Prime Number Races
Youness Lamzouri
University of Illinois at Urbana-Champaign
2011 CMS Winter Meeting, Toronto
December 11, 2011
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
1 / 29
How many prime numbers are there up to x?
π(x) is the number of primes up to x.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
2 / 29
How many prime numbers are there up to x?
π(x) is the number of primes up to x.
Gauss conjectured in 1792 or 1793 (he was then 15 or 16), that the
primes occur with density 1/ log x at around x.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
2 / 29
How many prime numbers are there up to x?
π(x) is the number of primes up to x.
Gauss conjectured in 1792 or 1793 (he was then 15 or 16), that the
primes occur with density 1/ log x at around x. In other words
Z x
dt
.
π(x) ≈ Li(x) =
2 log t
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
2 / 29
How many prime numbers are there up to x?
π(x) is the number of primes up to x.
Gauss conjectured in 1792 or 1793 (he was then 15 or 16), that the
primes occur with density 1/ log x at around x. In other words
Z x
dt
.
π(x) ≈ Li(x) =
2 log t
Table: The number of primes up to x
x
108
1010
1012
1014
1016
1018
Youness Lamzouri (Illinois)
π(x)
5761455
455052511
37607912018
3204941750802
279238341033925
24739954287740860
Prime Number Races
Li(x) − π(x)
753
3103
38262
314889
3214631
21949554
December 11, 2011
2 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
Youness Lamzouri (Illinois)
p prime
Prime Number Races
December 11, 2011
3 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
p prime
ζ(s) has an analytic continuation to the complex plan except at s = 1
where it has a simple pole.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
3 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
p prime
ζ(s) has an analytic continuation to the complex plan except at s = 1
where it has a simple pole.
Notation: f (x) ∼ g (x) if limx→∞ f (x)/g (x) = 1.
Prime Number Theorem (Hadamard and de la Vallée Poussin, 1898)
π(x) ∼ Li(x)
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
3 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
p prime
ζ(s) has an analytic continuation to the complex plan except at s = 1
where it has a simple pole.
Notation: f (x) ∼ g (x) if limx→∞ f (x)/g (x) = 1.
Prime Number Theorem (Hadamard and de la Vallée Poussin, 1898)
π(x) ∼ Li(x)∼
Youness Lamzouri (Illinois)
x
.
log x
Prime Number Races
December 11, 2011
3 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
p prime
ζ(s) has an analytic continuation to the complex plan except at s = 1
where it has a simple pole.
Notation: f (x) ∼ g (x) if limx→∞ f (x)/g (x) = 1.
Prime Number Theorem (Hadamard and de la Vallée Poussin, 1898)
π(x) ∼ Li(x)∼
x
.
log x
The Riemann Hypothesis (1,000,000$)
1. If σ + it is a complex number with 0 ≤ σ ≤ 1 and ζ(σ + it) = 0, then
σ = 1/2.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
3 / 29
Riemann’s Memoir (1859): π(x) is connected with the zeros of the
“Riemann Zeta function” ζ(s), a function of a complex variable defined by
∞
X
Y 1
1 −1
for Re(s) > 1.
ζ(s) =
=
1− s
ns
p
n=1
p prime
ζ(s) has an analytic continuation to the complex plan except at s = 1
where it has a simple pole.
Notation: f (x) ∼ g (x) if limx→∞ f (x)/g (x) = 1.
Prime Number Theorem (Hadamard and de la Vallée Poussin, 1898)
π(x) ∼ Li(x)∼
x
.
log x
The Riemann Hypothesis (1,000,000$)
1. If σ + it is a complex number with 0 ≤ σ ≤ 1 and ζ(σ + it) = 0, then
σ = 1/2.
2.
π(x) = Li(x) + O x 1/2 log x .
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
3 / 29
Table: The number of primes up to x
x
108
1010
1012
1014
1016
1018
Youness Lamzouri (Illinois)
π(x)
5761455
455052511
37607912018
3204941750802
279238341033925
24739954287740860
Prime Number Races
Li(x) − π(x)
753
3103
38262
314889
3214631
21949554
December 11, 2011
4 / 29
Table: The number of primes up to x
x
108
1010
1012
1014
1016
1018
π(x)
5761455
455052511
37607912018
3204941750802
279238341033925
24739954287740860
Li(x) − π(x)
753
3103
38262
314889
3214631
21949554
For all x < 1023 we have π(x) < Li(x).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
4 / 29
Table: The number of primes up to x
x
108
1010
1012
1014
1016
1018
π(x)
5761455
455052511
37607912018
3204941750802
279238341033925
24739954287740860
Li(x) − π(x)
753
3103
38262
314889
3214631
21949554
For all x < 1023 we have π(x) < Li(x).
Theorem (Littlewood 1914)
The difference π(x) − Li(x) changes sign for infinitely many integers x.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
4 / 29
Table: The number of primes up to x
x
108
1010
1012
1014
1016
1018
π(x)
5761455
455052511
37607912018
3204941750802
279238341033925
24739954287740860
Li(x) − π(x)
753
3103
38262
314889
3214631
21949554
For all x < 1023 we have π(x) < Li(x).
Theorem (Littlewood 1914)
The difference π(x) − Li(x) changes sign for infinitely many integers x.
Bays and Hudson (1999): The first x for which π(x) > Li(x) is expected
to be around 1.3982 × 10316 .
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
4 / 29
Primes in arithmetic progressions
π(x; q, a) is the number of primes p ≤ x such that p ≡ a mod q
(primes of the form qn + a).
φ(q) is the number of integers 1 ≤ a ≤ q such that (a, q) = 1.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
5 / 29
Primes in arithmetic progressions
π(x; q, a) is the number of primes p ≤ x such that p ≡ a mod q
(primes of the form qn + a).
φ(q) is the number of integers 1 ≤ a ≤ q such that (a, q) = 1.
Theorem (Dirichlet 1837)
If (a, q) = 1, then there are infinitely many primes p ≡ a mod q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
5 / 29
Primes in arithmetic progressions
π(x; q, a) is the number of primes p ≤ x such that p ≡ a mod q
(primes of the form qn + a).
