Last Cases of Dejean's Conjecture - Semantic Scholar
Last Cases of Dejean's Conjecture
Michaël Rao CNRS - LaBRI - Université Bordeaux 1, France
[email protected] http://www.labri.fr/rao
Michaël Rao
Last Cases of Dejean's Conjecture
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Dejean's Conjecture
k)
Let RT(
be the smallest
word over a
k -letter
x
such that there is an in nite
k ≥ 2).
alphabet (
Conjecture (Dejean's conjecture, 1972)
RT(
k) =
7 4
7
5 k
k−
Michaël Rao
1
if k = 3 if k = 4 otherwise.
Last Cases of Dejean's Conjecture
x + -free
Dejean's Conjecture
Already proved for:
k =2 k =3 k =4 5 ≤ k ≤ 11 12 ≤ k ≤ 14 k ≥ 33 k ≥ 27 8 ≤ k ≤ 38 15 ≤ k ≤ 26
[Thue 1906] [Dejean 1972] [Pansiot 1984] [Moulin Ollagnier 1992] [Currie, Mohammad-Noori 2004] [Carpi 2007] [Currie, Rampersad 2008,2009] [R. 2009] [Currie, Rampersad 2009]
Michaël Rao
Last Cases of Dejean's Conjecture
Dejean's Conjecture
Already proved for:
k =2 k =3 k =4 5 ≤ k ≤ 11 12 ≤ k ≤ 14 k ≥ 33 k ≥ 27 8 ≤ k ≤ 38 15 ≤ k ≤ 26
[Thue 1906] [Dejean 1972] [Pansiot 1984] [Moulin Ollagnier 1992] [Currie, Mohammad-Noori 2004] [Carpi 2007] [Currie, Rampersad 2008,2009] [R. 2009] [Currie, Rampersad 2009]
Michaël Rao
Last Cases of Dejean's Conjecture
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
Let
Mk ()
i.e.
Pk (Mk (w )) = w
Let
w4
be inverse of
Pk ().
(s.t.
Mk (w )[ ..k − 1
for every binary word
be a xed point of
] = 1 ...
1
w.
h4 : 0 → 101101, 1 → 10.
Theorem (Pansiot 1984)
M4 (w4 ) is
7 5
+
-free.
Michaël Rao
Last Cases of Dejean's Conjecture
k−
1)
Pansiot's Coding
Let
Mk ()
i.e.
Pk (Mk (w )) = w
Let
w4
be inverse of
Pk ().
(s.t.
Mk (w )[ ..k − 1
for every binary word
be a xed point of
] = 1 ...
1
w.
h4 : 0 → 101101, 1 → 10.
Theorem (Pansiot 1984)
M4 (w4 ) is
7 5
+
-free.
Michaël Rao
Last Cases of Dejean's Conjecture
k−
1)
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
A picture...
|u | = k − 1.
w:
u
u
z }| {
z }| {
Pk (w ): | Then
v
{z
Ψ(v ) = Idk .
Michaël Rao
Last Cases of Dejean's Conjecture
}
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
To prove Dejean's conjecture for
Mk (w ) w where
nd a morphism
has no forbidden short repetition,
has no
w
k ≥ 5,
Ψ-kernel
repetition
is a xed point of
(p , q )
with
|
pq|+k − |p |
1
h
>
h.
Michaël Rao
Last Cases of Dejean's Conjecture
s.t.:
k k−
1
.
Moulin Ollagnier's ideas
Idea: Limit us to
h
such that:
Ψ(h(0)) = σ · Ψ(0) · σ −1 Ψ(h(1)) = σ · Ψ(1) · for a
and
σ −1 ,
σ ∈ Sk .
Lemma (Moulin Ollagnier)
Let (p , q ) be a Ψ-kernel repetition in w . If q is long enough, then (p , q ) is an image by h of a smaller Ψ-kernel repetition in w .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Idea: Limit us to
h
such that:
Ψ(h(0)) = σ · Ψ(0) · σ −1 Ψ(h(1)) = σ · Ψ(1) · for a
and
σ −1 ,
σ ∈ Sk .
