PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
CHRISTIAN MAUDUIT AND JOËL RIVAT
P
Abstract. For a large class of digital functions f , we estimate the sums n≤x Λ(n)f (n) (and
P
n≤x µ(n)f (n)) where Λ denotes the von Mangoldt function (and µ the Möbius function). We
deduce from these estimates a Prime Number Theorem (and a Möbius randomness principle)
for sequences of integers with digit properties including the Rudin-Shapiro sequence and some
of its generalizations.
Contents
1. Introduction
2. Statement of the results
3. Notations and preliminary lemmas
4. Carry propagation lemmas
5. Sums of type I
6. Sums of type II
7. Distribution of the Discrete Fourier Transform
8. End of the estimate of the sums of type II
9. Proof of Theorems 1 and 2
10. Proof of Corollaries 1, 2 and 3
11. Application to Rudin-Shapiro sequences
References
1
3
4
7
9
13
22
26
27
27
29
37
1. Introduction
We denote by N the set of integers n ≥ 0, by U the set of complex numbers of modulus 1,
by P the set of prime numbers and for any a ∈ Z and m ∈ N with m ≥ 1, by P(a, m) the set
of prime numbers p ≡ a mod m. For n ∈ N, n ≥ 1, we denote by τ (n) the number of divisors
of n, by ω(n) the number of distinct prime factors of n, by Λ(n) the von Mangoldt function
(defined by Λ(n) = log p if n = pk with k ∈ N, k ≥ 1 and Λ(n) = 0 otherwise) and by µ(n) the
Möbius function (defined by µ(n) = (−1)ω(n) if n is squarefree and µ(n) = 0 otherwise). For
x ∈ R we denote by kxk the distance of x to the nearest integer, by π(x) the number of prime
numbers less or equal to x and we set e(x) = exp(2iπx).
Throughout this work P
we denote by q an integer greater or equal to 2. Any n ∈ N can be
written in base q as n = j≥0 εj (n)q j with εj (n) ∈ {0, . . . , q − 1} for all j ∈ N.
Let (un )n∈N be a sequence of complex numbers of modulus at most 1 generated by a simple
algorithm. Many recent works are devoted to P
the proof that special sequences (un )n∈N satisfy
the Möbius randomness principle (i.e. that
n≤x µ(n)un = o(x), see [13, page 338]) or a
Prime
Number Theorem (i.e. an asymptotic formula for the more difficult to handle sum
P
n≤x Λ(n)un ), see [6], [7], [12], [20]. These works are related to the Sarnak conjecture (see
[25]) which
asserts that if (un )n∈N is produced by a zero topological entropy dynamical system,
P
then n≤x µ(n)un = o(x). In the case of sequences (un )n∈N such that un is defined by a digital
property of the integer n, Dartyge and Tenenbaum proved in [7], using Daboussi’s convolution
Date: June 8, 2013.
2010 Mathematics Subject Classification. Primary: 11A63, 11B85, 11N05, Secondary: 11L20,11N60.
Key words and phrases. Rudin-Shapiro sequence, prime numbers, Möbius function, exponential sums.
This work was supported by the Agence Nationale de la Recherche project ANR-10-BLAN 0103 MUNUM.
1
2
CHRISTIAN MAUDUIT AND JOËL RIVAT
method, that for any real number α we have
!
X
µ(n) e α
n≤x
X
εj (n)
=O
j≥0
x
log log x
.
In [20] we proved that, for any real number α such that (q − 1)α 6∈ Z, there exists a real number
η(α) < 1 such that
!
X
X
(1)
Λ(n) e α
εj (n) = O xη(α) ,
n≤x
j≥0
answering a question due to Gelfond in [10] (see [9] for an explicit value of η(α)) and [18] for an
extension to more general digital functions). The proof of (1), based on Vaughan’s identity ([13,
(13.39)]) and the estimate of type I and type II bilinear sums, can be applied to the Möbius
function µ using [13, (13.40)] and this shows that, for any real number α such that (q −1)α 6∈ Z,
there exists a real number η(α) < 1 such that
!
X
X
µ(n) e α
εj (n) = O xη(α) .
n≤x
j≥0
Kalai asked in [14] and [15] a series of questions concerning the computational complexity
of µ that can be translated in proving a Möbius randomness principle for some specific binary
sequences. In [3] Bourgain proved that
(2)
max
S⊆{0,...,ν−1}
X
P
µ(n)(−1)
i∈S
εi (n)
= O(2ν−ν
1/10
),
n<2ν
showing both a Möbius randomness principle and a Prime Number Theorem for these sequences
(see [11] for a related result showing that µ is orthogonal to any Boolean function computable
by constant depth and polynomial
size circuits).
Studying more precisely the distribution of the
P
P
Fourier-Walsh coefficients n<2ν µ(n)(−1) i∈S εi (n) , Bourgain proved in [5] that µ is orthogonal
to any monotone Boolean function (see [4] for a lower bound for the number of primes captured
by these functions). The estimate (2) means that for any polynomial P ∈ Z[X0 , . . . , Xν−1 ] of
degree at most 1 we have
X
1/10
µ(n)(−1)P (ε0 (n),...,εν−1 (n)) = O(2ν−ν ),
n<2ν
but the question asked by Kalai in [16] concerning the case of polynomials of degree greater
than 1 is open. The simplest case of polynomial of degree 2 is given by the Rudin-Shapiro
sequence
P
(3)
(−1) i≥1 εi−1 (n)εi (n)
n∈N
introduced independently by Shapiro in [26] and by Rudin in [24] for which Tao suggests in [16]
a strategy to prove a Möbius randomness principle, i.e.
X
P
µ(n)(−1) i≥1 εi−1 (n)εi (n) = o(x).
n≤x
In this paper we will obtain as a special case in Theorem 3 a quantitative Prime Number
Theorem (and a Möbius randomness principle) for the sequences
P
(−1) i≥δ+1 εi−δ−1 (n)εi (n)
n∈N
for all integers δ ≥ 0 (including the Rudin-Shapiro sequence for δ = 0), and in Theorem 4 a
quantitative Prime Number Theorem (and a Möbius randomness principle) for the sequences
P
εi−d+1 (n)···εi−1 (n)εi (n)
i≥d−1
(−1)
n∈N
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
3
for all integers d ≥ 2, providing an answer to Kalai’s question for the simplest case of polynomial
of degree d.
2. Statement of the results
One of
Theorem for the sequence
ingredients in our proof of a Prime Number
Pthe main
1
exp α i≥0 εi (n) n∈N in [20] was to establish that the L norm of the Discrete Fourier
Transform of this sequence is very small. Unfortunately this property is generally not true
for other digital sequences and in particular for the Rudin-Shapiro sequence (3). Such a difference in the behaviour of the Fourier
transforms is not
surprising if we remember that the
P
P
sequences (−1) i≥0 εi (n)
and (−1) i≥1 εi−1 (n)εi (n)
have quite different spectral propn∈N
n∈N
erties:
the correlation measure of the first one is a singular measure, namely the Riesz product
Q
n
n≥0 (1 − cos 2 t) (see [23, section 3.3.3] or [17]), while for the second one it is the Lebesgue
measure (see [23, corollary 8.5]).
For f : N → U and any λ ∈ N, let us denote by fλ the q λ -periodic function defined by
∀n ∈ {0, . . . , q λ − 1}, ∀k ∈ Z, fλ (n + kq λ ) = f (n).
(4)
Definition 1. A function f : N → U has the carry property if, uniformly for (λ, κ, ρ) ∈ N3
with ρ < λ, the number of integers 0 ≤ ` < q λ such that there exists (k1 , k2 ) ∈ {0, . . . , q κ − 1}2
with
(5)
f (`q κ + k1 + k2 ) f (`q κ + k1 ) 6= fκ+ρ (`q κ + k1 + k2 ) fκ+ρ (`q κ + k1 )
is at most O(q λ−ρ ) where the implied constant may depend only on q and f .