φ(q) is the number of integers 1 ≤ a ≤ q such that (a, q) = 1.
Theorem (Dirichlet 1837)
If (a, q) = 1, then there are infinitely many primes p ≡ a mod q.
How many primes congruent to a mod q are there up to x ?
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
5 / 29
Primes in arithmetic progressions
π(x; q, a) is the number of primes p ≤ x such that p ≡ a mod q
(primes of the form qn + a).
φ(q) is the number of integers 1 ≤ a ≤ q such that (a, q) = 1.
Theorem (Dirichlet 1837)
If (a, q) = 1, then there are infinitely many primes p ≡ a mod q.
How many primes congruent to a mod q are there up to x ?
Prime Number Theorem for Arithmetic Progressions
Let (a, q) = 1. Then, as x → ∞, we have
π(x; q, a) ∼
Youness Lamzouri (Illinois)
Li(x)
π(x)
∼
.
φ(q)
φ(q)
Prime Number Races
December 11, 2011
5 / 29
Chebyshev’s Bias
In 1853, Chebyshev wrote a letter to M. Fuss with the following statement
There is a notable difference in the splitting of the primes between the two
forms 4n + 3, 4n + 1: the first form contains a lot more than the second.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
6 / 29
Chebyshev’s Bias
In 1853, Chebyshev wrote a letter to M. Fuss with the following statement
There is a notable difference in the splitting of the primes between the two
forms 4n + 3, 4n + 1: the first form contains a lot more than the second.
Recall that π(x; 4, 1) ∼ π(x)/2 ∼ π(x; 4, 3) as x → ∞.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
6 / 29
Chebyshev’s Bias
In 1853, Chebyshev wrote a letter to M. Fuss with the following statement
There is a notable difference in the splitting of the primes between the two
forms 4n + 3, 4n + 1: the first form contains a lot more than the second.
Recall that π(x; 4, 1) ∼ π(x)/2 ∼ π(x; 4, 3) as x → ∞.
Table: The number of primes of the form 4n + 3 and 4n + 1 up to x
x
100
500
1000
5000
10,000
50,000
100,000
Youness Lamzouri (Illinois)
π(x; 4, 3)
13
50
87
339
619
2583
4808
Prime Number Races
π(x; 4, 1)
11
44
80
329
609
2549
4783
December 11, 2011
6 / 29
The race modulo 4
π(x; 4, 1) > π(x; 4, 3) for the first time when x = 26,861.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
7 / 29
The race modulo 4
π(x; 4, 1) > π(x; 4, 3) for the first time when x = 26,861.
For x ≥ 26,863, π(x; 4, 1) > π(x; 4, 3) occurs for the first time when
x = 616,841, and also at various numbers until 633,798.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
7 / 29
The race modulo 4
π(x; 4, 1) > π(x; 4, 3) for the first time when x = 26,861.
For x ≥ 26,863, π(x; 4, 1) > π(x; 4, 3) occurs for the first time when
x = 616,841, and also at various numbers until 633,798.
The first x > 633,798 for which π(x; 4, 1) > π(x; 4, 3) is x =
12,306,137.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
7 / 29
The race modulo 4
π(x; 4, 1) > π(x; 4, 3) for the first time when x = 26,861.
For x ≥ 26,863, π(x; 4, 1) > π(x; 4, 3) occurs for the first time when
x = 616,841, and also at various numbers until 633,798.
The first x > 633,798 for which π(x; 4, 1) > π(x; 4, 3) is x =
12,306,137.
Theorem (Littlewood 1914)
The difference π(x; 4, 1) − π(x; 4, 3) changes sign for infinitely many
integers x.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
7 / 29
The race modulo 3
Table: The number of primes of the form 3n + 2 and 3n + 1 up to x
x
100
1000
10,000
100,000
1,000,000
Youness Lamzouri (Illinois)
π(x; 3, 2)
13
87
617
4807
39,266
Prime Number Races
π(x; 3, 1)
11
80
611
4784
39,231
December 11, 2011
8 / 29
The race modulo 3
Table: The number of primes of the form 3n + 2 and 3n + 1 up to x
x
100
1000
10,000
100,000
1,000,000
π(x; 3, 2)
13
87
617
4807
39,266
π(x; 3, 1)
11
80
611
4784
39,231
Bays and Hudson (Christmas Day of 1976): π(x; 3, 1) > π(x; 3, 2) for the
first time when x = 608,981,813,029 ≈ 6 × 1011 .
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
8 / 29
The race modulo 3
Table: The number of primes of the form 3n + 2 and 3n + 1 up to x
x
100
1000
10,000
100,000
1,000,000
π(x; 3, 2)
13
87
617
4807
39,266
π(x; 3, 1)
11
80
611
4784
39,231
Bays and Hudson (Christmas Day of 1976): π(x; 3, 1) > π(x; 3, 2) for the
first time when x = 608,981,813,029 ≈ 6 × 1011 .
Theorem (Littlewood 1914)
π(x; 3, 1) − π(x; 3, 2) changes sign for infinitely many integers x.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
8 / 29
Two more races, modulo 10 and modulo 8
Table: The number of primes of the form 10n + j up to x
x
1000
10,000
100,000
1,000,000
Last Digit 1
40
306
2387
19,617
Youness Lamzouri (Illinois)
Last Digit 3
42
310
2402
19,665
Prime Number Races
Last Digit 7
46
308
2411
19,621
Last Digit 9
38
303
2390
19,593
December 11, 2011
9 / 29
Two more races, modulo 10 and modulo 8
Table: The number of primes of the form 10n + j up to x
x
1000
10,000
100,000
1,000,000
Last Digit 1
40
306
2387
19,617
Last Digit 3
42
310
2402
19,665
Last Digit 7
46
308
2411
19,621
Last Digit 9
38
303
2390
19,593
Table: The number of primes of the form 8n + j up to x
x
1000
10,000
100,000
1,000,000
Youness Lamzouri (Illinois)
π(x; 8, 1)
37
295
2384
19,552
π(x; 8, 3)
44
311
2409
19,653
Prime Number Races
π(x; 8, 5)
43
314
2399
19,623
π(x; 8, 7)
43
308
2399
19,669
December 11, 2011
9 / 29
Dirichlet characters and Dirichlet L-functions
A Dirichlet character modulo q is a function χ : Z → C such that
1. χ is periodic with period q.