Lemma (Moulin Ollagnier)
Let (p , q ) be a Ψ-kernel repetition in w . If q is long enough, then (p , q ) is an image by h of a smaller Ψ-kernel repetition in w .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's results
Let
w
be a xed point of
Checking if check
h.
k -free is decidable: Mk (w ) is k − 1 if Mk (w ) has no small forbidden repetition, +
check if iterated images of small kernel repetitions in
w
not forbidden.
Moulin Ollagnier gives morphisms
Mk (wk ) →
is
k
k−
+
1
-free (where
wk
for 5
≤ k ≤ 11
is a xed point of
Dejean's conjecture holds for 5
Michaël Rao
hk
≤ k ≤ 11.
such that
hk ).
Last Cases of Dejean's Conjecture
are
Moulin Ollagnier's results
Let
w
be a xed point of
Checking if check
h.
k -free is decidable: Mk (w ) is k − 1 if Mk (w ) has no small forbidden repetition, +
check if iterated images of small kernel repetitions in
w
not forbidden.
Moulin Ollagnier gives morphisms
Mk (wk ) →
is
k
k−
+
1
-free (where
wk
for 5
≤ k ≤ 11
is a xed point of
Dejean's conjecture holds for 5
Michaël Rao
hk
≤ k ≤ 11.
such that
hk ).
Last Cases of Dejean's Conjecture
are
Moulin Ollagnier's morphisms
h5 : h6 : h7 : h8 : h9 : h10 : h11 :
( 0 → 010101101101010110110 1 → 101010101101101101101 ( 0 → 010101101101011010110 1 → 101011010110110101101 ( 0 → 0110110110110101101101101010 1 → 1010110110110101101101101101 ( 0 → 1011010101101011010110101010 1 → 1011010101011011011010101101 ( 0 → 101011010110110101011010110110101010 1 → 101010110110110101011011010110101101 ( 0 → 1010101011011011011010101011011011010110 1 → 1010101011011011010110101010101011010101 ( 0 → 1010101010101101101011010110101010110110 1 → 1010101010101011011011011011011010101101
Michaël Rao
Last Cases of Dejean's Conjecture
New results
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Let
σa = Ψ(h(a))
Let
Ψ0 : {a, b}∗ → Sk
and
σb = Ψ(h(b)), s.t.
Ψ0 (a) = σa
and
Ψ0 (b) = σb .
Idea behind this:
h such that: h(wTM ) has to avoid forbidden short Ψ-kernel repetitions wTM has to avoid forbidden long Ψ0 -kernel repetitions.
Find a
We obtain results for smaller
Michaël Rao
h.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Results
Theorem
For every k ∈ {8, . . . 38}, there is a uniform morphism hk such that k + -free. Mk (hk (wTM )) is k − 1 Corollary
Dejean's conjecture holds for
8
≤ k ≤ 38.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
Example:
k = 18.
(
h18 :
Size of
a → 10101101010110101101010110110101011010110 b → 10101011010110101101011010110101011010101.
hk (x )
From 30 (for
k = 8)
up to 74 (for
k = 38).
Computation time on a 2.4 Ghz processor Few seconds/minutes for small Couple of hours for
k.
