Computer assisted proof for Apwenian sequences related to Hankel determinants

0.32 MB·

by  Hao Fu

COMPUTER ASSISTED PROOF FOR APWENIAN
SEQUENCES RELATED TO HANKEL
DETERMINANTS
HAO FU AND GUO-NIU HAN
6
1 Abstract. Aninfinite±1-sequenceiscalledApwenianifitsHan-
0 kel determinant of order n divided by 2n−1 is an odd number for
2 everypositiveintegern. In1998,Allouche,Peyrière,WenandWen
n discovered and proved that the Thue–Morse sequence is an Ap-
a wenian sequence by direct determinant manipulations. Recently,
J BugeaudandHanre-provedthelatterresultbymeansofanappro-

8 priate combinatorialmethod. By significantly improving the com-
1
binatorialmethod,weprovethatseveralotherApweniansequences

related to the Hankel determinants with Computer Assistance.
]
T
N
.
h
1. introduction
t
a
m Foreachinfinitesequence c = (c ) andeachnonnegativeintegern
k k≥0
[ the Hankel determinant of order n of the sequence c is defined by

1
c c ··· c
v 0 1 n−1
70 (1.1) Hn(c) := (cid:12)(cid:12) c...1 c...2 ·.·.·. c...n (cid:12)(cid:12).
3 (cid:12) (cid:12)
4 (cid:12)c c ··· c (cid:12)
(cid:12) n−1 n 2n−2(cid:12)
0
(cid:12) (cid:12)
. We also speak of the Hankel(cid:12)determinants of the(cid:12)power series c˜(x) =
1
(cid:12) (cid:12)
0 c xk and write H (c˜(x)) = H (c). The Hankel determinants are
k≥0 k n n
6
widely studied in Mathematics and, in several cases, can be evaluated
1
P
: by basic determinant manipulation, LU-decomposition, or Jacobi con-
v
tinued fraction (see, e.g., [15, 16, 7, 18, 17]). However, the Hankel de-
i
X
terminants studied inthepresent paperapparently have noclosed-form
r
a expressions, and require additional efforts to obtain specific arithmeti-
cal properties.
An infinite ±1-sequence c = (c ) is called Apwenian if its Han-
k k≥0
kel determinant of order n divided by 2n−1 is an odd number, i.e.,
H (c)/2n−1 ≡ 1 (mod 2), for all positive integer n. The corresponding
n
Date: May 21, 2015.
2010 Mathematics Subject Classification. 05A05, 05A15, 11B50, 11B85,11C20.
Key words and phrases. Hankel determinant, Thue–Morse sequence, Apwenian
sequence, permutation, computer assisted proof.
ThefirstauthorwassupportedbytheNationalBasicResearchProgramofChina
Grant 2011CBA00300, 2011CBA00301, the National Natural Science Foundation
of China Grant 61033001,61361136003.
1
2 HAOFU AND GUO-NIUHAN
generatingfunctionorthepowerseries˜c(x)isalsosaidtobeApwenian.
Recall that the Thue–Morse sequence, denoted by
e = (e ) = (1,−1,−1,1,−1,1,1,−1,−1,1,1,−1...),
k k≥0
is a special ±1-sequence [10], defined by the generating function
∞ ∞
(1.2) e˜(x) = e xk = (1−x2k),
k
k=0 k=0
X Y
or equivalently, by the recurrence relations
(1.3) e = 1, e = e and e = −e for k ≥ 0.
0 2k k 2k+1 k
TheThue–MorsesequenceisalsocalledProuhet–Thue–Morsesequence.
For other equivalent definitions and properties related to the sequence,
see [2, 3, 11, 9, 8]. In 1998, Allouche, Peyrière, Wen and Wen estab-
lished a congruence relation concerning the Hankel determinants of the
Thue–Morse sequence [1].
Theorem 1.1 (APWW). The Thue–Morse sequence on {1,−1} is Ap-
wenian.