2. χ(n) = 0 if (n, q) > 1.
3. χ is completely multiplicative: χ(mn) = χ(m)χ(n).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
10 / 29
Dirichlet characters and Dirichlet L-functions
A Dirichlet character modulo q is a function χ : Z → C such that
1. χ is periodic with period q.
2. χ(n) = 0 if (n, q) > 1.
3. χ is completely multiplicative: χ(mn) = χ(m)χ(n).
There are φ(q) Dirichlet characters modulo q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
10 / 29
Dirichlet characters and Dirichlet L-functions
A Dirichlet character modulo q is a function χ : Z → C such that
1. χ is periodic with period q.
2. χ(n) = 0 if (n, q) > 1.
3. χ is completely multiplicative: χ(mn) = χ(m)χ(n).
There are φ(q) Dirichlet characters modulo q.
Orthogonality relation: If (a, q) = 1 then
(
X
1 if n ≡ a mod q,
1
χ(a)χ(n) =
φ(q)
0 otherwise.
χ mod q
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
10 / 29
Dirichlet characters and Dirichlet L-functions
A Dirichlet character modulo q is a function χ : Z → C such that
1. χ is periodic with period q.
2. χ(n) = 0 if (n, q) > 1.
3. χ is completely multiplicative: χ(mn) = χ(m)χ(n).
There are φ(q) Dirichlet characters modulo q.
Orthogonality relation: If (a, q) = 1 then
(
X
1 if n ≡ a mod q,
1
χ(a)χ(n) =
φ(q)
0 otherwise.
χ mod q
Examples of Dirichlet characters
The principal character modulo q: χ0 (n) = 1 if (n, q) = 1.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
10 / 29
Dirichlet characters and Dirichlet L-functions
A Dirichlet character modulo q is a function χ : Z → C such that
1. χ is periodic with period q.
2. χ(n) = 0 if (n, q) > 1.
3. χ is completely multiplicative: χ(mn) = χ(m)χ(n).
There are φ(q) Dirichlet characters modulo q.
Orthogonality relation: If (a, q) = 1 then
(
X
1 if n ≡ a mod q,
1
χ(a)χ(n) =
φ(q)
0 otherwise.
χ mod q
Examples of Dirichlet characters
The principal character modulo q: χ0 (n) = 1 if (n, q) = 1.
The non-principal character modulo 4: χ1 (n) = 1 if n ≡ 1 mod 4 and
χ1 (n) = −1 if n ≡ 3 mod 4.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
10 / 29
Let χ be a Dirichlet character modulo q. The Dirichlet L-function L(s, χ)
is defined by
∞
X
Y χ(n)
χ(p) −1
for Re(s) > 1.
L(s, χ) =
=
1− s
ns
p
n=1
Youness Lamzouri (Illinois)
p prime
Prime Number Races
December 11, 2011
11 / 29
Let χ be a Dirichlet character modulo q. The Dirichlet L-function L(s, χ)
is defined by
∞
X
Y χ(n)
χ(p) −1
for Re(s) > 1.
L(s, χ) =
=
1− s
ns
p
n=1
p prime
L(s, χ) can be analytically continued to an entire function over the
complex plan.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
11 / 29
Let χ be a Dirichlet character modulo q. The Dirichlet L-function L(s, χ)
is defined by
∞
X
Y χ(n)
χ(p) −1
for Re(s) > 1.
L(s, χ) =
=
1− s
ns
p
n=1
p prime
L(s, χ) can be analytically continued to an entire function over the
complex plan.
Dirichlet (1837): L(1, χ) 6= 0 for all χ 6= χ0 mod q. This implies that
there are infinitely many primes p ≡ a mod q if (a, q) = 1.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
11 / 29
Let χ be a Dirichlet character modulo q. The Dirichlet L-function L(s, χ)
is defined by
∞
X
Y χ(n)
χ(p) −1
for Re(s) > 1.
L(s, χ) =
=
1− s
ns
p
n=1
p prime
L(s, χ) can be analytically continued to an entire function over the
complex plan.
Dirichlet (1837): L(1, χ) 6= 0 for all χ 6= χ0 mod q. This implies that
there are infinitely many primes p ≡ a mod q if (a, q) = 1.
If χ1 is the non-principal Dirichlet character modulo 4. Then
L(s, χ1 ) =
Youness Lamzouri (Illinois)
1
1
1
1
1
1
− s + s − s + s − s + · · · for Re(s) > 1.
1s
3
5
7
9
11
Prime Number Races
December 11, 2011
11 / 29
Let χ be a Dirichlet character modulo q. The Dirichlet L-function L(s, χ)
is defined by
∞
X
Y χ(n)
χ(p) −1
for Re(s) > 1.
L(s, χ) =
=
1− s
ns
p
n=1
p prime
L(s, χ) can be analytically continued to an entire function over the
complex plan.
Dirichlet (1837): L(1, χ) 6= 0 for all χ 6= χ0 mod q. This implies that
there are infinitely many primes p ≡ a mod q if (a, q) = 1.
If χ1 is the non-principal Dirichlet character modulo 4. Then
L(s, χ1 ) =
1
1
1
1
1
1
− s + s − s + s − s + · · · for Re(s) > 1.
1s
3
5
7
9
11
The Generalized Riemann Hypothesis GRH
Let χ be a Dirichlet character modulo q. If σ + it is a complex number
with 0 ≤ σ ≤ 1 and L(σ + it, χ) = 0, then σ = 21 .
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
11 / 29
The origin of Chebyshev’s Bias
More primes of the form 4n + 3 than of the form 4n + 1.
More primes of the forms 10n + 3 and 10n + 7, than of the forms
10n + 9 and 10n + 1.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
12 / 29
The origin of Chebyshev’s Bias
More primes of the form 4n + 3 than of the form 4n + 1.