k = 38.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Morphisms
h4 :
h8 : h9 : h10 : h11 : h12 : h13 : h14 : h15 : h16 :
8 a→ > > >
b → 1011010101101101011011010101101010110110 . . . > > : . . . 1010110110101011011010110110101011010101 ( a → 101101101101010101011010101101 b → 101101010101011011011011011010 ( a → 1011011010101010101011011010110101 b → 1010110110101010101101101011010110 ( a → 10101101010101101010101101010110101101 b → 10101101010110101101101011010101010110 ( a → 101010110101101010110101101011010110101010110110 b → 101011010101101011010110101011010110101010110101 ( a → 1011010110110110101010101011010110110110110 b → 1011010101010110101101101101011010110101101 ( a → 10101101010110101010110101101010101011011010101 b → 10101011011010101011010101101101010101011010110 ( a → 1010110110101101101011010110101011010110110110 b → 1011010110101101101011010110101011010110110101 ( a → 101101101101010110101010101101011011011011010101010110 b → 101011011011010110110101010101011011010110110101010101 ( a → 101010110101101101011010110101101101011010110101011011010110 b → 101010110110101101011010101101101101011011010101011011010101 1011010101101101011011010101101101010110
Michaël Rao
Last Cases of Dejean's Conjecture
(
a → 10110101101101010110110110110101010110110110101 b → 10110101101101010110110110101101010110110110110
(
a → 10101101010110101101010110110101011010110 b → 10101011010110101101011010110101011010101
(
a → 101010110110110101101011010101010110101011010110110110 b → 101010110110110101101011010110110110101010110101010101
(
a → 101011010101011010110110101010110101101011011011010110101 b → 101011010101011011010110101010110101011011011011010110110
(
a → 1010101101010110110110110110101101011010101010101 b → 1010110101010110110110110101101101011010101010110
(
a → 10101101011010110110110101101011010101101010110 b → 10101101011010110110110110101011010101101010101
(
a → 10101010101101101101101010110110110101010101 b → 10101010101101101101011010110110110101010110
(
a → 1010101010101011011011010101010110110101010101 b → 1010101010101011011010110101010110110101010110
(
a → 10101011010101101011010101101010101101010110101 b → 10101011010101101010110101101010101101010110110
(
a → 101101010101010101010101101101101010101101101101 b → 101101010101010101101101101101101010101101101010
(
a → 1010101010101101101011011010101010101011011011010101 b → 1010101010101101101011010110101010101011011011010110
(
a → 101010101010101011010110110101010101010101011011010101 b → 101010101010101011010110101101010101010101011011010110
h17 : h18 : h19 : h20 : h21 : h22 : h23 : h24 : h25 : h26 : h27 : h28 :
Michaël Rao
Last Cases of Dejean's Conjecture
(
a → 101010110101010101101010101010101101010101011010101101 b → 101010110101010101101101101010101101010101011010101010
(
a → 1010101010101011011011011011010101010110110110110101010101 b → 1010101010101011011011011010110101010110110110110101010110
(
a → 10110110110110110110101011010110110101101101010110110110101 b → 10110110110110110110101010110110110101101101010110110110110
(
a → 101101011011011010101101010101101101011011010101011010101101 b → 101101011011011010101101101101101101011011010101011010101010
(
a → 1010101010101011010101101011011010101010101010101101010110110101 b → 1010101010101011010101101011010110101010101010101101010110110110
(
a → 101011011010101010101010101010110101101010110101010101010101010101 b → 101011011010101010101010101010101101101010110101010101010101010110
h29 : h30 : h31 : h32 : h33 : h34 : . . .
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
If the Pansiot's code of
w
If the Pansiot's code of
w
w
has a 0 at position
has a letter with frequency
w
k− 1
1
has a 0 at position
has a letter with frequency
k+ 1
1
i (mod k − 1),
then
i (mod k + 1),
then
.
.
Theorem
For every
9
≤ k ≤ 38, Ochem's conjecture holds.
(Does not work for
k = 8.)
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
If the Pansiot's code of
w
If the Pansiot's code of
w
w
has a 0 at position
has a letter with frequency
w
k− 1
1
has a 0 at position
has a letter with frequency
k+ 1
1
i (mod k − 1),
then
i (mod k + 1),
then
.
.
Theorem
For every
9
≤ k ≤ 38, Ochem's conjecture holds.
(Does not work for
k = 8.)