We introduce a set of functions with uniformly small Discrete Fourier Transforms:
Definition 2. Given a non decreasing function γ : R → R satisfying limλ→+∞ γ(λ) = +∞ and
c > 0 we denote by Fγ,c the set of functions f : N → U such that for (κ, λ) ∈ N2 with κ ≤ cλ
and t ∈ R:
q −λ
(6)
X
f (uq κ ) e (−ut) ≤ q −γ(λ) .
0≤u<q λ
For example, for any
P α such that (q − 1)α ∈ R \ Z, it follows from [19, Lemme 9] that the
function f (n) = e(α i≥0 εi (n)) verifies Definition 1 and f ∈ Fγ,c in Definition 2 for any c > 0
and γ such that for λ ≥ 2
π2
2
π2
γ(λ) =
1−
k(q − 1)αk2 λ −
.
12 log q
q+1
48 log q
The goal of this paper is to present a new method which allows to prove a Prime Number
Theorem for a large class of sequences with digit properties including the Rudin-Shapiro sequence and some of its generalizations. Roughly speaking, we prove that if we control the carry
properties of a function f : N → U (Definition 1) for which the discrete Fourier transform is
uniformly small (Definition 2), then we have a Prime Number Theorem (Theorem 1) (and a
Möbius randomness principle (Theorem 2)) for f . This general result can be applied in many
situations. In part 11 we will apply it to the case of Rudin-Shapiro sequences.
Theorem 1. Let γ : R → R be a non decreasing function satisfying limλ→+∞ γ(λ) = +∞, and
f : N → U be a function satisfying Definition 1 and f ∈ Fγ,c for some c ≥ 10 in Definition 2.
Then for any ϑ ∈ R we have
(7)
X
n≤x
Λ(n)f (n) e (ϑn) c1 (q)(log x)c2 (q) x q −γ(2b(log x)/80 log qc)/20 ,
4
CHRISTIAN MAUDUIT AND JOËL RIVAT
1
with c1 (q) = max(τ (q), log2 q)1/4 (log q)−2− 4 max(ω(q),2) and c2 (q) =
9
4
+ 41 max(ω(q), 2).
Remark 1. Theorem 1 gives a non trivial result if
γ(λ)
20 c2 (q)
(8)
lim inf
>
.
λ→∞ log λ
log q
Corollary 1. Let b : N → N be such that, for any α ∈ R \ Q, the function f (n) = e(αb(n))
satisfies Definition 1 and f ∈ Fγ,c in Definition 2 for some c ≥ 10 and γ satisfying (8). Then
for any a ∈ Z, m ∈ N, m ≥ 1 with gcd(a, m) = 1, the sequence (αb(p))p∈P(a,m) is uniformly
distributed modulo 1 if and only if α ∈ R \ Q.
Corollary 2. Let b : N → N and (m, m0 ) ∈ N2 , m, m0 ≥ 1 be such that, for any integer j 0 ,
0
1 ≤ j 0 < m0 , the function f (n) = e( mj 0 b(n)) satisfies Definition 1 and f ∈ Fγ,c in Definition 2
for some c ≥ 10 and γ satisfying (8). Then for any (a, a0 ) ∈ Z2 such that gcd(a, m) = 1 we
have for x → +∞
π(x; a, m)
card{p ≤ x, p ∈ P(a, m), b(p) ≡ a0 mod m0 } = (1 + o(1))
.
m0
Corollary 3. Let b : N → N and (m, m0 ) ∈ N2 , m, m0 ≥ 1 be such that, for any integer j 0 ,
0
1 ≤ j 0 < m0 , the function f (n) = e( mj 0 b(n)) satisfies Definition 1 and f ∈ Fγ,c in Definition
2 for some c ≥ 10 and γ satisfying (8). Then for any (a, a0 ) ∈ Z2 such that gcd(a, m) = 1
the sequence (ϑp)p∈B(a,m,a0 ,m0 ) is uniformly distributed modulo 1 if and only if ϑ ∈ R \ Q, where
B(a, m, a0 , m0 ) = {p ∈ P(a, m), b(p) ≡ a0 mod m0 }.
P
In order to estimate sums of the form n Λ(n)F (n) in Theorem 1 by using a combinatorial
identity like Vaughan’s identity (see (13.39) of [13]), it is sufficient to estimate bilinear sums of
the form
XX
am bn F (mn)
m
n
(we have described this method in details in [20]). These sums are said of type I if bn is a
smooth function of n. Otherwise they are said of type II. The key of this approach is that for
type I sums the summation over the smooth variable n is of significant length, while for type
II sums both summations have a significant length.
Using (13.40) instead of (13.39) of [13] we obtain a similar result for the Möbius function µ
(a better exponent of the factor log x might be obtained with some extra work):
Theorem 2. Let γ : R → R be a non decreasing function satisfying limλ→+∞ γ(λ) = +∞,
c ≥ 10 and f : N → U be a function satisfying Definition 1 and f ∈ Fγ,c in Definition 2. Then
for any ϑ ∈ R we have
X
9
1
(9)
µ(n)f (n) e (ϑn) c1 (q) (log x) 4 + 4 max(ω(q),2) x q −γ(2b(log x)/80 log qc)/20 ,
n≤x
with c1 (q) and c2 (q) defined in Theorem 1.
3. Notations and preliminary lemmas
For a ∈ Z and κ ∈ N we denote by rκ (a) the unique integer r ∈ {0, . . . , q κ − 1} such that
a ≡ r mod q κ . More generally for integers 0 ≤ κ1 ≤ κ2 we denote by rκ1 ,κ2 (a) the unique integer
u ∈ {0, . . . , q κ2 −κ1 − 1} such that a j= kq κ2k+ uq κ1 + v for some v ∈ {0, . . . , q κ1 − 1} and k ∈ Z.
r (a)
We notice that we have rκ1 ,κ2 (a) = κq2κ1 and for any u ∈ {0, . . . , q κ2 −κ1 − 1},
a
u
u+1
(10)
rκ1 ,κ2 (a) = u ⇐⇒ κ2 ∈ κ2 −κ1 , κ2 −κ1 + Z.
q
q
q
For a ≥ 0, rκ (a) is the integer obtained from the κ least significant digits of a, while rκ1 ,κ2 (a) is
the integer obtained using the digits of a of indexes κ1 , . . . , κ2 − 1.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
5
The following lemma is a classical method to detect real numbers in an interval modulo 1
by means of exponential sums. For α ∈ R with 0 ≤ α < 1 we denote by χα the characteristic
function of the interval [0, α) modulo 1:
χα (x) = bxc − bx − αc .
(11)
Lemma 1. For all α ∈ R with 0 ≤ α < 1 and all integer H ≥ 1 there exist real valued
trigonometric polynomials Aα,H (x) and Bα,H (x) such that for all x ∈ R
|χα (x) − Aα,H (x)| ≤ Bα,H (x),
(12)
where
(13)
Aα,H (x) =
X
X
ah (α, H) e(hx), Bα,H (x) =
|h|≤H
bh (α, H) e(hx),
|h|≤H
with coefficients ah (α, H) and bh (α, H) satisfying
1
, |bh (α, H)| ≤
(14)
a0 (α, H) = α, |ah (α, H)| ≤ min α, π|h|
1
.
H+1
Proof. In order to apply Theorem 19 of [27] we need to normalize χα : let us define for all x ∈ R
1
t→0+ 2
χ
fα (x) = lim
(χα (x − t) + χα (x + t)) .