More primes of the forms 10n + 3 and 10n + 7, than of the forms
10n + 9 and 10n + 1.
Chebyshev’s Bias: More primes of the form qn + a than of the form
qn + b if a is non-square and b is a square modulo q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
12 / 29
The origin of Chebyshev’s Bias
More primes of the form 4n + 3 than of the form 4n + 1.
More primes of the forms 10n + 3 and 10n + 7, than of the forms
10n + 9 and 10n + 1.
Chebyshev’s Bias: More primes of the form qn + a than of the form
qn + b if a is non-square and b is a square modulo q.
Explicit formula (goes back to Riemann, 1859)
Sum over primes in arithmetic progressions ⇐⇒ Sum over zeros of
Dirichlet L-functions L(s, χ)
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
12 / 29
The origin of Chebyshev’s Bias
More primes of the form 4n + 3 than of the form 4n + 1.
More primes of the forms 10n + 3 and 10n + 7, than of the forms
10n + 9 and 10n + 1.
Chebyshev’s Bias: More primes of the form qn + a than of the form
qn + b if a is non-square and b is a square modulo q.
Explicit formula (goes back to Riemann, 1859)
Sum over primes in arithmetic progressions ⇐⇒ Sum over zeros of
Dirichlet L-functions L(s, χ)
Our sum over primes in arithmetic progressions is
X X
√
1
1
= π(x; q, a) + |{p ≤ x : p 2 ≡ a mod q}| + error.
k
2
k
k≥1
p ≤x
p k ≡a mod q
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
12 / 29
The origin of Chebyshev’s Bias
More primes of the form 4n + 3 than of the form 4n + 1.
More primes of the forms 10n + 3 and 10n + 7, than of the forms
10n + 9 and 10n + 1.
Chebyshev’s Bias: More primes of the form qn + a than of the form
qn + b if a is non-square and b is a square modulo q.
Explicit formula (goes back to Riemann, 1859)
Sum over primes in arithmetic progressions ⇐⇒ Sum over zeros of
Dirichlet L-functions L(s, χ)
Our sum over primes in arithmetic progressions is
X X
√
1
1
= π(x; q, a) + |{p ≤ x : p 2 ≡ a mod q}| + error.
k
2
k
k≥1
p ≤x
p k ≡a mod q
The second term is the source of Chebychev’s Bias.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
12 / 29
Can we measure the bias?
Conjecture (Knapowski-Turàn 1962)
As X → ∞, the percentage of integers x ≤ X for which
π(x; 4, 1) > π(x; 4, 3) goes to to 0%.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
13 / 29
Can we measure the bias?
Conjecture (Knapowski-Turàn 1962)
As X → ∞, the percentage of integers x ≤ X for which
π(x; 4, 1) > π(x; 4, 3) goes to to 0%.
The following table gives, for X in various ranges, the maximum
percentage of values of x ≤ X for which π(x; 4, 1) > π(x; 4, 3):
Table:
For X in the range
0 − 107
107 − 108
107 − 108
108 − 109
109 − 1010
1010 − 1011
Youness Lamzouri (Illinois)
Maximum percentage of such x ≤ X
2.6%
0.6%
0.6%
0.1%
1.6%
2.8%
Prime Number Races
December 11, 2011
13 / 29
We need assumptions to quantify these biases !
Theorem (Kaczorowski, Rubinstein-Sarnak 1994)
If the Generalized Riemann Hypothesis GRH is true, then the
Knapowski-Turàn Conjecture is false.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
14 / 29
We need assumptions to quantify these biases !
Theorem (Kaczorowski, Rubinstein-Sarnak 1994)
If the Generalized Riemann Hypothesis GRH is true, then the
Knapowski-Turàn Conjecture is false.
Theorem (Kaczorowski 1994)
Assume GRH. For any (a, q) = 1 such that a 6≡ 1 mod q, the set of
positive integers x for which π(x; q, 1) > π(x; q, a) has a positive lower
density.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
14 / 29
We need assumptions to quantify these biases !
Theorem (Kaczorowski, Rubinstein-Sarnak 1994)
If the Generalized Riemann Hypothesis GRH is true, then the
Knapowski-Turàn Conjecture is false.
Theorem (Kaczorowski 1994)
Assume GRH. For any (a, q) = 1 such that a 6≡ 1 mod q, the set of
positive integers x for which π(x; q, 1) > π(x; q, a) has a positive lower
density.
Theorem 1 (Ford, Konyagin and L. 2011+)
Let (a, q) = 1 be such that a2 6≡ 1 mod q. Then we can construct a set
S = {σ + it : 0 ≤ σ ≤ 1 and σ 6= 1/2} such that if a certain L(s, χ) has
zeros in S, then the percentage of integers x ≤ X for which
π(x; q, 1) > π(x; q, a) goes to 0%.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
14 / 29
Measuring the bias: the work of Rubinstein-Sarnak
Theorem (Kaczorowski 1994)
Assume the Generalized Riemann Hypothesis. The quantity
1
|{x ≤ X : π(x; 4, 3) > π(x; 4, 1)}
X
does not tend to any limit as X → ∞.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
15 / 29
Measuring the bias: the work of Rubinstein-Sarnak
Theorem (Kaczorowski 1994)
Assume the Generalized Riemann Hypothesis. The quantity
1
|{x ≤ X : π(x; 4, 3) > π(x; 4, 1)}
X
does not tend to any limit as X → ∞.
Rubinstein and Sarnak (1994)
The natural density is not the correct way to measure the bias!
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
15 / 29
Measuring the bias: the work of Rubinstein-Sarnak
Theorem (Kaczorowski 1994)
Assume the Generalized Riemann Hypothesis. The quantity
1
|{x ≤ X : π(x; 4, 3) > π(x; 4, 1)}
X
does not tend to any limit as X → ∞.
Rubinstein and Sarnak (1994)
The natural density is not the correct way to measure the bias!
The adequate measure to use is the logarithmic measure.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
15 / 29
Measuring the bias: the work of Rubinstein-Sarnak
Theorem (Kaczorowski 1994)
Assume the Generalized Riemann Hypothesis. The quantity
1
|{x ≤ X : π(x; 4, 3) > π(x; 4, 1)}
X
does not tend to any limit as X → ∞.