Michaël Rao
Last Cases of Dejean's Conjecture
Further Researchs
Generalized Repetition Threshold
The
growth rate
of
L
is
(next two talks).
g (L) = limn→∞
p n |L ∩ Σn |.
Question
Compute g (Dk ) for k ≥ 3. Good lower and upper bounds by Kolpakov and Shur. E.g.: 1.245
≤ g (D3 ) ≤ 1.2456148
[Kolpakov 06, Shur 08].
Michaël Rao
Last Cases of Dejean's Conjecture
Further Researchs
Generalized Repetition Threshold
The
growth rate
of
L
is
(next two talks).
g (L) = limn→∞
p n |L ∩ Σn |.
Question
Compute g (Dk ) for k ≥ 3. Good lower and upper bounds by Kolpakov and Shur. E.g.: 1.245
≤ g (D3 ) ≤ 1.2456148
[Kolpakov 06, Shur 08].
Michaël Rao
Last Cases of Dejean's Conjecture
Thank you !
Michaël Rao
Last Cases of Dejean's Conjecture
Michaël Rao CNRS - LaBRI - Université Bordeaux 1, France
[email protected] http://www.labri.fr/rao
Michaël Rao
Last Cases of Dejean's Conjecture
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Repetitions
(p , q )
is a
repetition
in a word
w
if :
pq is a factor of w , p 6= and q is a pre x of pq . The
exponent
of the repetition is
pq p
| | | | .
Squares are repetitions of exponent 2. A word is said
x-free
x + -free) if it does not contain y with y ≥ x (resp. y > x ).
(resp.
repetition of exponent
Michaël Rao
Last Cases of Dejean's Conjecture
a
Dejean's Conjecture
k)
Let RT(
be the smallest
word over a
k -letter
x
such that there is an in nite
k ≥ 2).
alphabet (
Conjecture (Dejean's conjecture, 1972)
RT(
k) =
7 4
7
5 k
k−
Michaël Rao
1
if k = 3 if k = 4 otherwise.
Last Cases of Dejean's Conjecture
x + -free
Dejean's Conjecture
Already proved for:
k =2 k =3 k =4 5 ≤ k ≤ 11 12 ≤ k ≤ 14 k ≥ 33 k ≥ 27 8 ≤ k ≤ 38 15 ≤ k ≤ 26
[Thue 1906] [Dejean 1972] [Pansiot 1984] [Moulin Ollagnier 1992] [Currie, Mohammad-Noori 2004] [Carpi 2007] [Currie, Rampersad 2008,2009] [R. 2009] [Currie, Rampersad 2009]
Michaël Rao
Last Cases of Dejean's Conjecture
Dejean's Conjecture
Already proved for:
k =2 k =3 k =4 5 ≤ k ≤ 11 12 ≤ k ≤ 14 k ≥ 33 k ≥ 27 8 ≤ k ≤ 38 15 ≤ k ≤ 26
[Thue 1906] [Dejean 1972] [Pansiot 1984] [Moulin Ollagnier 1992] [Currie, Mohammad-Noori 2004] [Carpi 2007] [Currie, Rampersad 2008,2009] [R. 2009] [Currie, Rampersad 2009]
Michaël Rao
Last Cases of Dejean's Conjecture
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
k=2
and
k=3
Theorem (Thue 1906)
Thue-Morse word (i.e. xed point of
0
→ 01,
1
→ 10) is
2
f (a) = abcacbcabcbacbcacba f (b) = bcabacabcacbacabacb f (c ) = cabcbabcabacbabcbac Theorem (Dejean 1972)
A xed point of f is
7 4
+
-free.
Theorem (Brandenburg 1983)
Fixed point method does not work for k ≥ 4. Michaël Rao
Last Cases of Dejean's Conjecture
+ -free.
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
−1 -free, then k -letter alphabet is kk − 2 k − 1 consists of k − 1 dierent letters.