Theorem 1.1 has an important application to Number Theory. As a
consequence of Theorem 1.1, all the Hankel determinants of the Thue–
Morse sequence are nonzero. This property allowed Bugeaud [5] to
prove that theirrationality exponents ofthe Thue–Morse–Mahler num-
bers are exactly 2.
The goal of the paper is to find more Apwenian sequences. Let d be
a positive integer and v = (v ,v ,v ,...,v ) a finite ±1-sequence of
0 1 2 d−1
length d such that v = 1. The generating polynomial of v is denoted
0
by v˜(x) = d−1v xi. It is clear that the following power series
i=0 i
∞
P
(1.4) Φ(v˜(x)) = v˜(xdk)
k=0
Y
defines a ±1-sequence. Thus, the power series displayed in (1.2) is
equal to Φ(1−x). Our main result is stated next.
Theorem 1.2. The following power series are all Apwenian:
F (x) = Φ(1−x),
2
F (x) = Φ(1−x−x2),
3
F (x) = Φ(1−x−x2 −x3 +x4),
5
F (x) = Φ(1−x−x2 +x3 −x4 +x5 +x6 +x7 +x8 −x9
11
−x10),
F (x) = Φ(1−x−x2 +x3 −x4 −x5 −x6 −x7 −x8 +x9
13
−x10 −x11 +x12),
COMPUTER ASSISTED PROOF FOR APWENIAN SEQUENCES 3
F (x) = Φ(1−x−x2 +x3 −x4 +x5 +x6 +x7 +x8 +x9
17a
+x10 +x11 −x12 +x13 −x14 −x15 +x16),
F (x) = Φ(1−x−x2 −x3 +x4 +x5 −x6 +x7 +x8 +x9
17b
−x10 +x11 +x12 −x13 −x14 −x15 +x16).
Remarks. Let us make some useful comments about the above
Theorem.
(1) The fact that thegenerating functionF (x) for the Thue–Morse
2
sequence is Apwenian has already been proved in [1].
(2) By using the Jacobi continued fraction expansion of a power
series F(x), we know that H (F(x)) = H (F(−x)). See, for
n n
example, [15, 7, 12, 13]. Hence, Theorem 1.2 implies that
F (−x) = Φ(1+x−x2), F (−x) = Φ(1+x−x2+x3+x4), etc.
3 5
are all Apwenian.
(3) There is no F in Theorem 1.2, but two F (we mean F and
7 17 17a
F ).
17b
(4) Φ(1−x−x2 +x3) is Apwenian since it is equal to Φ(1−x).
Actually, Theorem 1.1 has three proofs. The original proof of The-
orem 1.1 is based on determinant manipulation by using the so-called
sudoku method [1, 14]. The second one is a combinatorial proof derived
byBugeaudandHan[6]. Thethirdproofisveryshort andmakesuseof
Jacobi continued fraction algebra [13]. Unfortunately, the method de-
velopedintheshortproofcannotbeusedforprovingourmaintheorem,
because the underlying Jacobi continued fractions are not ultimately
periodic [13, 12]. However, another analogous result for the sequence
F (x) when dealing with modulo 3 (instead of modulo 2) is established
3
using the short method, as stated in the next theorem [12].
Theorem 1.3. For every positive integer n the Hankel determinant
H (F (x)) of the sequence F (x) verifies the following relation
n 3 3
1 (mod 3) if n ≡ 1,2 (mod 4);
(1.5) H (F (x)) ≡
n 3
2 (mod 3) if n ≡ 3,0 (mod 4).
(
Combining Theorem 1.3 and Theorem 1.2 yields the following result.
Corollary 1.4. For every positive integer n the Hankel determinant
H (F (x)) verifies the following relation
n 3
H (F (x)) 1 (mod 6) if n ≡ 0,1 (mod 4);
n 3
(1.6) ≡
2n−1 5 (mod 6) if n ≡ 2,3 (mod 4).
(
In the following table we reproduce the first few values of the Hankel
determinants of the sequence F (x) for illustrating Theorems 1.2, 1.3
3
and Corollary 1.4.