Rubinstein and Sarnak (1994)
The natural density is not the correct way to measure the bias!
The adequate measure to use is the logarithmic measure.
Need to use the Generalized Riemann Hypothesis and the Linear
Independence Conjecture LI, which is the assumption that the
positive imaginary parts of the zeros of the associated Dirichlet
L-functions are linearly independent over the rational numbers.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
15 / 29
From now on we assume the truth of the Generalized Riemann Hypothesis,
and the Linear Independence Conjecture.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
16 / 29
From now on we assume the truth of the Generalized Riemann Hypothesis,
and the Linear Independence Conjecture.
Theorem (Rubinstein and Sarnak, 1994)
As X → ∞,
Youness Lamzouri (Illinois)
1
log X
X
x≤X
π(x;4,3)>π(x;4,1)
1
→ 0.9959 . . .
x
Prime Number Races
December 11, 2011
16 / 29
From now on we assume the truth of the Generalized Riemann Hypothesis,
and the Linear Independence Conjecture.
Theorem (Rubinstein and Sarnak, 1994)
As X → ∞,
1
log X
X
x≤X
π(x;4,3)>π(x;4,1)
1
→ 0.9959 . . .
x
Chebyshev was correct 99.95% of the time!
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
16 / 29
From now on we assume the truth of the Generalized Riemann Hypothesis,
and the Linear Independence Conjecture.
Theorem (Rubinstein and Sarnak, 1994)
As X → ∞,
1
log X
X
x≤X
π(x;4,3)>π(x;4,1)
1
→ 0.9959 . . .
x
Chebyshev was correct 99.95% of the time!
Theorem (Rubinstein and Sarnak, 1994)
Let (a1 , q) = (a2 , q) = 1 such that a1 6≡ a2 mod q. The logarithmic density
Z
1
dt
,
δ(q; a1 , a2 ) := lim
t∈[2,X ]
X →∞ log X
t
π(t;q,a1 )>π(t;q,a2 )
exists and is positive.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
16 / 29
δ(q; a1 , a2 ) is “the probability” that π(x; q, a1 ) > π(x; q, a2 ).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
17 / 29
δ(q; a1 , a2 ) is “the probability” that π(x; q, a1 ) > π(x; q, a2 ).
Theorem (Rubinstein-Sarnak, 1994)
δ(q; a1 , a2 ) =
modulo q.
Youness Lamzouri (Illinois)
1
2
if a1 and a2 are both squares or both non-squares
Prime Number Races
December 11, 2011
17 / 29
δ(q; a1 , a2 ) is “the probability” that π(x; q, a1 ) > π(x; q, a2 ).
Theorem (Rubinstein-Sarnak, 1994)
δ(q; a1 , a2 ) =
modulo q.
1
2
if a1 and a2 are both squares or both non-squares
Chebyshev’s bias: δ(q; a1 , a2 ) > 1/2 if a1 is a non-square and a2 is a
square modulo q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
17 / 29
δ(q; a1 , a2 ) is “the probability” that π(x; q, a1 ) > π(x; q, a2 ).
Theorem (Rubinstein-Sarnak, 1994)
δ(q; a1 , a2 ) =
modulo q.
1
2
if a1 and a2 are both squares or both non-squares
Chebyshev’s bias: δ(q; a1 , a2 ) > 1/2 if a1 is a non-square and a2 is a
square modulo q.
δ(4; 3, 1) = 0.9959 . . . and δ(3; 2, 1) = 0.9990 . . . .
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
17 / 29
Prime number races with three or more competitors
Let 2 ≤ r ≤ φ(q) and a1 , . . . , ar be distinct residue classes modulo q
which are coprime to q.
Let P(q; a1 , . . . , ar ) be the set of positive integer x such that
π(x; q, a1 ) > π(a; q, a2 ) > · · · > π(x; q, ar ).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
18 / 29
Prime number races with three or more competitors
Let 2 ≤ r ≤ φ(q) and a1 , . . . , ar be distinct residue classes modulo q
which are coprime to q.
Let P(q; a1 , . . . , ar ) be the set of positive integer x such that
π(x; q, a1 ) > π(a; q, a2 ) > · · · > π(x; q, ar ).
Theorem (Rubinstein and Sarnak, 1994)
The logarithmic density of P(q; a1 , . . . , ar ) defined by
Z
1
dt
,
δ(q; a1 , . . . , ar ) := lim
X →∞ log X t∈P(q;a1 ,...,ar )∩[2,X ] t
exists and is positive.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
18 / 29
Prime number races with three or more competitors
Let 2 ≤ r ≤ φ(q) and a1 , . . . , ar be distinct residue classes modulo q
which are coprime to q.
Let P(q; a1 , . . . , ar ) be the set of positive integer x such that
π(x; q, a1 ) > π(a; q, a2 ) > · · · > π(x; q, ar ).
Theorem (Rubinstein and Sarnak, 1994)
The logarithmic density of P(q; a1 , . . . , ar ) defined by
Z
1
dt
,
δ(q; a1 , . . . , ar ) := lim
X →∞ log X t∈P(q;a1 ,...,ar )∩[2,X ] t
exists and is positive.
δ(q; a1 , . . . , ar ) is the “probability” that π(x; q, a1 ) > · · · > π(x; q, ar ).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
18 / 29
δ(q; a1 , . . . , ar ) is the “probability” that π(x; q, a1 ) > · · · > π(x; q, ar ).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
19 / 29
δ(q; a1 , . . . , ar ) is the “probability” that π(x; q, a1 ) > · · · > π(x; q, ar ).
The race {q; a1 , . . . , ar } is said to be unbiased if for every
permutation σ of the set {1, 2, . . . , r } we have
δ(q; aσ(1) , . . . , aσ(r ) ) = δ(q; a1 , . . . , ar ) =
Youness Lamzouri (Illinois)
Prime Number Races
1
.
r!