If a word length
→w
w
on a
can be encoded by a binary word
Pk (w )[i ] = w P6 (w ) Remark: If
= =
w
(
1
0
if
1
if
2
3
every factor of
Pk (w ):
w [i + k − 1] = w [i ] w [i + k − 1] 6∈ {w [i ], . . . , w [i + k − 2]}. 4
5
1
6
3
2
4
1
5
0
1
0
1
1
0
1
validates Dejean's conjecture then
Michaël Rao
Pk (w )
is
... ...
{00, 111}-free.
Last Cases of Dejean's Conjecture
Pansiot's Coding
Let
Mk ()
i.e.
Pk (Mk (w )) = w
Let
w4
be inverse of
Pk ().
(s.t.
Mk (w )[ ..k − 1
for every binary word
be a xed point of
] = 1 ...
1
w.
h4 : 0 → 101101, 1 → 10.
Theorem (Pansiot 1984)
M4 (w4 ) is
7 5
+
-free.
Michaël Rao
Last Cases of Dejean's Conjecture
k−
1)
Pansiot's Coding
Let
Mk ()
i.e.
Pk (Mk (w )) = w
Let
w4
be inverse of
Pk ().
(s.t.
Mk (w )[ ..k − 1
for every binary word
be a xed point of
] = 1 ...
1
w.
h4 : 0 → 101101, 1 → 10.
Theorem (Pansiot 1984)
M4 (w4 ) is
7 5
+
-free.
Michaël Rao
Last Cases of Dejean's Conjecture
k−
1)
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Pansiot's coding can also be viewed by the way of an action on the symmetric group Let
Ψ
be the morphism between
Ψ(0) = (1 For all
Sk :
i ≥0
2...
and 1
k − 1)
and
{0, 1}∗
and
Ψ(1) = (1
Sk
such that:
2...
k − 1 k)
≤ j ≤ k − 1, Mk (w )[i + j ] = Ψ(w [1..i ])(j ).
Mk (w )[i .. i + k − 1] = Mk (w )[j .. j + k − 1] Ψ(w [i + 1 .. j ]) = Idk .
(for
j > i)
i
Michaël Rao
Last Cases of Dejean's Conjecture
A picture...
|u | = k − 1.
w:
u
u
z }| {
z }| {
Pk (w ): | Then
v
{z
Ψ(v ) = Idk .
Michaël Rao
Last Cases of Dejean's Conjecture
}
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
A repetition
(p , q )
(Note: if forbidden A repetition is a
is
short
⇒
if
|q | < k − 1.
bounded size.)
kernel repetition
if
|q | ≥ k − 1.
On Pansiot's codes:
(p , q ) is a Ψ-kernel repetition Ψ(p ) = Idk .
if
(p , q )
is a repetition, and
Lemma (Moulin Ollagnier)
Mk (w ) has a kernel repetition (p , q ) ⇔ w has a Ψ-kernel repetition (p 0 , q 0 ) with |p | = |p 0 | and |q 0 | = |q | − k + 1 .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
To prove Dejean's conjecture for
Mk (w ) w where
nd a morphism
has no forbidden short repetition,
has no
w
k ≥ 5,
Ψ-kernel
repetition
is a xed point of
(p , q )
with
|
pq|+k − |p |
1
h
>
h.
Michaël Rao
Last Cases of Dejean's Conjecture
s.t.:
k k−
1
.
Moulin Ollagnier's ideas
Idea: Limit us to
h
such that:
Ψ(h(0)) = σ · Ψ(0) · σ −1 Ψ(h(1)) = σ · Ψ(1) · for a
and
σ −1 ,
σ ∈ Sk .
Lemma (Moulin Ollagnier)
Let (p , q ) be a Ψ-kernel repetition in w . If q is long enough, then (p , q ) is an image by h of a smaller Ψ-kernel repetition in w .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's ideas
Idea: Limit us to
h
such that:
Ψ(h(0)) = σ · Ψ(0) · σ −1 Ψ(h(1)) = σ · Ψ(1) · for a
and
σ −1 ,
σ ∈ Sk .