Applying (7.24) of [27] we get
|f
χα (x) − Aα,H (x)| ≤ Bα,H (x),
with the coefficients ah (α, H) and bh (α, H) defined by a0 (α, H) = α,
|h|
|h|
|h|
∗
∗
sin πhα
−hα
+
(15) ah (α, H) = ah (α, H) e 2 , ah (α, H) = πh
π H+1 1 − H+1 cot π H+1
|h|
H+1
and
(16)
bh (α, H) =
b∗h (α, H) e
−hα
2
b∗h (α, H)
,
=
1
H+1
1−
|h|
H+1
cos(πhα).
In order to see that Aα,H (x) is real valued we notice that a∗−h (α, H) = a∗h (α, H) and
Aα,H (x) = a0 (α, H) +
H
X
a∗h (α, H) e h x −
α
2
+ e −h x −
α
2
h=1
= a0 (α, H) + 2
H
X
a∗h (α, H) cos 2πh x −
α
2
.
h=1
Since Bα,H (x) ≥ 0, obviously Bα,H (x) is real valued. By the argument above we have
Bα,H (x) = b0 (α, H) + 2
H
X
b∗h (α, H) cos 2πh x −
α
2
.
h=1
Observing that for all x ∈ R we have χα (x) = lim χ
fα (x + t), we obtain (12).
t→0+
The upper bound of |ah (α, H)| given by (14) follows from Theorem 6 of [27] and the upper
bound of |bh (α, H)| given by (14) follows from (16).
In dimension 2, we can detect points in a square (modulo 1) using the following:
Lemma 2. For α1 , α2 ∈ [0, 1) and integers H1 ≥ 1, H2 ≥ 1, we have for all (x, y) ∈ R2
(17)
|χα1 (x)χα2 (y) − Aα1 ,H1 (x)Aα2 ,H2 (y)|
≤ χα1 (x)Bα2 ,H2 (y) + Bα1 ,H1 (x)χα2 (y) + Bα1 ,H1 (x)Bα2 ,H2 (y)
where Aα,H (.) and Bα,H (.) are the real valued trigonometric polynomials defined by (13).
6
CHRISTIAN MAUDUIT AND JOËL RIVAT
Proof. For (x, y) ∈ R2 we have
χα1 (x)χα2 (y) − Aα1 ,H1 (x)Aα2 ,H2 (y)
= χα1 (x)(χα2 (y) − Aα2 ,H2 (y)) + (χα1 (x) − Aα1 ,H1 (x))χα2 (y)
−(χα1 (x) − Aα1 ,H1 (x))(χα2 (y) − Aα2 ,H2 (y)).
Since χα1 (x) ≥ 0 and χα2 (y) ≥ 0, by (12) we get (17).
The following lemma is a generalization of van der Corput’s inequality.
Lemma 3. For all complex numbers z1 , . . . , zN and all integers k ≥ 1 and R ≥ 1 we have
!
2
X
X
X X
r
N + kR − k
zn ≤
(18)
|zn |2 + 2
1−
< (zn+kr zn )
R
R
1≤n≤N
1≤r<R
1≤n≤N
1≤n≤N −kr
where <(z) denotes the real part of z.
Proof. See for example Lemma 17 of [19].
We will often make use of the following upper bound of geometric series of ratio e(ξ) for
(L1 , L2 ) ∈ Z2 , L1 ≤ L2 and ξ ∈ R:
X
(19)
e(`ξ) ≤ min(L2 − L1 , |sin πξ|−1 ).
L1 <`≤L2
Lemma 4 and Lemma 5 allow to estimate on average the minimums arising from (19).
Lemma 4. Let (a, m) ∈ Z2 with m ≥ 1 and b ∈ R. For any real number U > 0 we have
X
−1
(20)
min U, sin π an+b
gcd(a, m) U + m log m.
m
0≤n≤m−1
Proof. It follows from Lemma 6 of [20].
Lemma 5. Let m ≥ 1 and A ≥ 1 be integers and b ∈ R. For any real number U > 0 we have
1 X X
−1
(21)
min U, sin π an+b
τ (m) U + m log m.
m
A 1≤a≤A 0≤n<m
Proof. By (20) it is enough to observe that
X
X X
X X
X A
gcd(a, m) =
d
1≤
d
1=
d
≤ A τ (m),
d
1≤a≤A
1≤a≤A
1≤a≤A
d|m
d≤A
(a,m)=d
d|m
d≤A
d|m
d≤A
d|a
which implies (21).
The following lemma is a classical application of the large sieve inequality:
Lemma 6. For any complex numbers z1 ,. . . ,zN and Q > 0 we have
2
q
N
N
X X
X
X
an
2
(22)
zn e
≤ (N − 1 + Q )
|zn |2 .
q
n=1
n=1
q≤Q a=1
(a,q)=1
Proof. See Theorem 3 and Section 8 of [21].
Let f : N → U, λ ∈ N and fλ defined by (4). The Discrete Fourier Transform of fλ is defined
for t ∈ R by
X
X
ut
1
ut
1
fλ (u) e − λ = λ
f (u) e − λ .
(23)
fbλ (t) = λ
q
q
q
q
λ
λ
0≤u<q
0≤u<q
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
7
For λ ∈ N and t ∈ R we have
2
X
(24)
fbλ (h + t)
= 1.
0≤h<q λ
so that, if f satisfies (6), then
2
1=
X
0≤h<q λ
X
q −λ
f (uq κ ) e −
0≤u<q λ
u(h + t)
qλ
X
≤
q −2γ(λ) = q λ−2γ(λ)
0≤h<q λ
and
γ(λ) ≤
(25)
λ
.
2
4. Carry propagation lemmas
0
Lemma 7. Let µ ≥ 1, ν ≥ 1, µ0 ≥ 1 be integers with µ0 ≤ µ + ν. For B ⊆ {0, . . . , q µ+ν−µ − 1},
0
the number N of pairs (m, n) ∈ {q µ−1 , . . . , q µ − 1} × {q ν−1 , . . . , q ν − 1} such that mn = a + q µ b
0
with 0 ≤ a < q µ and b ∈ B satisfies
0
0
N ≤ q µ log q + q µ − q µ−1 + q µ −µ+1 card B.
0
Proof. For each m ∈ {q µ−1 , . . . , q µ − 1}, the number Nm of n such that mn = a + q µ b with
0
0 ≤ a < q µ and b ∈ B satisfies
X
0
0
card{a : 0 ≤ a < q µ , a + q µ b ≡ 0 mod m}.
Nm ≤
b∈B
This gives
Nm ≤
X
b∈B
0
qµ
1+
m
0
qµ
= 1+
card B.
m
It follows
X
N =
Nm ≤
q µ−1 ≤m<q µ
X
q µ−1 ≤m<q µ
0
qµ
1+
m
so that
µ
N ≤ (q − q
µ−1
) card B + q
µ0
1
q µ−1
Z
qµ
+
and the result follows.
q µ−1
card B,
dt
t
card B
Lemma 8. If f : N → U satifies Definition 1 then for (µ, ν, ρ) ∈ N3 with 2ρ < ν the set E
of pairs (m, n) ∈ {q µ−1 , . . . , q µ − 1} × {q ν−1 , . . . , q ν − 1} such that there exists k < q µ+ρ with
f (mn + k) f (mn) 6= fµ+2ρ (mn + k) fµ+2ρ (mn) satisfies
(26)
card E (log q) q µ+ν−ρ .