December 11, 2011
19 / 29
δ(q; a1 , . . . , ar ) is the “probability” that π(x; q, a1 ) > · · · > π(x; q, ar ).
The race {q; a1 , . . . , ar } is said to be unbiased if for every
permutation σ of the set {1, 2, . . . , r } we have
δ(q; aσ(1) , . . . , aσ(r ) ) = δ(q; a1 , . . . , ar ) =
1
.
r!
If a race {q; a1 , . . . , ar } is unbiased, then all the ai must be either squares
or either non-squares modulo q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
19 / 29
δ(q; a1 , . . . , ar ) is the “probability” that π(x; q, a1 ) > · · · > π(x; q, ar ).
The race {q; a1 , . . . , ar } is said to be unbiased if for every
permutation σ of the set {1, 2, . . . , r } we have
δ(q; aσ(1) , . . . , aσ(r ) ) = δ(q; a1 , . . . , ar ) =
1
.
r!
If a race {q; a1 , . . . , ar } is unbiased, then all the ai must be either squares
or either non-squares modulo q.
Theorem (Rubinstein and Sarnak, 1994)
Let q ≥ 2. If a1 , a2 and a3 are distinct residue classes modulo q which are
coprime to q, such that
a2 ≡ a1 ρ mod q, a3 ≡ a1 ρ2 mod q,
for some ρ 6= 1 with ρ3 ≡ 1 mod q, then the race {q; a1 , a2 , a3 } is
unbiased.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
19 / 29
Guess: If a1 , . . . , ar are all squares or all non-squares modulo q, then
the race {q; a1 , . . . , ar } is unbiased.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
20 / 29
Guess: If a1 , . . . , ar are all squares or all non-squares modulo q, then
the race {q; a1 , . . . , ar } is unbiased.
Feuerverger and Martin (2000): Our guess is wrong!
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
20 / 29
Guess: If a1 , . . . , ar are all squares or all non-squares modulo q, then
the race {q; a1 , . . . , ar } is unbiased.
Feuerverger and Martin (2000): Our guess is wrong!
Theorem (Feuerverger and Martin, 2000)
The races {8; 3, 5, 7} and {12; 5, 7, 11} are biased.
1
6
1
δ(8; 3, 7, 5) = δ(8; 5, 7, 3) = 0.1664262 · · · <
6
1
δ(8; 5, 3, 7) = δ(8; 7, 3, 5) = 0.1407724 · · · <
6
δ(8; 3, 5, 7) = δ(8; 7, 5, 3) = 0.1928013 · · · >
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
20 / 29
Guess: If a1 , . . . , ar are all squares or all non-squares modulo q, then
the race {q; a1 , . . . , ar } is unbiased.
Feuerverger and Martin (2000): Our guess is wrong!
Theorem (Feuerverger and Martin, 2000)
The races {8; 3, 5, 7} and {12; 5, 7, 11} are biased.
1
6
1
δ(8; 3, 7, 5) = δ(8; 5, 7, 3) = 0.1664262 · · · <
6
1
δ(8; 5, 3, 7) = δ(8; 7, 3, 5) = 0.1407724 · · · <
6
δ(8; 3, 5, 7) = δ(8; 7, 5, 3) = 0.1928013 · · · >
All the densities δ(q; a1 , . . . , ar ) they computed are such that r ≤ 4
and q ≤ 12.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
20 / 29
Guess: If a1 , . . . , ar are all squares or all non-squares modulo q, then
the race {q; a1 , . . . , ar } is unbiased.
Feuerverger and Martin (2000): Our guess is wrong!
Theorem (Feuerverger and Martin, 2000)
The races {8; 3, 5, 7} and {12; 5, 7, 11} are biased.
1
6
1
δ(8; 3, 7, 5) = δ(8; 5, 7, 3) = 0.1664262 · · · <
6
1
δ(8; 5, 3, 7) = δ(8; 7, 3, 5) = 0.1407724 · · · <
6
δ(8; 3, 5, 7) = δ(8; 7, 5, 3) = 0.1928013 · · · >
All the densities δ(q; a1 , . . . , ar ) they computed are such that r ≤ 4
and q ≤ 12.
It is difficult to compute these densities, since we need to use many
zeros of Dirichlet L-functions!
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
20 / 29
Theorem 2 (L, 2011)
Fix r ≥ 3. There exists a constant q0 (r ) such that if q ≥ q0 (r ) is is a
positive integer, then
There exist distinct residue classes a1 , . . . , ar mod q, with (ai , q) = 1,
a1 , . . . , ar are squares modulo q and the race {q; a1 , . . . , ar } is biased.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
21 / 29
Theorem 2 (L, 2011)
Fix r ≥ 3. There exists a constant q0 (r ) such that if q ≥ q0 (r ) is is a
positive integer, then
There exist distinct residue classes a1 , . . . , ar mod q, with (ai , q) = 1,
a1 , . . . , ar are squares modulo q and the race {q; a1 , . . . , ar } is biased.
There exist distinct residue classes b1 , . . . , br mod q, with (bi , q) = 1,
b1 , . . . , br are non-squares modulo q and the race {q; b1 , . . . , br } is
biased.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
21 / 29
Theorem 2 (L, 2011)
Fix r ≥ 3. There exists a constant q0 (r ) such that if q ≥ q0 (r ) is is a
positive integer, then
There exist distinct residue classes a1 , . . . , ar mod q, with (ai , q) = 1,
a1 , . . . , ar are squares modulo q and the race {q; a1 , . . . , ar } is biased.
There exist distinct residue classes b1 , . . . , br mod q, with (bi , q) = 1,
b1 , . . . , br are non-squares modulo q and the race {q; b1 , . . . , br } is
biased.
Explicit constructions
Biased races with three squares:
Let q ≡ 1 mod 8 be a prime. Then 1, −1 and 2 are squares modulo
q. Consider the race {q; 1, −1, 2}. If q is large then
δ(q; 1, 2, −1) >
Youness Lamzouri (Illinois)
1
> δ(q; −1, 1, 2).
