b
1a
11
b
2a
21 b
3a
31
. .. ... ...
.
The non-degeneracy condition on µ
0, µ
1, . . . shows that the (infinite) Hankel matrix H with coefficients h
i,j= µ
i+j, 0 ≤ i, j has an LU −decomposition H = LU where L is lower triangular unipotent and U = DL
tis upper tri- angular with D diagonal and invertible.
The diagonal entries of D are given by d
i,i=
i
Y
k=1
b
kand we have thus
det(H(n)) =
n−1
Y
k=1
(b
k)
n−kwhere H(n) is the n−th Hankel matrix with coefficients h
i,j= µ
i+j, 0 ≤ i, j < n. The i−th row vector
(m
i,0, m
i,1, . . .)
of M = L
−1encodes the coefficients of the i−th orthogonal polynomial Q
i=
i
X
k=0
m
i,kx
k.
The Stieltjes matrix S satisfies
LS = L
−=
l
1,0l
1,1l
2,0l
2,1l
2,2.. . . .. ...
where L
−is obtained by removing the first row of the unipotent lower trian- gular matrix L with coefficients l
i,j, 0 ≤ i, j involved in the LU −decomposition of H.
The generating function g = P
∞k=0
µ
kx
kcan be expressed as a continued fraction of Jacobi type
g(x) = 1/(1 − a
0x − b
1x
2/(1 − a
1x − b
2x
2/(1 − a
2x − b
3x
2/(1 − . . . )))) where a
0, a
1, . . . and b
1, b
2, . . . are as above and are encoded in the Stieltjes matrix.
4 The Catalan numbers
The Catalan numbers C
n=
2nn 1n+1
are the coefficients of the algebraic generating function
c(x) = P
∞n=0
C
nx
n=
1−√2x1−4x= 1 + x + 2x
2+ 5x
3+ 14x
4+ . . .
= 1|
|1 − x|
|1 − x|
|1 − x|
|1 − . . .
= 1|
|1 − x − x
2|
|1 − 2x − x
2|
|1 − 2x − x
2|
|1 − 2x − . . . and satisfies c = 1 + x c
2.
We have g(x) ≡ c(s) (mod 2) where g(x) = P
∞j=0
x
2j−1, cf. the last lines of [14]. A different proof can be given by remarking that c
ncounts the number of planar binary rooted trees with n + 1 leaves. Call two such trees equivalent if they are equivalent as abstract (non-planar) rooted trees. The cardinality of each equivalence class is then a power of two. An equivalence classe containing only one element is given by a rooted 2−regular trees having 2
nleaves which are all at the same distance n from the root.
Many interesting mathematical objects associated with g(x) are in fact obtained by reducing the corresponding objects of c(x) modulo 2. Mirac- ulously, all this works over Z by choosing suitable signs given by a few recursively-defined sequences (which are in fact always 2−automatic, see [2]
for a definition).
Consider the unipotent lower triangular matrices
L =
1
1 1
2 3 1
5 9 5 1
14 28 20 7 1 42 90 75 35 9 1
and
L ˜ =
1
2 1
5 4 1
14 14 6 1
42 48 27 8 1
132 165 10 44 10 1
with coefficients l
i,j=
i−2ij−
i−2ij−1, respectively ˜ l
i,j=
2i+1i−j−
i2i+1−j−1for 0 ≤ i, j obtained by taking all even (respectively odd) lines (starting with line number zero) of the so-called Catalan-triangle
1 1
1 1
2 1
2 3 1
5 4 1
5 9 5 1
14 14 6 1
14 28 20 7 1
The matrix L is associated with the LU −decomposition of the Hankel matrix H = L L
thaving coefficients h
i,j= C
i+j=
2(i+j)i+j/(i + j + 1), 0 ≤ i, j the Catalan numbers. The matrix ˜ L yields the LU −decomposition of the Hankel matrix ˜ H = ˜ L L ˜
thaving coefficients ˜ h
i,j= C
i+j+1=
2(i+j+1)i+j+1/(i + j+2), 0 ≤ i, j associated with the shifted Catalan sequence 1, 2, 5, 14, 42, . . ..
The inverse matrices L
−1and ˜ L
−1are given by D
aM D
aand D
aM D ˜
awhere D
ais the diagonal matrix with 2-periodic alternating diagonal coef- ficients 1, −1, 1, −1, and where M and ˜ M have coefficients m
i,j=
i+j2jand
˜
m
i,j=
i+j+12j+1for 0 ≤ i, j.