Proof. Applying Definition 1 with λ = ν − ρ and κ = µ + ρ, let B be the set of ` < q ν−ρ
such that there exists (k1 , k2 ) ∈ {0, . . . , q κ − 1}2 for which (5) is true. By Definition 1 we have
card B q ν−2ρ . We need to count the pairs (m, n) ∈ {q µ−1 , . . . , q µ − 1} × {q ν−1 , . . . , q ν − 1}
0
such that mn is of the form mn = k1 + q µ ` with ` ∈ B. Applying Lemma 7 with µ0 = µ + ρ we
get
card E q µ+ρ log q + q µ − q µ−1 + q ρ+1 card B (log q) q µ+ν−ρ ,
which gives (26).
8
CHRISTIAN MAUDUIT AND JOËL RIVAT
Lemma 9. Let f : N → U satisfying Definition 1 and (µ, ν, µ0 , µ1 , µ2 ) ∈ N5 with µ0 ≤ µ1 ≤
µ ≤ µ2 , µ ≤ ν and 2 (µ2 − µ) ≤ µ0 . For (a, b, c) ∈ N3 the set E(a, b, c) of pairs (m, n) ∈
{q µ−1 , . . . , q µ − 1} × {q ν−1 , . . . , q ν − 1} such that
fµ2 (mn + am + bn + c) fµ2 (q µ0 rµ0 ,µ2 (mn + am + bn + c))
6= fµ1 (mn + am + bn + c) fµ1 (q µ0 rµ0 ,µ2 (mn + am + bn + c))
satisfies
ω(q) µ+ν+µ0 −µ1
card E(a, b, c) max(τ (q), log q) µ2
(27)
q
.
Proof. Let B be the set of ` ∈ {0, . . . , q µ2 −µ0 −1} for which there exists (k1 , k2 ) ∈ {0, . . . , q µ0 −1}2
with
fµ2 (q µ0 ` + k1 + k2 ) fµ2 (q µ0 ` + k1 ) 6= fµ1 (q µ0 ` + k1 + k2 ) fµ1 (q µ0 ` + k1 ).
For 0 ≤ ` ≤ q µ2 −µ0 −2 we have 0 ≤ q µ0 `+k1 +k2 ≤ q µ2 −2. Therefore we have fµ2 (q µ0 `+k1 +k2 ) =
f (q µ0 ` + k1 + k2 ) and fµ2 (q µ0 ` + k1 ) = f (q µ0 ` + k1 ) except possibly if ` = q µ2 −µ0 − 1. Since
f satisfies Definition 1 it follows that card B = O(q µ2 −µ0 −(µ1 −µ0 ) ) = O(q µ2 −µ1 ). Observing for
k = mn+am+bn+c that k = r0,µ0 (k)+q µ0 rµ0 ,µ2 (k)+q µ2 k 0 , we notice that E(a, b, c) ⊆ E 0 (a, b, c)
where E 0 (a, b, c) is the set of pairs (m, n) such that rµ0 ,µ2 (mn + am + bn + c) ∈ B. Then we can
write
X
card E 0 (a, b, c) =
card{(m, n), rµ0 ,µ2 (mn + am + bn + c) = `},
`∈B
which by (10) and (11) can be written
XXX
mn + am + bn + c
`
0
card E (a, b, c) =
χqµ0 −µ2
− µ2 −µ0 .
q µ2
q
n
`∈B m
Using Lemma 1 it follows that for any integer H ≥ 1 there exists ah (q µ0 −µ2 , H) and bh (q µ0 −µ2 , H)
satisfying (14) such that
XXX X
h(mn + am + bn + c)
h`
0
µ0 −µ2
card E (a, b, c) ≤
ah (q
, H) e
− µ2 −µ0
µ2
q
q
m
n
`∈B
|h|≤H
XXX X
h`
h(mn + am + bn + c)
µ0 −µ2
+
bh (q
− µ2 −µ0 .
, H) e
q µ2
q
n
`∈B m
|h|≤H
Taking H = q
µ2 −µ0
the contribution of the terms h = 0 in both sums is bounded by
q µ+ν+µ0 −µ2 card B q µ+ν+µ0 −µ1 .
We handle both sums over h similarly, exchanging the order of summations and using the
bounds |ah (q µ0 −µ2 , H)| ≤ q µ0 −µ2 and |bh (q µ0 −µ2 , H)| ≤ H −1 = q µ0 −µ2 . We obtain the upper
bound
X
X X h(mn + am + bn + c) card B
0
µ+ν+µ0 −µ1
e
.
card E (a, b, c) q
+ µ2 −µ0
q
q µ2
µ −µ
n
m
1≤|h|≤q
2
0
This gives
card E 0 (a, b, c) q µ+ν+µ0 −µ1
q µ2 −µ1
+ µ2 −µ0
q
X
X
1≤|h|≤q µ2 −µ0
n
h(n + a)
min q µ , sin π
q µ2
−1
!
The summation on n runs over at most dq ν−µ2 e periods modulo q µ2 . By (21) it follows
card E 0 (a, b, c) q µ+ν+µ0 −µ1 + q µ2 −µ1 (q ν−µ2 + 1) (τ (q µ2 )q µ + q µ2 log q µ2 ) .
ω(q)
By multiplicativity of the function τ we have τ (q µ2 ) ≤ τ (q)µ2 and we obtain
ω(q)
card E 0 (a, b, c) q µ+ν+µ0 −µ1 1 + q −µ0 (1 + q µ2 −ν ) τ (q)µ2 + q µ2 −µ log q µ2
.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
9
and using µ1 ≤ µ ≤ ν we may replace µ and ν by µ1 in the parentheses and this leads to
ω(q)
card E 0 (a, b, c) q µ+ν+µ0 −µ1 1 + µ2 max(τ (q), log q)q 2µ2 −2µ−µ0
which, using the hypothesis 2 (µ2 − µ) ≤ µ0 , gives (27).
5. Sums of type I
We take a non decreasing function γ : R → R satisfying limλ→+∞ γ(λ) = +∞, c ≥ 2 and
f : N → U be a function satisfying Definition 1 and f ∈ Fγ,c in Definition 2. Let
1 ≤ M ≤ N such that M ≤ (M N )1/3 .
(28)
Let µ and ν be the unique integers such that
q µ−1 ≤ M < q µ
and q ν−1 ≤ N < q ν .
Let ϑ ∈ R, an interval I(M, N ) ⊆ [0, M N ] and
X
SI (ϑ) =
M
q
X
f (mn) e(ϑmn) .
n
<m≤M mn∈I(M,N
)
Proposition 1. Assuming (28) and with c ≥ 2, we have uniformly for ϑ ∈ R
1
SI (ϑ) (log q)5/2 (µ + ν)2 q µ+ν− 2 γ(
(29)
µ+ν
)
3
.
Proof. Let 0 ≤ ` < q µ+ν . For Mq < m ≤ M , we have ` = mn with mn ∈ I(M, N ) if and only if
` ∈ I(M, N ) and ` ≡ 0 mod m. Therefore the inner sum (over n) in SI (ϑ) is
X
X
X
h(u − `) 1 X
k`
1
e
e
f (`) e(ϑ`)
µ+ν
µ+ν
q
q
m 0≤k<m
m
µ+ν
µ+ν
u∈I(M,N )
0≤`<q
0≤h<q
i.e.

X
X
e

0≤h<q µ+ν
u∈I(M,N )

hu  1 X 1
q µ+ν
m 0≤k<m q µ+ν
X
f (`) e ϑ` −
0≤`<q µ+ν
h`
q µ+ν
This gives
SI (ϑ) ≤
X
min q
µ+ν
πh
−1
, sin qµ+ν
SI0 h − ϑq µ+ν ,
0≤h<q µ+ν
where
(30)
X
SI0 (ϑ0 ) =
M
q
<m≤M
1 X d 0
fµ+ν ϑ −
m 0≤k<m
Then we have uniformly for ϑ ∈ R
0
0
SI (ϑ) ≤ max
SI (ϑ )
0
ϑ ∈R
X
min q
µ+ν
k µ+ν
q
m
πh
.