4 HAOFU AND GUO-NIUHAN
n 1 2 3 4 5 6 7 8 9 10
H (f) 1 −2 −4 8 16 −32 −64 128 4864 −9728
n
H (f) (mod 3) 1 1 2 2 1 1 2 2 1 1
n
H (f)/2n−1 1 −1 −1 1 1 −1 −1 1 19 −19
n
Hn(f) (mod 2) 1 1 1 1 1 1 1 1 1 1
2n−1
Hn(f) (mod 6) 1 5 5 1 1 5 5 1 1 5
2n−1
Recently, Bugeaud and Han re-proved Theorem 1.1 by means of an
appropriate combinatorial method [6]. The latter method has been sig-
nificantly upgraded to prove that F (x) is Apwenian. As can be seen,
3
in Section 3 Step 2, a family of cases (called types) is considered for
proving the various recurrence relations. Roughly speaking, the types
are indexed by words s s s ···s of length d+1 over a d-letter alpha-
0 1 2 d
bet. Comparing to the original combinatorial method, the upgrading
does not provide a shorter proof; however, it involves of a systematic
proof by exhaustion that only consists of checking all the types. The
proof of Theorem 1.2 is then achieved with Computer Assistance.
In practice, the number of types is very large. For example, as de-
scribed in §2.3 for the study of F (x), there are 2274558 types! Fortu-
11
nately, theset ofpermutations ofeach typecanbedecomposed into the
Cartesian product of so-called atoms (see Substep 3(d) in the sequel),
and moreover, the cardinality of each atom can be rapidly evaluated
by a sequence of tests (see Definition 4.1 and Table 4.1).
Problem 1.5. Is the following power series Apwenian:
F (x) = Φ(1−x−x2 −x3 +x4 −x5 +x6 −x7 −x8 +x9
19
+x10 −x11 −x12 −x13 −x14 −x15 +x16 −x17 −x18)?
Find a fast computer assisted proof for Theorem 1.2 to answerthe above
question.
For proving that F (x) is Apwenian, our C program has taken
17a
about one week by using 24 CPU cores. No hope for F (x).
19
Problem 1.6. Find a human proof of Theorem 1.2 without computer
assistance.
Problem 1.7. Characterize all the finite ±1-sequences v such that
Φ(v˜(x)) is Apwenian.
As an application of Theorem 1.2 in Number Theory, the irrational-
ity exponents of F (1/b),F (1/b),F (1/b),F (1/b) are proved to be
5 11 17a 17b
equal to 2 (see [4]).
COMPUTER ASSISTED PROOF FOR APWENIAN SEQUENCES 5
2. Proof of Theorem 1.2
Let d be a positive integer and v = (v ,v ,v ,...,v ) be a finite
0 1 2 d−1
±1-sequence of length d with v = 1. Let f = (f ) be the ±1-
0 k k≥0
sequence defined by the following generating function
∞
(2.1) ˜f(x) = Φ(v˜(x)) = v˜(xdk),
k=0
Y
where v˜(x) = d−1v xi. The above power series satisfies the following
i=0 i
functional equation
P
∞
(2.2) ˜f(x) = v˜(x) v˜ xdk = v˜(x)˜f(xd).
kY=1 (cid:16) (cid:17)
The sequence f can also be defined by the recurrence relations
(2.3) f = 1, f = v f for n ≥ 0 and 0 ≤ i ≤ d−1.
0 dn+i i n
We divide the set {1,2,...,d−1} into two disjoint subsets
P = {1 ≤ i ≤ d−1 | v 6= v },
i−1 i
Q = {1 ≤ i ≤ d−1 | v = v }.
i−1 i
Two disjoint infinite sets of integers J and K play an important in the
proof of Theorem 1.2.
Definition 2.1. If v = −1, define
d−1
J = {(dn+p)d2k −1 | n,k ∈ N,p ∈ P}
{(dn+q)d2k+1 −1 | n,k ∈ N,q ∈ Q},
K = {([dn+q)d2k −1 | n,k ∈ N,q ∈ Q}
{(dn+p)d2k+1 −1 | n,k ∈ N,p ∈ P}.