6
Prime Number Races
December 11, 2011
21 / 29
Biased races with three non-squares:
Let q ≡ 11 mod 24 be a prime. Then −1, 2 and 6 are non-squares
modulo q. Consider the race {q; −1, 2, 6}. If q is large, then
δ(q; 2, −1, 6) >
Youness Lamzouri (Illinois)
1
> δ(q; −1, 2, 6).
6
Prime Number Races
December 11, 2011
22 / 29
Biased races with three non-squares:
Let q ≡ 11 mod 24 be a prime. Then −1, 2 and 6 are non-squares
modulo q. Consider the race {q; −1, 2, 6}. If q is large, then
δ(q; 2, −1, 6) >
1
> δ(q; −1, 2, 6).
6
Biased races with r squares:
Let q be positive integer with (q, 6) = 1. Consider the race
{q; 1, 64 , 66 , . . . , 62(r −1) , 4}. If q is large, then
δ(q; 1, 64 , 66 , . . . , 62(r −1) , 4) >
Youness Lamzouri (Illinois)
1
> δ q; 64 , 66 , . . . , 62(r −1) , 1, 4 .
r!
Prime Number Races
December 11, 2011
22 / 29
Biased races with three non-squares:
Let q ≡ 11 mod 24 be a prime. Then −1, 2 and 6 are non-squares
modulo q. Consider the race {q; −1, 2, 6}. If q is large, then
δ(q; 2, −1, 6) >
1
> δ(q; −1, 2, 6).
6
Biased races with r squares:
Let q be positive integer with (q, 6) = 1. Consider the race
{q; 1, 64 , 66 , . . . , 62(r −1) , 4}. If q is large, then
δ(q; 1, 64 , 66 , . . . , 62(r −1) , 4) >
1
> δ q; 64 , 66 , . . . , 62(r −1) , 1, 4 .
r!
Biased races with r non-squares:
Let q ≡ 3 mod 4 be a prime. Then −1 is a non-square modulo q.
Consider the race {q; −1, −64 , −66 , . . . , −62(r −1) , −4}. If q is large,
δ(q; −1, . . . , −62(r −1) , −4) >
Youness Lamzouri (Illinois)
1
> δ(q; −64 , . . . , −62(r −1) , −1, −4).
r!
Prime Number Races
December 11, 2011
22 / 29
The size of δ(q; a1 , . . . , ar ) when r is fixed and q is large
δ(4; 3, 1) = 0.9959 . . . (Rubinstein-Sarnak, 1994)
δ(420; 17, 1) = 0.7956 . . . (Fiorilli-Martin, 2009)
δ(997; 11, 1) = 0.5082 . . . (Fiorilli-Martin, 2009)
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
23 / 29
The size of δ(q; a1 , . . . , ar ) when r is fixed and q is large
δ(4; 3, 1) = 0.9959 . . . (Rubinstein-Sarnak, 1994)
δ(420; 17, 1) = 0.7956 . . . (Fiorilli-Martin, 2009)
δ(997; 11, 1) = 0.5082 . . . (Fiorilli-Martin, 2009)
The bias towards non-squares becomes “less pronounced” when q grows.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
23 / 29
The size of δ(q; a1 , . . . , ar ) when r is fixed and q is large
δ(4; 3, 1) = 0.9959 . . . (Rubinstein-Sarnak, 1994)
δ(420; 17, 1) = 0.7956 . . . (Fiorilli-Martin, 2009)
δ(997; 11, 1) = 0.5082 . . . (Fiorilli-Martin, 2009)
The bias towards non-squares becomes “less pronounced” when q grows.
Theorem (Rubinstein and Sarnak, 1994)
max
a1 ,a2 mod q
Youness Lamzouri (Illinois)
δ(q; a1 , a2 ) −
1 → 0 as q → ∞.
2
Prime Number Races
December 11, 2011
23 / 29
The size of δ(q; a1 , . . . , ar ) when r is fixed and q is large
δ(4; 3, 1) = 0.9959 . . . (Rubinstein-Sarnak, 1994)
δ(420; 17, 1) = 0.7956 . . . (Fiorilli-Martin, 2009)
δ(997; 11, 1) = 0.5082 . . . (Fiorilli-Martin, 2009)
The bias towards non-squares becomes “less pronounced” when q grows.
Theorem (Rubinstein and Sarnak, 1994)
max
a1 ,a2 mod q
δ(q; a1 , a2 ) −
1 → 0 as q → ∞.
2
More generally if r ≥ 2 is fixed, then
1
δ(q; a1 , . . . , ar ) − → 0 as q → ∞.
max
a1 ,a2 ,...,ar mod q
r!
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
23 / 29
Let
1
δ(q; a1 , . . . , ar ) − .
∆r (q) :=
max
a1 ,a2 ,...,ar mod q
r!
Theorem (Fiorilli and Martin, 2009)
ρ(q)
1
∆2 (q) ∼ p
= 1/2+o(1) ,
q
2 πφ(q) log q
where ρ(q) is the number of solutions of t 2 ≡ 1 mod q, and o(1) is a
quantity which goes to 0 as q → ∞.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
24 / 29
Let
1
δ(q; a1 , . . . , ar ) − .
∆r (q) :=
max
a1 ,a2 ,...,ar mod q
r!
Theorem (Fiorilli and Martin, 2009)
ρ(q)
1
∆2 (q) ∼ p
= 1/2+o(1) ,
q
2 πφ(q) log q
where ρ(q) is the number of solutions of t 2 ≡ 1 mod q, and o(1) is a
quantity which goes to 0 as q → ∞.