Let us mention that the products P = LM and ˜ P = ˜ L M ˜ have non-zero coefficients p
i,j=
i! (2j)! (i(2i)! j!−j)!and ˜ p
i,j= 4
(i−j) jifor 0 ≤ j < i. They can
also be given by
P = L M = exp(
0 2 0
6 0 10 0
. .. ...
)
and
P ˜ = ˜ L M ˜ = exp(
0 4 0
8 0 12 0
. .. ...
) .
Computations suggest that the products M L and ˜ M L ˜ have also nice logarithms:
M L = exp(
0 2 0 0 6 0 2 0 10 0
0 6 0 14 0
2 0 10 0 18 0
0 6 0 14 0 22 0
.. . .. . . .. ...
)
and
M ˜ L ˜ = exp(
0 4 0 0 8 0 4 0 12 0
0 8 0 16 0
4 0 12 0 20 0 .. . .. . . .. ...
) .
5 Binomial coefficients modulo 2
Chercher ref. (Concrete Maths? Graham, Knuth, Patashnik) Using the Frobenius automorphism we have
(1 + x)
Piνi2i≡ Y
i
(1 + x
2i)
νi(mod 2)
which shows the formula n
k
≡ Y
i
ν
iκ
i(mod 2) where n = P
i≥0
ν
i2
i, ν
i∈ {0, 1} and k = P
i≥0
κ
i2
i, κ
i∈ {0, 1} are two natural binary integers. Kummer showed that the 2−valuation v
2of
nk(defined by 2
v2|
nkand 2
v2+16 |
nk) equals the number of carry-overs when adding the two natural numbers k and n − k, written in base 2 (cf. [16]).
As an application, one (re)proves easily that C
n=
2nn 1n+1
≡ 1 (mod 2) if and only if n+1 ∈ {2
k}
k∈Nis a power of two. Indeed,
2nn 1n+1
≡
n!(2n+1)!′n+1)!=
2n+1 n
(mod 2) and addition of the binary integers n and (n + 1) needs always a carry except if n = P
k−1i=0
2
kand n + 1 = 2
k.
6 Proof of Theorem 1.2
The aim of this section is to prove Theorem 1.2 which describes the LU −de- composition of the Hankel matrix H with coefficients h
i,j= µ
i+j, 0 ≤ i, j where
X
∞k=0
µ
ix
i= X
∞j=0
x
2j−1= 1 + x + x
3+ x
7+ x
15+ . . . .
We give in fact 3/2 proofs: We first prove assertion (i) by using the definition l
i,j≡
2i+1i−j(mod 2) of L.
We prove then assertion (ii) showing that the matrix L satisfies the BA0BAB−construction.
Finally, we reprove assertion (i) using the recursive BA0BAB−structure of L.
Proof of assertion (i) in Theorem 1.2. We denote by
nk∈ {0, 1}
the reduction (mod 2) with value in {0, 1} of the binomial coefficient
nk. The main ingredient of the proof is the fact that
2n + 1 k
= n
⌊k/2⌋
where ⌊k/2⌋ = k/2 if k is even and ⌊k/2⌋ = (k − 1)/2 if k is odd.
We compute first the coefficient r
2i,2jof the product R = D
sLD
aL
tD
s(recall that L has coefficients l
i,j=
2i+1i−j):
r
2i,2j= s
2is
2jP
k
(−1)
k 4i+12i−k 4j+12j−k
= s
2is
2jP
k=0 4i+1 2i−2k
4j+12j−2k
− P
k=1 4i+1 2i−2k+1
4j+12j−2k+1
= s
2is
2jP
k=0 2i i−k
2jj−k
− P
k=1 2i i−k
2jj−k
= s
2is
2j 2ii 2jj
=
1 if i = j = 0
0 otherwise.
Similarly, we get
r
2i+1,2j+1= s
2i+1s
2j+1P
k
(−1)
k 2i+14i+3−k 4j+32j+1−k
= s
2i+1s
2j+1P
k=0 4i+3 2i+1−2k
4j+32j+1−2k
− P
k=0 4i+3 2i−2k
4j+32j−2k
= s
2i+1s
2j+1P
k=0 2i+1 i−k
2j+1j−k
− P
k=0 2i+1 i−k
2j+1j−k
= 0 For r
2i,2j+1= r
2j+1,2iwe have
r
2i,2j+1= s
2is
2j+1P
k 4i+1 2i−2k
4j+32j+1−2k
− P
k 4i+1 2i−2k−1
4j+32j−2k
= s
2is
2j+1P
k 2i i−k
2j+1j−k
− P
k 2i
i−k−1
2j+1j−k
= s
2is
2j+1P
k
2i i−k+
i−2ik−12j+1 j−k
− 2 P
k 2i
i−k−1
2j+1j−k
= s
2is
2j+1P
k 2i+1 i−k
2j+1j−k
− 2 P
k 2i
i−k−1
2j+1j−k
.