If vd−1 = 1, defin[e
J = {(dn+p)dk −1 | n,k ∈ N,p ∈ P},
K = {(dn+q)dk −1 | n,k ∈ N,q ∈ Q}.
From the above definition it is easy to see that N = J ∪K.
Lemma 2.1. For each t ≥ 0 the integer δ := |(f −f )/2| is equal
t t t+1
to 1 if and only if t is in J.
Proof. Let t = (dn+ℓ)dk −1. By (2.3) we have
f = f = v f ,
(dn+ℓ)dk−1 d[(dn+ℓ)dk−1−1]+(d−1) d−1 (dn+ℓ)dk−1−1
and
f = v f = ··· = vk f = vk v f .
(dn+ℓ)dk−1 d−1 (dn+ℓ)dk−1−1 d−1 dn+ℓ−1 d−1 ℓ−1 n
In the same manner,
f(dn+ℓ)dk = f(dn+ℓ)dk−1 = ··· = fdn+ℓ = vℓfn.
6 HAOFU AND GUO-NIUHAN
Hence,
1 1
δ = (f −f ) = (v −vk v )
t 2 (dn+ℓ)dk−1 (dn+ℓ)dk 2 ℓ d−1 ℓ−1
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
is odd if and o(cid:12)nly if (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(2.4) vk ·v v = −1.
d−1 ℓ−1 ℓ
There are two cases are to be considered: (i) If v = 1, condition
d−1
(2.4) is equivalent to v v = −1, or ℓ ∈ P, or t ∈ J by Definition 2.1.
ℓ−1 ℓ
(ii) If v = −1, condition (2.4) becomes v v = (−1)k+1, which is
d−1 ℓ−1 ℓ
equivalent to ℓ ∈ P when k is even and ℓ ∈ Q when k is odd. In other
words, t ∈ J. (cid:3)
LetS = S bethesetofallpermutationson{0,1,...,m−
m {0,1,...,m−1}
1}. The following Theorem may be viewed as the combinatorial inter-
pretation of Theorem 1.2.
Theorem 2.2. Let v be a ±1-sequence of length d with v = 1. The
0
sequence f and the set J associated with v are defined by (2.1) and
Definition 2.1 respectively. Then, the sequence f is Apwenian if, and
only if, the number of permutations σ ∈ S such that i+σ(i) ∈ J for
m
i = 0,1,...,m−2 (no constraint on m−1+σ(m−1) ∈ N) is an odd
integer for every integer m ≥ 1.
Proof. Let m be a positive integer. By means of elementary transfor-
mations the Hankel determinant H (f) is equal to
m
f f ··· f
0 1 m−1
f f ··· f
Hm(f) = (cid:12)(cid:12) ...1 ...2 ... ...m (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12)f f ··· f (cid:12)
(cid:12) m−1 m 2m−2(cid:12)
(cid:12) (cid:12)
(cid:12) f0 −f1 f1 −f2 (cid:12) ··· fm−2 −fm−1 fm−1
(cid:12) (cid:12)
f −f f −f ··· f −f f
1 2 2 3 m−1 m m
= (cid:12)(cid:12) ... ... ... ... ... (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12)f −f f −f ··· f −f f (cid:12)
(cid:12) m−1 m m m+1 2m−3 2m−2 2m−2(cid:12)
(cid:12) (cid:12)
(cid:12) f0−f1 f1−f2 ··· fm−2−fm−1 f (cid:12)
(cid:12) 2 2 2 m−1 (cid:12)
f1−f2 f2−f3 ··· fm−1−fm f
= 2m−1 ×(cid:12)(cid:12)(cid:12) 2... 2... ... 2... ...m (cid:12)(cid:12)(cid:12).
(cid:12) (cid:12)
(cid:12)fm−1−fm fm−fm+1 ··· f2m−3−f2m−2 f (cid:12)
(cid:12) 2 2 2 2m−2(cid:12)
(cid:12) (cid:12)
By Lemma 2.1, we h(cid:12)ave (cid:12)
(cid:12) (cid:12)
δ δ ··· δ 1
0 1 m−2
Hm(f) δ1 δ2 ··· δm−1 1
(2.5) 2m−1 ≡ (cid:12)(cid:12) ... ... ... ... ...(cid:12)(cid:12) (mod 2).