−1
, sin qµ+ν
0≤h<q µ+ν
which gives
(31)
SI (ϑ) max
SI0 (ϑ0 )
ϑ0 ∈R
q µ+ν log q µ+ν .
It remains to estimate SI0 (ϑ0 ) uniformly for ϑ0 ∈ R. For any κ such that
(32)
1 ≤ κ ≤ 32 (µ + ν),
,
k`
+
m
.
10
CHRISTIAN MAUDUIT AND JOËL RIVAT
by (23) we can write
fd
µ+ν (t) =
1
X
q µ+ν
(u + vq κ )t
f (u + vq ) e −
q µ+ν
µ+ν−κ
X
0≤u<q κ 0≤v<q
κ
.
This gives
1
fd
µ+ν (t) =
f (vq ) e −
X
q µ+ν−κ
vt
κ
q µ+ν−κ
ut
1 X
κ
κ
f (u + vq )f (vq ) e − µ+ν .
q κ 0≤u<qκ
q
0≤v<q µ+ν−κ
Given ρ1 such that
1 ≤ ρ1 ≤ µ + ν − κ,
(33)
by Definition 1 the number of v ∈ {0, . . . , q µ+ν−κ − 1} such that there exists u ∈ {0, . . . , q κ − 1}
for which
f (u + vq κ )f (vq κ ) 6= fκ+ρ1 (u + vq κ )fκ+ρ1 (vq κ )
fκ of pairs (u, v) with this property satisfies
is at most O(q µ+ν−κ−ρ1 ). Hence the set W
fκ q µ+ν−ρ1 .
card W
(34)
Therefore for all t ∈ R, all κ satisfying (32), all ρ1 satisfying (33) we have
(35)
fd
µ+ν (t) = Gκ,1 (t) + Gκ,2 (t),
with
Gκ,1 (t) =
1
f (vq ) e −
X
q µ+ν−κ
κ
0≤v<q µ+ν−κ
vt
q µ+ν−κ
1 X
ut
κ
κ
fκ+ρ1 (u + vq )fκ+ρ1 (vq ) e − µ+ν
q κ 0≤u<qκ
q
and
Gκ,2 (t) =
1
X
(u + vq κ )t
f (vq ) e −
q µ+ν
κ
q µ+ν
fκ
(u,v)∈W
f (u + vq κ )f (vq κ ) − fκ+ρ1 (u + vq κ )fκ+ρ1 (vq κ ) .
Let us introduce in Gκ,1 (t) the residue w of v mod q ρ1 in order to make the variables u and v
independent:
X
X
1
vt
1 X
v−w
κ
Gκ,1 (t) =
f (vq ) e − µ+ν−κ
e h ρ1
q µ+ν−κ
q
q ρ1 0≤h<qρ1
q
0≤w<q ρ1
0≤v<q µ+ν−κ
1 X
ut
fκ+ρ1 (u + wq κ )fκ+ρ1 (wq κ ) e − µ+ν .
κ
q 0≤u<qκ
q
Writing
(36)
1
cκ,ρ1 (u, h) = ρ1
q
hw
fκ+ρ1 (u + wq )fκ+ρ1 (wq κ ) e − ρ1 ,
q
ρ
0≤w<q 1
X
κ
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
11
this leads to
!
1 X
−ut
cκ,ρ1 (u, h) e µ+ν
q κ 0≤u<qκ
q
X
Gκ,1 (t) =
0≤h<q ρ1

1

κ
f (vq ) e −
X
q µ+ν−κ
0≤v<q µ+ν−κ
vt
q µ+ν−κ

hv
+ ρ1  .
q
By (32) we have κ ≤ 2(µ + ν − κ) and we may use (6) (with c ≥ 2) for the sum over v with
κ = κ and λ = µ + ν − κ. We get
X
1 X
−ut
−γ(µ+ν−κ)
|Gκ,1 (t)| q
cκ,ρ1 (u, h) e µ+ν .
q κ 0≤u<qκ
q
0≤h<q ρ1
From (30) we can write
X
SI0 (ϑ0 ) ≤
X
1≤d≤M
SI0 (ϑ0 ),
In order to estimate
unique integer such that
M
q
<m≤M
1 X d 0
fµ+ν ϑ −
m 0≤k<m
k µ+ν
q
m
.
(k,m)=d
for each 1 ≤ d ≤ M , we will use (35) with κd defined to be the
q κd −1 < M 2 /d2 ≤ q κd .
(37)
Hence
0
0
SI0 (ϑ0 ) ≤ SI,1
(ϑ0 ) + SI,2
(ϑ0 ),
with
X
0
SI,1
(ϑ0 ) =
1≤d≤M
X
M
q
<m≤M
and
0
SI,2
(ϑ0 ) =
X
1≤d≤M
X
M
q
<m≤M
1 X
Gκd ,1 ϑ0 −
m 0≤k<m
k µ+ν
q
m
(k,m)=d
1 X
Gκd ,2 ϑ0 −
m 0≤k<m
k µ+ν
q
m
.
(k,m)=d
Then
k µ+ν
0
≤ q −γ(µ+ν−κd )
Gκd ,1 ϑ − q
m
X
0≤h<q ρ1
1
q κd
uϑ0
uk
cκd ,ρ1 (u, h) e − µ+ν +
.
q
m
κ
0≤u<q d
X
Since κd is decreasing with d, by (37) and (28) we can check that (32) is satisfied:
2
(38)
1 ≤ κd ≤ κ1 ≤ 2µ ≤ (µ + ν).
3
But γ is non decreasing, which implies
00
X SI,1
(M, d)
0
0
−γ( µ+ν
)
3
(39)
SI,1 (ϑ ) ≤ q
,
d q κd
1≤d≤M
where
00
SI,1
(M, d)
=
X
X
0≤h<q ρ1
M
<m0 ≤ M
qd
d
1
m0
X
0≤k0 <m0
(k0 ,m0 )=1
uϑ0
uk 0
cκd ,ρ1 (u, h) e − µ+ν + 0 .
q
m
κ
0≤u<q d
X
Since
(40)
X
X
M
<m0 ≤ M
qd
d
0≤k0 <m0
(k0 ,m0 )=1
1
≤
m02
X
M
<m0 ≤ M
qd
d
1
log q,
m0
12
CHRISTIAN MAUDUIT AND JOËL RIVAT
by the Cauchy-Schwarz inequality we get
2
00
SI,1
(M, d)
(log q) q
X
ρ1
X
0≤h<q ρ1 M <m0 ≤ M
qd
d
uϑ0
uk 0
cκd ,ρ1 (u, h) e − µ+ν + 0
q
m
0≤u<q κd
X
X
0≤k0 <m0
(k0 ,m0 )=1
2
.
By (22) it follows that
00
SI,1
(M, d)
2
X (log q)q ρ1
M2
d2
q κd +
X
|cκd ,ρ1 (u, h)|2 .