Theorem 3 (L, 2011)
Let r ≥ 3 be a fixed integer. There exist positive constants c1 (r ), c2 (r )
and a positive integer q0 such that, if q ≥ q0 then
c1 (r )
c2 (r )
≤ ∆r (q) ≤
.
log q
log q
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
24 / 29
Extremely biased races
Definition
Fix r ≥ 3 and let q be a large positive integer. We call a race
{q; a1 , . . . , ar } “extremely biased” if for some permutation σ of the set
{1, . . . , r } we have
δ(q; aσ(1) , . . . , aσ(r ) ) − 1 ≥ c(r ) .
r ! log q
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
25 / 29
Extremely biased races
Definition
Fix r ≥ 3 and let q be a large positive integer. We call a race
{q; a1 , . . . , ar } “extremely biased” if for some permutation σ of the set
{1, . . . , r } we have
δ(q; aσ(1) , . . . , aσ(r ) ) − 1 ≥ c(r ) .
r ! log q
Theorem 4 (L, 2011)
If a1 , . . . , ar are fixed and q is large, then the race {q; a1 , . . . , ar } is
extremely biased if and only if there exist 1 ≤ j 6= k ≤ r such that aj /ak
equals −1 or a prime power.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
25 / 29
Why is the case r = 2 different from r ≥ 3
Theorem (Rubinstein and Sarnak, 1994)
Z
δ(q; a1 , . . . , ar ) =
dµq;a1 ,...,ar (x1 , . . . , xr ),
x1 >x2 >···>xr
where µq;a1 ,...,ar is a probability measure corresponding to a certain
random vector Xq;a1 ,...,ar = (X (q, a1 ), . . . , X (q, ar )).
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
26 / 29
Why is the case r = 2 different from r ≥ 3
Theorem (Rubinstein and Sarnak, 1994)
Z
δ(q; a1 , . . . , ar ) =
dµq;a1 ,...,ar (x1 , . . . , xr ),
x1 >x2 >···>xr
where µq;a1 ,...,ar is a probability measure corresponding to a certain
random vector Xq;a1 ,...,ar = (X (q, a1 ), . . . , X (q, ar )).
The mean vector of Xq;a1 ,...,ar is (−cq (a1 ), . . . , −cq (ar )), where
cq (a) = −1 + |{b mod q : b 2 ≡ a mod q}|.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
26 / 29
Why is the case r = 2 different from r ≥ 3
Theorem (Rubinstein and Sarnak, 1994)
Z
δ(q; a1 , . . . , ar ) =
dµq;a1 ,...,ar (x1 , . . . , xr ),
x1 >x2 >···>xr
where µq;a1 ,...,ar is a probability measure corresponding to a certain
random vector Xq;a1 ,...,ar = (X (q, a1 ), . . . , X (q, ar )).
The mean vector of Xq;a1 ,...,ar is (−cq (a1 ), . . . , −cq (ar )), where
cq (a) = −1 + |{b mod q : b 2 ≡ a mod q}|. This mean vector is
responsible for Chebyshev’s Bias.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
26 / 29
Why is the case r = 2 different from r ≥ 3
Theorem (Rubinstein and Sarnak, 1994)
Z
δ(q; a1 , . . . , ar ) =
dµq;a1 ,...,ar (x1 , . . . , xr ),
x1 >x2 >···>xr
where µq;a1 ,...,ar is a probability measure corresponding to a certain
random vector Xq;a1 ,...,ar = (X (q, a1 ), . . . , X (q, ar )).
The mean vector of Xq;a1 ,...,ar is (−cq (a1 ), . . . , −cq (ar )), where
cq (a) = −1 + |{b mod q : b 2 ≡ a mod q}|. This mean vector is
responsible for Chebyshev’s Bias.
The covariance matrix of Xq;a1 ,...,ar , noted Aq;a1 ,...,ar (j, k) satisfies
(
D(q) ∼ φ(q) log q
if j = k
Aq;a1 ,...,ar (j, k) =
Aq (aj , ak ) = O(φ(q)) if j 6= k.
The off-diagonal terms of the covariance matrix are responsible for
the different behavior of δ(q; a1 , . . . , ar ) when r ≥ 3.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
26 / 29
The size of δ(q; a1 , . . . , ar ) when r → ∞ as q → ∞
Conjecture 1 (Feuerverger-Martin, 2000)
We have
lim
max
q→∞ a1 ,...,ar mod q
δ(q; a1 , . . . , ar ) = 0,
for any arbitrary function r = r (q) tending to infinity with q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
27 / 29
The size of δ(q; a1 , . . . , ar ) when r → ∞ as q → ∞
Conjecture 1 (Feuerverger-Martin, 2000)
We have
lim
max
q→∞ a1 ,...,ar mod q
δ(q; a1 , . . . , ar ) = 0,
for any arbitrary function r = r (q) tending to infinity with q.
Conjecture 2 (Feuerverger-Martin, 2000)
There exist a function r0 (q) with r0 (q) → ∞ as q → ∞, such that for any
integer r ≤ r0 (q) we have
lim
max
q→∞ a1 ,...,ar mod q
|r !δ(q; a1 , . . . , ar ) − 1| = 0,
Conjecture 2 implies Conjecture 1.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
27 / 29
Theorem 5 (L, 2011+)
√
Conjecture 2 holds for any r0 (q) = o( log q) as q → ∞.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
28 / 29
Theorem 5 (L, 2011+)
√
Conjecture 2 holds for any r0 (q) = o( log q) as q → ∞.
√
For any integer r such that 2 ≤ r ≤ log q we have
2 1
r
δ(q; a1 , . . . , ar ) =
1+O
,
r!
log q
uniformly for all r -tuples (a1 , . . . , ar ) of distinct residue classes
modulo q that are coprime to q.
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
28 / 29
Theorem 5 (L, 2011+)
√
Conjecture 2 holds for any r0 (q) = o( log q) as q → ∞.
√
For any integer r such that 2 ≤ r ≤ log q we have
2 1
r
δ(q; a1 , . . . , ar ) =
1+O
,
r!
log q
uniformly for all r -tuples (a1 , . . . , ar ) of distinct residue classes
modulo q that are coprime to q.
Conjecture (Ford, L.)
If r = r (q) is such that r / log(q) → ∞ as q → ∞, then
lim
max
q→∞ a1 ,...,ar mod q
r !δ(q; a1 , . . . , ar ) = +∞,
and
lim
min
q→∞ a1 ,...,ar mod q
Youness Lamzouri (Illinois)
r !δ(q; a1 , . . . , ar ) = 0.
Prime Number Races
December 11, 2011
28 / 29
Thank you for your attention !
Youness Lamzouri (Illinois)
Prime Number Races
December 11, 2011
29 / 29