Using parity arguments and the recursion s
2i= (−1)
is
i, s
2j+1= s
jwe get for i even
r
2i,2j+1= s
2is
2j+1P
k
(−1)
k 2i+1i−k 2j+1j−k
+ +2 P
k
2i+1 i−2k−1−
i−2k2i−22j+1 j−2k−1
= s
is
jP
k
(−1)
k 2i+1i−k 2j+1j−k
+ 0
= r
i,j. Similarly, for i odd we have
r
2i,2j+1= −s
2is
2j+1P
k
(−1)
k 2i+1i−k 2j+1j−k
+
−2 P
k
2i+1 i−2k−
i−2k2i−12j+1 j−2k
= s
is
jP
k
(−1)
k 2i+1i−k 2j+1j−k
+ 0
= r
i,j.
Finally, the above computations together with induction on i + j show that r
i,j= 0 except if i + j = 2
k− 1 for some k ∈ N . In this case we get r
i,j= 1
which proves the result. 2
Proof of assertion (ii) in Theorem 1.2. Assume that L(2
n+1) is given by
A B A
. (This holds clearly for n = 0.) We have for 0 ≤ i, j < 2
n+12(i + 2
n+1) + 1 (i + 2
n+1) − (j + 2
n+1)
=
2i + 2
n+2+ 1 i − j
≡
2i + 1 i − j
(mod 2) since 2i + 1 < 2
n+2. This shows already that L(2
n+2) is of the form A ˜
B ˜ A ˜
with ˜ A = L(2
n+1).
Consider now a binomial coefficient of the form 2(2
n+1+ i) + 1
(2
n+1+ i) − j
=
2
n+2+ 2i + 1 2
n+1+ i − j
with 0 ≤ i, j < 2
n+1which is odd. This implies i + j + 1 ≥ 2
n+1since otherwise (2
n+1+i) − j < (2
n+1+ i) +j + 1 < 2
n+2< 2
n+2+ 2i+ 1 and there must therefore be a carry when adding the binary integers ((2
n+1+ i) − j) and ((2
n+1+ i) + j + 1).
Two such binary integers ((2
n+1+ i) − j) and ((2
n+1+ i) + j + 1) add thus without carry if and only if the binary integers (2
n+2+ 2
n+1+ i − j) and (2
n+1+ i + j + 1 − 2
n+2) add without carry. This shows the equality
2(2
n+1+ i) + 1 (2
n+1+ i) − j
≡
2(2
n+1+ i) + 1 (2
n+1+ i) − (2
n+2− 1 − j)
for 0 ≤ i, j < 2
n+1. Geometrically, this amounts to the fact that the block B ˜ is the vertical mirror of the block ˜ A and this symmetry is preserved by
the BA0BAB−construction. 2
BA0BAB−proof of assertion (i). We have
1 1 1 0 1 1 1 1 1 1
1
−1 1
−1
1 1 0 1 1 1 1 1 1 1
=
1 1 0 1
1 0 −1 0
0 −1 0 0
1 0 0 0
and conjugation by
1
1
−1 1
yields
1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0
.
The proof is now by induction using the BA0BAB−construction of L.
Writing ˜ A = D
aA
tand ˜ B = D
aB
t(with D
adenoting the obvious subma- trix of finite size of the previously infinite matrix D
a) we have
A B A
0 B A
B A B A
A ˜ B ˜ 0 B ˜ A ˜ B ˜ A ˜ A ˜ B ˜ A ˜
=
A A ˜ A B ˜ 0 A B ˜
B A B ˜ B ˜ + A A ˜ A B ˜ B B ˜ + A A ˜ 0 B A ˜ B B ˜ + A A ˜ B A ˜ + A B ˜ B A B ˜ B ˜ + A A A ˜ B ˜ + B A ˜ 2B B ˜ + 2A A ˜
.