Lemma (Moulin Ollagnier)
Let (p , q ) be a Ψ-kernel repetition in w . If q is long enough, then (p , q ) is an image by h of a smaller Ψ-kernel repetition in w .
Michaël Rao
Last Cases of Dejean's Conjecture
Moulin Ollagnier's results
Let
w
be a xed point of
Checking if check
h.
k -free is decidable: Mk (w ) is k − 1 if Mk (w ) has no small forbidden repetition, +
check if iterated images of small kernel repetitions in
w
not forbidden.
Moulin Ollagnier gives morphisms
Mk (wk ) →
is
k
k−
+
1
-free (where
wk
for 5
≤ k ≤ 11
is a xed point of
Dejean's conjecture holds for 5
Michaël Rao
hk
≤ k ≤ 11.
such that
hk ).
Last Cases of Dejean's Conjecture
are
Moulin Ollagnier's results
Let
w
be a xed point of
Checking if check
h.
k -free is decidable: Mk (w ) is k − 1 if Mk (w ) has no small forbidden repetition, +
check if iterated images of small kernel repetitions in
w
not forbidden.
Moulin Ollagnier gives morphisms
Mk (wk ) →
is
k
k−
+
1
-free (where
wk
for 5
≤ k ≤ 11
is a xed point of
Dejean's conjecture holds for 5
Michaël Rao
hk
≤ k ≤ 11.
such that
hk ).
Last Cases of Dejean's Conjecture
are
Moulin Ollagnier's morphisms
h5 : h6 : h7 : h8 : h9 : h10 : h11 :
( 0 → 010101101101010110110 1 → 101010101101101101101 ( 0 → 010101101101011010110 1 → 101011010110110101101 ( 0 → 0110110110110101101101101010 1 → 1010110110110101101101101101 ( 0 → 1011010101101011010110101010 1 → 1011010101011011011010101101 ( 0 → 101011010110110101011010110110101010 1 → 101010110110110101011011010110101101 ( 0 → 1010101011011011011010101011011011010110 1 → 1010101011011011010110101010101011010101 ( 0 → 1010101010101101101011010110101010110110 1 → 1010101010101011011011011011011010101101
Michaël Rao
Last Cases of Dejean's Conjecture
New results
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Moulin Ollagnier ideas can be extended to HDOLs.
It is su cient to work of a special case of HDOLs: Image by a morphism of the Thue-Morse sequence.
Let
wTM
be the Thue-Morse sequence on
g : a → ab, b → ba.
i.e. xed point of Let
h : {a, b}∗ → {0, 1}∗
Question Is
Mk (h(wTM ))
k k−
1
+
a, b
be a morphism.
-free ?
Michaël Rao
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Let
σa = Ψ(h(a))
Let
Ψ0 : {a, b}∗ → Sk
and
σb = Ψ(h(b)), s.t.
Ψ0 (a) = σa
and
Ψ0 (b) = σb .
Idea behind this:
h such that: h(wTM ) has to avoid forbidden short Ψ-kernel repetitions wTM has to avoid forbidden long Ψ0 -kernel repetitions.
Find a
We obtain results for smaller
Michaël Rao
h.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Image of the Thue-Morse sequence
Following Moulin Ollagnier's idea: Limit us to
h
such that:
Ψ0 (g (a)) = σa · σb = σ · σa · σ −1 Ψ0 (g (b)) for a
= σb · σa = σ · σb ·
and
σ −1
σ ∈ Sk .
Remark:
σa
Moreover:
h
and is
σb
have to be conjugate.
uniform, synchronizing,
Michaël Rao
and the last letters dier.
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Let
(p , q ) If
q
be a
Ψ-kernel
is long enough, then
repetition in
Let
(p , q ) If
q
repetition of
be a
wTM .