0≤u<q κd
0≤h<q ρ1
But by (36) and (24) we have
X
|cκd ,ρ1 (u, h)|2 = 1,
0≤h<q ρ1
hence summing over u and using (37) we obtain
ρ1
00
SI,1
(M, d) (log q)1/2 q κd + 2 ,
which lead by (39) to
0
(ϑ0 )
SI,1
(41)
(log q)
1/2 −γ( µ+ν
)
3
q
X q ρ21
ρ1
µ+ν
µ (log q)3/2 q 2 −γ( 3 ) .
d
1≤d≤M
0
In order to estimate SI,2
(ϑ0 ) we denote by Wκd the set of integers w = u + vq κd such that
fκ . Using the bijective correspondence between Wκ and W
fκ given by w 7→
(u, v) ∈ W
d
d
d
(rκd (w), rκd ,µ+ν (w)) we can write
X
1
wt
0
Gκd ,2 (t) = µ+ν
cκd ,ρ1 (w) e − µ+ν ,
q
q
µ+ν
0≤w<q
where
c0κd ,ρ1 (w) = f (q κd rκd ,µ+ν (w)) f (w)f (q κd rκd ,µ+ν (w)) − fκd +ρ1 (w)fκd +ρ1 (q κd rκd ,µ+ν (w))
satisfies c0κd ,ρ1 (w) ≤ 2 for 0 ≤ w < q µ+ν and c0κd ,ρ1 (w) = 0 for w 6∈ Wκd . Then we write
00
X SI,2
(M, d)
0
,
(42)
SI,2
(ϑ0 ) ≤
µ+ν
d
q
1≤d≤M
where
00
SI,2
(M, d)
=
X
M
<m0 ≤ M
qd
d
1
m0
X
X
c0κd ,ρ1 (w) e
0≤k0 <m0 0≤w<q µ+ν
(k0 ,m0 )=1
wk 0
wϑ0
− µ+ν + 0
.
q
m
It follows from the Cauchy-Schwarz inequality and (40) that
2
00
SI,2
(M, d)
2
(log q)
X
X
M
0≤k0 <m0
<m0 ≤ M
qd
d
(k0 ,m0 )=1
wϑ0
wk 0
c0κd ,ρ1 (w) e − µ+ν + 0
q
m
µ+ν
X
0≤w<q
By (22) we get
00
SI,2
(M, d)
2
(log q) q µ+ν +
M2
d2
M2
X
X
c0κd ,ρ1 (w)
0≤w<q µ+ν
(log q) q µ+ν +
d2
22 .
w∈Wκd
By (34) it follows that
ρ1
00
SI,2
(M, d) (log q)1/2 q µ+ν− 2 ,
2
.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
13
which lead by (39) to
0
(ϑ0 ) (log q)1/2
SI,2
(43)
X q −ρ1 /2
µ (log q)3/2 q −ρ1 /2 .
d
1≤d≤M
Taking
(44)
ρ1 = γ
µ+ν
3
,
by (25) we have ρ1 ≤ µ+ν
, so that by (38) ρ1 satisfies (33). By (31), (41) and (43) it follows
6
that uniformly for ϑ ∈ R we get (29).
6. Sums of type II
We take γ : R → R a non decreasing function satisfying limλ→+∞ γ(λ) = +∞, c ≥ 10 (this
condition appears in (95)) and f : N → U a function satisfying Definition 1 and f ∈ Fγ,c in
Definition 2. Let 1 ≤ M ≤ N . We denote by µ and ν the unique integers such that
q µ−1 ≤ M < q µ
and q ν−1 ≤ N < q ν .
Let us assume that
1
(µ
4
(45)
+ ν) ≤ µ ≤ ν ≤ 34 (µ + ν)
(replacing (1/4, 3/4) by (ξ, 1 − ξ) with 1/4 < ξ < 1/3 would provide a better exponent for
q in (46)). We assume also that the multiplicative dependence of the variables in the type II
sums has been removed by the classical method described (for example) in section 5 of [20].
Let ϑ ∈ R, am ∈ C, bn ∈ C with |am | ≤ 1, |bn | ≤ 1 and
XX
am bn f (mn) e(ϑmn)
SII (ϑ) =
m
n
where we sum over m ∈ (M/q, M ] and n ∈ (N/q, N ]. We will prove
Proposition 2. Assuming (45) and c ≥ 10, uniformly for |am | ≤ 1, |bn | ≤ 1 and ϑ ∈ R, we
have
(46)
1
|SII (ϑ)| max(τ (q) log q, log3 q)1/4 (µ + ν) 4 (1+max(ω(q),2)) q µ+ν−γ(2bµ/15c)/20 .
As often in this approach, the proof of this result is the most difficult part. The proof is quite
long and complicated and will be developped over several sections and completed at formula
(96). By the Cauchy-Schwarz inequality
2
2
|SII (ϑ)| ≤ M
X X
m
bn f (mn) e(ϑmn) .
n
Let ρ be an integer such that
(47)
1 ≤ 7ρ ≤ µ
and let
(48)
R = qρ
so that
(49)
1 ≤ R N.
Applying Lemma 3 to the summation over n with k = 1 and then summing over m we get
M 2N 2 M N X r
2
|SII (ϑ)| +
1−
<(S1 (r)),
R
R 1≤r<R
R
14
CHRISTIAN MAUDUIT AND JOËL RIVAT
with
S1 (r) =
X X
bn+r bn f (mn + mr)f (mn) e(ϑmr),
m n∈I(N,r)
where I(N, r) = (N/q, N − r]. Let
(50)
µ2 = µ + 2ρ.
If f satisfies the carry property explained in Definition 1, then by Lemma 8 the number of pairs
(m, n) for which f (mn + mr)f (mn) 6= fµ2 (mn + mr)fµ2 (mn) is O(q µ+ν−ρ ). Hence
S1 (r) = S10 (r) + O(q µ+ν−ρ ),
where
S10 (r) =
X X
bn+r bn fµ2 (mn + mr)fµ2 (mn) e(ϑmr).
m n∈I(N,r)
Using again the Cauchy-Schwarz inequality for the summation over r, this leads to
X
M 4N 4 M 2N 2
2
(51)
|SII (ϑ)|4 +
R
|S10 (r)| .
2
2
R
R
1≤r<R
It remains to give an upper bound for |S10 (r)|2 . We reverse the order of summation in S10 (r)
and obtain:
X X
fµ2 (mn + mr)fµ2 (mn) e(ϑmr) .
|S10 (r)| ≤
n∈I(N,r)
m
We may extend the summation over n to (N/q, N ] and apply the Cauchy-Schwarz inequality:
2
2
|S10 (r)| N
X
X
N/Q<n≤N
m
fµ2 (mn + mr)fµ2 (mn) e(ϑmr) .
Applying to the summation over m the Lemma 3 with positive integers k = q µ1 and S such
that
1 ≤ q µ1 S M
(52)
and then summing over n and r we get
X
M 2N 2R M N
2
0
(53)
|S1 (r)| +
<(S2 ),
S
S
1≤r<R
with
S2 =
X
X 1≤r<R 1≤s<S
1−
s
e(ϑq µ1 rs) S20 (r, s)
S
and
S20 (r, s) =
XX
m
fµ2 ((m + sq µ1 )(n + r))fµ2 (m(n + r))fµ2 ((m + sq µ1 )n)fµ2 (mn).
n
Using (51) and (53) we obtain uniformly for ϑ ∈ R:
|SII (ϑ)|4 (54)
M 4N 4 M 4N 4 M 3N 3 X X
+
+
|S 0 (r, s)| .
R2
S
RS 1≤r<R 1≤s<S 2
Writing fµ2 = fµ1 (fµ2 fµ1 ) and observing that fµ1 ((m + sq µ1 )(n + r)) = fµ1 (m(n + r)) and
fµ1 ((m + sq µ1 )n) = fµ1 (mn)) we get
XX
S20 (r, s) =
fµ1 ,µ2 (mn+mr+q µ1 sn+q µ1 rs)fµ1 ,µ2 (mn + mr)fµ1 ,µ2 (mn + q µ1 sn)fµ1 ,µ2 (mn),
m
n
with
(55)
fµ1 ,µ2 = fµ2 fµ1 .
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
15
We take
µ1 = µ − 2ρ,
(56)
and
S = R2 = q 2ρ ,
(57)
so that the condition (52) is fullfilled.