(cid:12) (cid:12)
(cid:12)δ δ ··· δ 1(cid:12)
(cid:12) m−1 m 2m−3 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
COMPUTER ASSISTED PROOF FOR APWENIAN SEQUENCES 7
By the very definition of a determinant or the Leibniz formula, the
determinant occurring on the right-hand side of the congruence (2.5)
is equal to
(2.6) S := (−1)inv(σ)δ δ ···δ ,
m 0+σ0 1+σ1 m−2+σm−2
σX∈Sm
where inv(σ) is the number of inversions of the permutation σ. By
Lemma 2.1 the product δ δ ···δ is equal to 1 if i+σ ∈
0+σ0 1+σ1 m−2+σm−2 i
J for i = 0,1,...,m− 2, and to 0 otherwise. Hence, the summation
S is congruent modulo 2 to the number of permutations σ ∈ S such
m m
that i+σ ∈ J for all i = 0,1,...,m−2. Hence, f is Apwenian if and
i
only if the number of permutations σ ∈ S such that i+σ(i) ∈ J for
m
i = 0,1,...,m−2 (no constraint on m−1+σ(m−1) ∈ N) is an odd
integer for every integer m ≥ 1. (cid:3)
For proving that the sequence f is Apwenian by means of Theo-
rem 2.2, it is convenient to introduce the following notations.
Definition 2.2. For m ≥ ℓ ≥ 0 let J (resp. K ) be the set of
m,ℓ m,ℓ
all permutations σ = σ σ ···σ ∈ S such that i + σ ∈ J (resp.
0 1 m−1 m i
i+ σ ∈ K) for i ∈ {0,1,...,m−1}\ {ℓ}. Let n ≥ 1; for simplicity,
i
write:
j := #J , k := #K ,
m,ℓ m,ℓ m,ℓ m,ℓ
n−1
X := j , Y := j , Z := j ,
n n,i n n,n n n,n−1
i=0
X
n−1
U := k , V := k , W := k ,
n n,i n n,n n n,n−1
i=0
X
T := X +X Y +Y ,
n n n n n
R := U +U V +V .
n n n n n
Notice that if ℓ = m, then {0,1,...,m−1}\{ℓ} = {0,1,...,m−1},
so that j (resp. k ) is the number of permutations σ ∈ S such
m,m m,m m
that i+σ(i) ∈ J (resp. ∈ K) for all i.
By Theorem 2.2 and Definition 2.2 the sequence f is Apwenian if
and only if Z ≡ 1 (mod 2). In Section 4 we describe an algorithm
n
enabling us to find and also prove a list of recurrence relations between
X ,Y ,X ,U ,V ,W . Then, it is routine to check whether Z ≡ 1
n n n n n n n
(mod 2) or not. Our program Apwen.py is an implementation of the
latter algorithm in Python.
We now produce the proof of Theorem 1.2 by means of the program
Apwen.py. Since F (x) has been proved to be Apwenian in [1], only the
2
three power series F (x),F (x) and F (x) require our attention. We
3 5 11
can also prove that F (x),F (x),F (x) are Apwenian in the same
13 17a 17b
8 HAOFU AND GUO-NIUHAN
manner. However, the full proofs are lengthy and are not reproduced
in the paper.
2.1. F (x) is Apwenian. Take v = (1,−1,−1) with d = 3 and v =
3 d−1
−1. Then, the corresponding infinite ±1-sequence f is equal to F (x).
3
We have P = {1}, Q = {2} and
J = {(3n+1)32k −1 | n,k ∈ N}∪{(3n+2)32k+1 −1 | n,k ∈ N}
= {0,3,5,6,8,9,12,14,15,18,...},
K = {(3n+2)32k −1 | n,k ∈ N}∪{(3n+1)32k+1 −1 | n,k ∈ N}
= {1,2,4,7,10,11,13,16,17,...} = N \J.
By enumerating a list of 24 types of permutations (see Section 3), the
program Apwen.py finds and proves the following recurrences.