Ψ0 -kernel
An image by
g
wTM .
of a
is an image by
repetition of
is long enough, then
repetition in
(p , q )
Ψ0 -kernel
(p , q )
h(wTM ). h
of a
Ψ0 -kernel
g
of a
Ψ0 -kernel
wTM .
is an image by
repetition has a smaller exponent.
This is decidable : Check only small
Ψ-kernel
Ψ0 -kernel repetitions in
repetitions in
wTM .
Michaël Rao
h(wTM )
and small
Last Cases of Dejean's Conjecture
Results
Theorem
For every k ∈ {8, . . . 38}, there is a uniform morphism hk such that k + -free. Mk (hk (wTM )) is k − 1 Corollary
Dejean's conjecture holds for
8
≤ k ≤ 38.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
Example:
k = 18.
(
h18 :
Size of
a → 10101101010110101101010110110101011010110 b → 10101011010110101101011010110101011010101.
hk (x )
From 30 (for
k = 8)
up to 74 (for
k = 38).
Computation time on a 2.4 Ghz processor Few seconds/minutes for small Couple of hours for
k.
k = 38.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Results
This technique does not work for
σa , σb , σ ∈ Sk
σa · σb = σ · σa · σ −1
wTM
σb · σa = σ · σb · has a
Works for
k ∈ {3, 5, 6, 7}
since for every
such that:
Ψ0 -kernel
and
σ −1 ,
k ).
repetition of exponent at least RT(
k = 4 (|h4 (x )| = 80).
Conjecture
For every k ≥ 8 there is a morphism hk such that Mk (hk (wTM )) is k + k −1 -free.
Michaël Rao
Last Cases of Dejean's Conjecture
Morphisms
h4 :
h8 : h9 : h10 : h11 : h12 : h13 : h14 : h15 : h16 :
8 a→ > > >
b → 1011010101101101011011010101101010110110 . . . > > : . . . 1010110110101011011010110110101011010101 ( a → 101101101101010101011010101101 b → 101101010101011011011011011010 ( a → 1011011010101010101011011010110101 b → 1010110110101010101101101011010110 ( a → 10101101010101101010101101010110101101 b → 10101101010110101101101011010101010110 ( a → 101010110101101010110101101011010110101010110110 b → 101011010101101011010110101011010110101010110101 ( a → 1011010110110110101010101011010110110110110 b → 1011010101010110101101101101011010110101101 ( a → 10101101010110101010110101101010101011011010101 b → 10101011011010101011010101101101010101011010110 ( a → 1010110110101101101011010110101011010110110110 b → 1011010110101101101011010110101011010110110101 ( a → 101101101101010110101010101101011011011011010101010110 b → 101011011011010110110101010101011011010110110101010101 ( a → 101010110101101101011010110101101101011010110101011011010110 b → 101010110110101101011010101101101101011011010101011011010101 1011010101101101011011010101101101010110
Michaël Rao
Last Cases of Dejean's Conjecture
(
a → 10110101101101010110110110110101010110110110101 b → 10110101101101010110110110101101010110110110110
(
a → 10101101010110101101010110110101011010110 b → 10101011010110101101011010110101011010101
(
a → 101010110110110101101011010101010110101011010110110110 b → 101010110110110101101011010110110110101010110101010101
(
a → 101011010101011010110110101010110101101011011011010110101 b → 101011010101011011010110101010110101011011011011010110110
(
a → 1010101101010110110110110110101101011010101010101 b → 1010110101010110110110110101101101011010101010110
(
a → 10101101011010110110110101101011010101101010110 b → 10101101011010110110110110101011010101101010101
(
a → 10101010101101101101101010110110110101010101 b → 10101010101101101101011010110110110101010110
(
a → 1010101010101011011011010101010110110101010101 b → 1010101010101011011010110101010110110101010110