For 0 ≤ r < R and 0 ≤ s < S and µ0 ≤ µ1 let us denote by Eµ0 ,µ1 ,µ2 (r, s) the set of pairs
(m, n) with M/q < m ≤ M and N/q < n ≤ N such that
fµ1 ,µ2 (mn + q µ1 sn + q µ1 rs) 6= fµ1 ,µ2 (q µ0 rµ0 ,µ2 (mn + q µ1 sn + q µ1 rs)).
The set Eµ0 ,µ1 ,µ2 (r, s) is a set of exceptions: if µ0 is taken sufficiently small, the function fµ1 ,µ2
will depend on the digits of index in µ0 , . . . , µ2 − 1, except for (m, n) ∈ Eµ0 ,µ1 ,µ2 (r, s). Of course
if µ0 = 0 we have Eµ0 ,µ1 ,µ2 (r, s) = ∅ but we want to choose µ0 more carefully so that this set is
still small enough. More precisely, let ρ0 ∈ N to be chosen later such that
0 ≤ ρ0 ≤ ρ.
(58)
Since f : N → U is a function satisfying Definition 1, we have by taking
µ0 = µ1 − 2ρ0
(59)
in Lemma 9:
0
card Eµ0 ,µ1 ,µ2 (r, s) max(τ (q), log q) (µ + ν)ω(q) q µ+ν−2ρ .
(60)
Remark 2. A direct argument depending on a better knowledge of f might permit to choose a
greater value of µ0 , leading to a sharper final estimate for such a more specific function f .
For k ∈ Z we define a q µ2 −µ0 -periodic function g by
g(k) = fµ1 ,µ2 (q µ0 k).
(61)
Let us put rµ0 ,µ2 (mn) = u0 so that
rµ0 ,µ2 (mn + q µ1 sn) = rµ0 ,µ2 (q µ0 u0 + q µ1 sn) = rµ2 −µ0 (u0 + q µ1 −µ0 sn)
and
fµ1 ,µ2 (q µ0 rµ0 ,µ2 (mn + q µ1 sn)) = g(u0 + q µ1 −µ0 sn).
Similarly if we put rµ0 ,µ2 (mn + mr) = u1 then we have
rµ0 ,µ2 (mn + mr + q µ1 sn + q µ1 sr) = rµ0 ,µ2 (q µ0 u1 + q µ1 sn + q µ1 sr)
= rµ2 −µ0 (u1 + q µ1 −µ0 sn + q µ1 −µ0 sr)
and
fµ1 ,µ2 (q µ0 rµ0 ,µ2 (mn + mr + q µ1 sn + q µ1 sr)) = g(u1 + q µ1 −µ0 sn + q µ1 −µ0 sr).
Using (60) and (10), we can write
0
S20 (r, s) = S3 (r, s) + O(max(τ (q), log q) (µ + ν)ω(q) q µ+ν−2ρ ),
(62)
where
S3 (r, s) =
XX
m
n
X
0≤u0 <q µ2 −µ0
0≤u1 <q µ2 −µ0
χqµ0 −µ2
u0
mn
−
µ
µ
q 2
q 2 −µ0
χqµ0 −µ2
u1
mn + mr
−
µ
µ
q 2
q 2 −µ0
g(u1 + q µ1 −µ0 sn + q µ1 −µ0 sr) g(u1 ) g(u0 + q µ1 −µ0 sn) g(u0 ),
with χqµ0 −µ2 defined by (11) and α = q µ0 −µ2 . Let H be an integer with q µ2 −µ0 ≤ H ≤ q µ to be
chosen later. Using (17) with α1 = α2 = q µ0 −µ2 we have
(63)
S3 (r, s) = S4 (r, s) + O(E4 (r, 0)) + O(E4 (0, r0 )) + O(E40 (r)),
16
CHRISTIAN MAUDUIT AND JOËL RIVAT
where
S4 (r, s) =
XX
m
n
X
g(u1 + q µ1 −µ0 sn + q µ1 −µ0 sr) g(u1 )g(u0 + q µ1 −µ0 sn) g(u0 )
0≤u0 <q µ2 −µ0
0≤u1 <q µ2 −µ0
E4 (r, r0 ) =
mn
u0
mn + mr
u1
Aqµ0 −µ2 ,H
− µ2 −µ0 Aqµ0 −µ2 ,H
− µ2 −µ0 ,
q µ2
q
q µ2
q
X
u0
mn + mr0
u1
mn + mr
− µ2 −µ0 χqµ0 −µ2
− µ2 −µ0 ,
Bqµ0 −µ2 ,H
q µ2
q
q µ2
q
µ −µ
XX
m
n
0≤u0 <q 2 0
0≤u1 <q µ2 −µ0
and
E40 (r)
=
XX
m
n
X
Bqµ0 −µ2 ,H
0≤u0 <q µ2 −µ0
0≤u1 <q µ2 −µ0
6.1. Estimate of E4 (r, r0 ). Since
X
u0
mn
−
µ
µ
q 2
q 2 −µ0
χqµ0 −µ2
mn+mr0
q µ2
−
Bqµ0 −µ2 ,H
u1
q µ2 −µ0
mn + mr
u1
−
µ
µ
q 2
q 2 −µ0
.
= 1,
0≤u1 <q µ2 −µ0
we have
E4 (r, r0 ) =
XX
m
X
Bqµ0 −µ2 ,H
mn+mr
q µ2
−
u0
q µ2 −µ0
,
0≤u0 <q µ2 −µ0
n
which by (13) gives
0
E4 (r, r ) =
X
bh0 (q
µ0 −µ2
, H)
XX
m
|h0 |≤H
By (14) we have |bh0 (q µ0 −µ2 , H)| ≤
1
.
H
n
X
e
h0 mn+mr
q µ2
u0
− h0 qµ2 −µ0 .
0≤u0 <q µ2 −µ0
For h0 6≡ 0 mod q µ2 −µ0 we have
0, while for h0 ≡ 0 mod q µ2 −µ0 this sum is equal to q µ2 −µ0 . Hence
P
e
with
(65)
q µ2 −µ0
E5 =
H
X
|h00 |≤H/qµ2 −µ0
X X h0 mn 0
e
.
µ0
q
m
n
After summation over n, we have
q µ2 −µ0
E5 H
X
X
0
min N, sin π hqµm0
−1
The summation over m runs over at most q µ−µ0 periods q µ0 , hence
µ2 −µ0
X
X
0 0
µ−µ0 q
min N, sin π hqµm0
E5 q
H
µ −µ 0≤m0 <q µ0
0
2
0
Using the trivial estimate for h0 = 0 and (21) when h0 6= 0, we obtain
E5 q µ+ν
q µ2 −µ0
+ q µ−µ0 (τ (q µ0 )N + q µ0 log q µ0 ) .
H
Choosing
(66)
.
|h0 |≤H/q µ2 −µ0 m
|h |≤H/q
−h0 u0
q µ2 −µ0
0≤u0 <q µ2 −µ0
writing h0 = h00 q µ2 −µ0
|E4 (r, r0 )| E5 ,
(64)
H = q µ2 −µ0 +2ρ ,
this gives
E5 q µ+ν−2ρ + q µ+ν−µ0 τ (q µ0 ) + q µ log q µ0 .
−1
.
=
we get
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
17
By (47), (59) and (45), we have µ0 ≥ µ − 4ρ ≥ 2ρ and ν ≥ 2ρ so that
E5 max(log q µ0 , τ (q µ0 )) q µ+ν−2ρ .