Lemma 2.3. For each n ≥ 1 we have
X ≡ U , Y ≡ U +V ,
3n+0 n 3n+0 n n
X ≡ W (U +V ), Y ≡ W V ,
3n+1 n+1 n n 3n+1 n+1 n
X ≡ W (U +V ), Y ≡ W V ,
3n+2 n+1 n+1 n+1 3n+2 n+1 n+1
Z ≡ W (U +U V +V ),
3n+0 n n n n n
Z ≡ W (U +U V +V ),
3n+1 n+1 n n n n
Z ≡ W .
3n+2 n+1
As explained in Section 3, the above relations express X,Y,Z in
function of U,V,W since v = −1. By exchanging the values of P
d−1
and Q, J and K, the program Apwen.py yields other relations which
express U,V,W in terms of X,Y,Z by enumerating a list of 26 types
of permutations.
Lemma 2.4. For each n ≥ 1 we have
U ≡ X , V ≡ X +Y ,
3n+0 n 3n+0 n n
U ≡ Z Y , V ≡ Z X ,
3n+1 n+1 n 3n+1 n+1 n
U ≡ Z Y , V ≡ Z X ,
3n+2 n+1 n+1 3n+2 n+1 n+1
W ≡ Z (X +X Y +Y ),
3n+0 n n n n n
W ≡ Z (X +X Y +Y ),
3n+1 n+1 n n n n
W ≡ Z .
3n+2 n+1
From Lemmas 2.3 and 2.4 we obtain the following “simplified” recur-
rence relations based on some elementary calculations.
Corollary 2.5. For each positive integer n we have
Z ≡ W R , W ≡ Z T ,
3n+0 n n 3n+0 n n
COMPUTER ASSISTED PROOF FOR APWENIAN SEQUENCES 9
Z ≡ W R , W ≡ Z T ,
3n+1 n+1 n 3n+1 n+1 n
Z ≡ W , W ≡ Z ,
3n+2 n+1 3n+2 n+1
T ≡ R , R ≡ T ,
3n+0 n 3n+0 n
T ≡ W R , R ≡ Z T ,
3n+1 n+1 n 3n+1 n+1 n
T ≡ W R , R ≡ Z T .
3n+2 n+1 n+1 3n+2 n+1 n+1
Since Z = 1,T = 3,W = 1,R = 1,Z = 1,T = 1,W = 1 and
1 1 1 1 2 2 2
R = 7, Corollary 2.5 yields Z ≡ T ≡ W ≡ R ≡ 1 (mod 2) for
2 m m m m
every positive integer m by induction. Hence, F (x) is Apwenian.
3
2.2. F (x) is Apwenian. Take v = (1,−1,−1,−1,1) with d = 5 and
5
v = 1. Then, the corresponding infinite ±1-sequence f is equal to
d−1
F (x). We have
5
P = {1,4},
Q = {2,3},
J = {(5n+1)5k −1 | n,k ∈ N}∪{(5n+4)5k −1 | n,k ∈ N}
= {0,3,4,5,8,10,13,15,18,19,20,23,24,25,28,29,30,33,...},
K = {(5n+2)5k −1 | n,k ∈ N}∪{(5n+3)5k −1 | n,k ∈ N}
= {1,2,6,7,9,11,12,14,16,17,21,22,26,27,31,32,34,36,...}.
Byenumeratingalistof225typesofpermutations,thePythonprogram
Apwen.py finds and proves the following recurrences.
Lemma 2.6. For each n ≥ 1 we have
X ≡ X , Y ≡ Y ,
5n+0 n 5n+0 n
X ≡ Z Y , Y ≡ Z (X +Y ),
5n+1 n+1 n 5n+1 n+1 n n
X ≡ Z (X +Y ), Y ≡ Z X ,
5n+2 n+1 n n 5n+2 n+1 n
X ≡ Z (X +Y ), Y ≡ Z X ,
5n+3 n+1 n+1 n+1 5n+3 n+1 n+1
X ≡ Z Y , Y ≡ Z (X +Y ),
5n+4 n+1 n+1 5n+4 n+1 n+1 n+1
Z ≡ Z (X +X Y +Y ),
5n+0 n n n n n
Z ≡ Z (X +X Y +Y ),
5n+1 n+1 n n n n
Z ≡ Z (X +X Y +Y ),
5n+2 n+1 n n n n
Z ≡ Z ,
5n+3 n+1
Z ≡ Z (X +X Y +Y ).