(
a → 10101011010101101011010101101010101101010110101 b → 10101011010101101010110101101010101101010110110
(
a → 101101010101010101010101101101101010101101101101 b → 101101010101010101101101101101101010101101101010
(
a → 1010101010101101101011011010101010101011011011010101 b → 1010101010101101101011010110101010101011011011010110
(
a → 101010101010101011010110110101010101010101011011010101 b → 101010101010101011010110101101010101010101011011010110
h17 : h18 : h19 : h20 : h21 : h22 : h23 : h24 : h25 : h26 : h27 : h28 :
Michaël Rao
Last Cases of Dejean's Conjecture
(
a → 101010110101010101101010101010101101010101011010101101 b → 101010110101010101101101101010101101010101011010101010
(
a → 1010101010101011011011011011010101010110110110110101010101 b → 1010101010101011011011011010110101010110110110110101010110
(
a → 10110110110110110110101011010110110101101101010110110110101 b → 10110110110110110110101010110110110101101101010110110110110
(
a → 101101011011011010101101010101101101011011010101011010101101 b → 101101011011011010101101101101101101011011010101011010101010
(
a → 1010101010101011010101101011011010101010101010101101010110110101 b → 1010101010101011010101101011010110101010101010101101010110110110
(
a → 101011011010101010101010101010110101101010110101010101010101010101 b → 101011011010101010101010101010101101101010110101010101010101010110
h29 : h30 : h31 : h32 : h33 : h34 : . . .
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
In a Dejean word (with
k+ 1
1
and at most
k− 1
1
k ≥ 5),
each letter has frequency at least
.
Conjecture (Ochem 2005) (1)
(2)
For every k ≥ 5, there exists an in nite k-letter with letter frequency k +1 1 . For every k ≥ 6, there exists an in nite k-letter with letter frequency k −1 1 .
k k− k k−
+
-free word over
+
-free word over
1
1
Theorem (Chalopin, Ochem 2006)
(1) holds for k = 5 and (2) holds for k = 6.
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
If the Pansiot's code of
w
If the Pansiot's code of
w
w
has a 0 at position
has a letter with frequency
w
k− 1
1
has a 0 at position
has a letter with frequency
k+ 1
1
i (mod k − 1),
then
i (mod k + 1),
then
.
.
Theorem
For every
9
≤ k ≤ 38, Ochem's conjecture holds.
(Does not work for
k = 8.)
Michaël Rao
Last Cases of Dejean's Conjecture
Ochem's stronger conjecture
If the Pansiot's code of
w
If the Pansiot's code of
w
w
has a 0 at position
has a letter with frequency
w
k− 1
1
has a 0 at position
has a letter with frequency
k+ 1
1
i (mod k − 1),
then
i (mod k + 1),
then
.
.
Theorem
For every
9
≤ k ≤ 38, Ochem's conjecture holds.
(Does not work for
k = 8.)
Michaël Rao
Last Cases of Dejean's Conjecture
Further Researchs
Generalized Repetition Threshold
The
growth rate
of
L
is
(next two talks).
g (L) = limn→∞
p n |L ∩ Σn |.
Question
Compute g (Dk ) for k ≥ 3. Good lower and upper bounds by Kolpakov and Shur. E.g.: 1.245
≤ g (D3 ) ≤ 1.2456148
[Kolpakov 06, Shur 08].
Michaël Rao
Last Cases of Dejean's Conjecture
Further Researchs
Generalized Repetition Threshold
The
growth rate
of
L
is
(next two talks).
g (L) = limn→∞
p n |L ∩ Σn |.
Question
Compute g (Dk ) for k ≥ 3. Good lower and upper bounds by Kolpakov and Shur. E.g.: 1.245
≤ g (D3 ) ≤ 1.2456148
[Kolpakov 06, Shur 08].
Michaël Rao
Last Cases of Dejean's Conjecture
Thank you !
Michaël Rao
Last Cases of Dejean's Conjecture