We have B B ˜ + A A ˜ = 0 by induction. The result is now correct up to signs if A B ˜ = −B A. ˜
We conjugate now this last matrix by the diagonal matrix with diagonal entries s
0, . . . , s
2n+2−1which we can write using blocks of size 2
nin the form:
D
αD
ω−D
αD
ω
.
We get (after simplification using induction and the assumption A B ˜ =
−B A) ˜
D
αA AD ˜
αD
αA BD ˜
ω0 D
αA BD ˜
ωD
ωB AD ˜
α0 −D
ωA BD ˜
α0 −D
αB AD ˜
ωD
ωB AD ˜
α
.
Supposing the identity A B ˜ = −B A ˜ we get
−D
αB AD ˜
ω= D
αA BD ˜
ωand − D
ωA BD ˜
α= D
ωB AD ˜
αwhich have the correct form (one’s on the antidiagonal and zeros everywhere else) by induction.
We have yet to prove that A B ˜ = −B A. Writing ˜ A = a
b a
, B = b
b a
and ˜ A = D
aA
t, B ˜ = D
aB
twe have
A B ˜ = a
b a
˜ b
˜ b ˜ a
=
0 a ˜ b a ˜ b 0
and
B A ˜ =
0 b b a
˜ a ˜ b 0 ˜ a
=
0 b˜ a b˜ a 0
which proves A B ˜ = −B A ˜ since a ˜ b = −b˜ a by induction. 2
7 Proof of Theorem 1.3
Denote by D
ethe diagonal matrix with diagonal coefficients 1, 0, 1, 0, 1, 0, . . . , 1, 0, and by D
o= 1−D
ethe diagonal matrix with coefficients 0, 1, 0, 1, 0, 1, . . . , 0, 1,.
We have thus 1 = D
e+D
oand D
a= D
e−D
o. The main ingredient for prov- ing Theorem 1.3 is the following proposition which is also of independent interest.
Proposition 7.1 We have
M D
eL = A + D
e, M D
oL = A + D
owhere A is the strictly lower triangular matrix with coefficients a
i,j= 1 if i > j and a
i,j= 0 otherwise.
Proof. We denote by α
i,j= (M D
eL)
i,jrespectively β
i,j= (M D
oL)
i,jthe corresponding coefficients of both products. The proof is by induction on i + j. Obviously, α
i,j= β
i,j= 0 for i < j. The proof for i = j is obvious since M and L are lower triangular unipotent matrices. Consider now for i > j
α
i,j= X
k
i + 2k 4k
4k + 1 2k − j
.
(recall that m
i,k=
i+k2k≡
i+k2k(mod 2) and l
k,j=
2k+1k−j≡
2k+1k−j(mod 2)).
Since 2i + 2k
4k
=
2i + 1 + 2k 4k
and
4k 2k − 2j
=
4k + 1 2k − 2j
=
4k + 1 2k − (2j − 1)
we have
α
i,j= X
k
⌊i/2⌋ + k 2k
2k k − ⌊(j + 1)/2⌋
= X
k, k≡⌊(j+1)/2⌋ (mod 2)
⌊i/2⌋ + k 2k
2k + 1 k − ⌊(j + 1)/2⌋
=
α
⌊i/2⌋,⌊(j+1)/2⌋if ⌊(j + 1)/2⌋ ≡ 0 (mod 2) , β
⌊i/2⌋,⌊(j+1)/2⌋if ⌊(j + 1)/2⌋ ≡ 1 (mod 2) . We have similarly
β
i,j= X
k
i + 2k + 1 4k + 2
4k + 3 2k + 1 − j
= X
2⌊(i − 1)/2⌋ + 1 + 2k + 1 4k + 2
4k + 3 2k + 1 − 2⌊j/2⌋
= X
k, k≡⌊(i−1)/2⌋ (mod 2)
⌊(i − 1)/2⌋ + k + 1 2k + 1
2k + 1 k − ⌊j/2⌋
= X
k, k≡⌊(i−1)/2⌋ (mod 2)
⌊(i + 1)/2⌋ + k 2k
2k + 1 k − ⌊j/2⌋
=
α
⌊(i+1)/2⌋,⌊j/2⌋if ⌊(i − 1)/2⌋ ≡ 0 (mod 2) , β
⌊(i+1)/2⌋,⌊j/2⌋if ⌊(i − 1)/2⌋ ≡ 1 (mod 2) . except if ⌊(i − 1)/2⌋ = ⌊j/2⌋ = s where
X
k, k≡s (mod 2)