5n+4 n+1 n+1 n+1 n+1 n+1
As explained in Section 4, the above relations are between X,Y,Z
without involving U,V,W, since v = 1. We obtain the following
d−1
“simplified” recurrencerelationsbasedonsomeelementarycalculations.
Corollary 2.7. For each positive integer n we have
Z ≡ Z T , T ≡ T ,
5n+0 n n 5n+0 n
10 HAOFU AND GUO-NIUHAN
Z ≡ Z T , T ≡ Z T ,
5n+1 n+1 n 5n+1 n+1 n
Z ≡ Z T , T ≡ Z T ,
5n+2 n+1 n 5n+2 n+1 n
Z ≡ Z , T ≡ Z T ,
5n+3 n+1 5n+3 n+1 n+1
Z ≡ Z T , T ≡ Z T .
5n+4 n+1 n+1 5n+4 n+1 n+1
Since Z = 1,T = 3,Z = 1,T = 1,Z = 1,T = 9,Z = 5,T =
1 1 2 2 3 3 4 4
129, Corollary 2.7 yields Z ≡ T ≡ 1 (mod 2) for each n ≥ 1 by
n n
induction. Hence, F (x) is Apwenian.
5
2.3. F (x) is Apwenian. Take
11
v = (1,−1,−1,1,−1,1,1,1,1,−1,−1)
with d = 11 and v = −1. Then, the corresponding infinite ±1-
d−1
sequence f is equal to F (x). We have
11
P = {1,3,4,5,9},
Q = {2,6,7,8,10},
J = {0,2,3,4,8,11,13,14,15,19,21,22,24,25,26,30,33,35,...},
K = {1,5,6,7,9,10,12,16,17,18,20,23,27,28,29,31,32,34,...}.
By enumerating a list of 2274558 types of permutations, the program
Apwen.py finds and proves the following recurrences.
Lemma 2.8. For each n ≥ 1 we have
X ≡ U , Y ≡ U +V ,
11n+0 n 11n+0 n n
X ≡ W (V +U ), Y ≡ V W ,
11n+1 n+1 n n 11n+1 n n+1
X ≡ U W , Y ≡ W (V +U ),
11n+2 n n+1 11n+2 n+1 n n
X ≡ W (V +U ), Y ≡ V W ,
11n+3 n+1 n n 11n+3 n n+1
X ≡ V W , Y ≡ U W ,
11n+4 n n+1 11n+4 n n+1
X ≡ U W , Y ≡ W (V +U ),
11n+5 n n+1 11n+5 n+1 n n
X ≡ U W , Y ≡ W (U +V ),
11n+6 n+1 n+1 11n+6 n+1 n+1 n+1
X ≡ V W , Y ≡ U W ,
11n+7 n+1 n+1 11n+7 n+1 n+1
X ≡ W (U +V ), Y ≡ V W ,
11n+8 n+1 n+1 n+1 11n+8 n+1 n+1
X ≡ U W , Y ≡ W (U +V ),
11n+9 n+1 n+1 11n+9 n+1 n+1 n+1
X ≡ W (U +V ), Y ≡ V W ,
11n+10 n+1 n+1 n+1 11n+10 n+1 n+1
Z ≡ W (V +U +U V ),
11n+0 n n n n n
Z ≡ W (U V +V +U ), (i = 1,2,3,4,5)
11n+i n+1 n n n n
Z ≡ W ,
11n+6 n+1
Z ≡ W (U +U V +V ). (i = 7,8,9,10)
11n+i n+1 n+1 n+1 n+1 n+1
As explained in Section 4, the above relations express X,Y,Z in
function of U,V,W, since v = −1. By exchanging the values of P
d−1