(67)
6.2. Estimate of E40 (r). We have
XX
X X
E40 (r) =
bh0 (q µ0 −µ2 , H)bh1 (q µ0 −µ2 , H)
m
|h0 |≤H |h1 |≤H
X
0≤u0 <q µ2 −µ0
0≤u1 <q µ2 −µ0
We observe that for h0 6≡ 0 mod q
n
mn
u0
mn + mr
u1
e h0 µ2 − h0 µ2 −µ0 e h1
− h1 µ2 −µ0 .
q
q
q µ2
q
µ2 −µ0
we have
X
u0
e −h0 qµ2 −µ0
= 0 and similarly for
0≤u0 <q µ2 −µ0
µ2 −µ0
h1 . Hence we may assume h0 ≡ h1 ≡ 0 mod q
. Writing h0 = h00 q µ2 −µ0 and h1 = h01 q µ2 −µ0
and using the upper bound |bh (q µ0 −µ2 , H)| ≤ H1 from (14) we get
X
X
X X (h0 + h0 )mn + h0 mr q 2(µ2 −µ0 )
0
1
1
0
e
.
|E4 (r)| µ0
H2
q
m
n
0
0
µ
−µ
µ
−µ
|h0 |≤H/q 2 0 |h1 |≤H/q 2 0
The contribution to E40 (r) of the terms for which h00 + h01 = 0, after summation over m, is
bounded by
X
−1
q 2(µ2 −µ0 )
h01 r
N
min M, sin π qµ0
.
H2
0
µ
−µ
|h1 |≤H/q 2 0
ρ
µ
Since 1 ≤ r < q and H ≤ q , we have |h01 r| < q µ−µ2 +µ0 +ρ = q µ0 −ρ (by (50)) so that the values
of h01 r are all distinct modulo q µ0 . Therefore we conclude that the contribution to E40 (r) of the
terms for which h00 + h01 = 0 is bounded by
q 2(µ2 −µ0 ) µ+ν
q 2(µ2 −µ0 )
µ0
µ0
N (M + q log q ) q (1 + q µ0 −µ log q µ0 ).
2
2
H
H
0
The contribution to E4 (r) of the terms for which h00 + h01 6= 0, after summation over n, is
bounded by
−1
q 2(µ2 −µ0 ) X X
(h00 +h01 )m
min N, sin π qµ0
H2
0
0
m
h0 +h1 6=0
which, writing h0 = h00 + h01 , is less than
X
q µ2 −µ0
H
X
0
min N, sin π hqµm0
−1
.
1≤|h0 |≤2H/q µ2 −µ0 m
The summation over m runs over at most q µ−µ0 periods q µ0 , which gives the upper bound
µ2 −µ0
X
X
−1
µ−µ0 q
h0 m0
min N, sin π qµ0
q
H
µ −µ 0≤m0 <q µ0
0
1≤h ≤2H/q
2
0
Using (21) this is bounded by
q µ−µ0 (τ (q µ0 )N + q µ0 log q µ0 ) q ν+µ−µ0 τ (q µ0 ) + q µ log q µ0 .
We conclude that
q 2(µ2 −µ0 ) µ+ν
q (1 + q µ0 −µ log q µ0 ) + q ν+µ−µ0 τ (q µ0 ) + q µ log q µ0 .
H2
With the choice (66) and by (47) and (59), we have µ0 ≥ µ − 4ρ ≥ 2ρ, so that
|E40 (r)| (68)
|E40 (r)| max(log q µ0 , τ (q µ0 )) q µ+ν−2ρ .
18
CHRISTIAN MAUDUIT AND JOËL RIVAT
By (63), (64), (67) and (68) we have
S3 (r, s) = S4 (r, s) + O(max(log q µ0 , τ (q µ0 )) q µ+ν−2ρ ).
(69)
6.3. Fourier analysis of S4 (r, s). We write
u0 + q µ1 −µ0 sn ≡ u2 mod q µ2 −µ0
and
u1 + q µ1 −µ0 sn + q µ1 −µ0 sr ≡ u3 mod q µ2 −µ0 .
This gives
S4 (r, s) =
X
X
ah0 (q µ0 −µ2 , H)ah1 (q µ0 −µ2 , H)
|h0 |≤H |h1 |≤H
1
X
q 2(µ2 −µ0 )
0≤h2 <q µ2 −µ0 0≤h3 <q µ2 −µ0
u0
h1 u1
g(u1 ) g(u0 )
e − qµh20−µ
e
−
µ
−µ
0
q 2 0
X
<q µ2 −µ0
X
X
g(u3 ) g(u2 )
0≤u2 <q µ2 −µ0
0≤u3 <q µ2 −µ0
0≤u0
0≤u1 <q µ2 −µ0
X X h0 mn + h1 mn + h1 mr u0 + q µ1 −µ0 sn − u2 e
e h2
µ2
q
q µ2 −µ0
m
n
u1 + q µ1 −µ0 sn + q µ1 −µ0 sr − u3
.
e h3
q µ2 −µ0
The discrete Fourier transform of g defined by (61) is
X
1
−uh
(70)
gb(h) = µ2 −µ0
g(u) e µ2 −µ0 .
q
q
µ −µ
0≤u<q
2
0
It follows that
S4 (r, s) = q
2(µ2 −µ0 )
X
X
ah0 (q
µ0 −µ2
, H)ah1 (q
µ0 −µ2
, H)
X
X
e
h3 sr
q µ2 −µ1
0≤h2 <q µ2 −µ0 0≤h3 <q µ2 −µ0
|h0 |≤H |h1 |≤H
gb(h0 − h2 ) gb(h3 − h1 ) gb(−h2 ) gb(h3 )
X X (h0 + h1 )mn + h1 mr + (h2 + h3 )q µ1 sn e
.
µ2
q
m
n
6.4. Estimate of S4 (r, s). We write
(71)
S4 (r, s) = S40 (r, s) + S400 (r, s),
where S40 (r, s) denotes the contribution of the terms for which h0 +h1 = 0, while S400 (r, s) denotes
the contribution of the terms for which h0 + h1 6= 0.
6.4.1. Contribution of S40 (r, s).
Since h0 + h1 = 0, the summations over m and n are independent. Noticing that by (48),
(66) (59) and (47) we have |h1 r| ≤ HR = q µ2 −µ0 +3ρ ≤ q µ2 /2, we deduce
!
−1
X h1 mr πh
r
q µ2
1
µ
µ
≤ min q , sin µ2
≤ min q ,
e
µ2
q
q
r |h1 |
m
and
X (h2 + h3 )q µ1 sn π(h2 + h3 )s
e
≤ min q ν , sin
µ
2
q
q µ2 −µ1
n
−1
Using the periodicity of gb modulo q µ2 −µ0 and writing h = h2 + h3 , we get
(72)
|S40 (r, s)| ≤ S5 (r, s),
!
.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
19
with
S5 (r, s) = q
2(µ2 −µ0 )
X
|h1 |≤H
X
q µ2
, H) min q ,
ah1 (q
r |h1 |
−1
min q ν , sin qµπhs
S6 (h, h1 )
2 −µ1
µ0 −µ2
2
µ
0≤h<q µ2 −µ0
and
X
S6 (h, h1 ) =
|b
g (h3 − h1 − h) gb(h3 − h1 ) gb(h3 − h) gb(h3 )| .
0≤h3 <q µ2 −µ0
By using the Cauchy-Schwarz inequality we have
!1/2
S6 (h, h1 ) ≤
X
!1/2
X
|b
g (h3 − h1 − h) gb(h3 − h)|2
h3
|b
g (h3 − h1 ) gb(h3 )|2
.
h3
The two quantities in the parentheses above are equal by periodicity modulo q µ2 −µ0 , hence
S6 (h, h1 ) ≤ S7 (h1 ),
with
(73)
X
S7 (h1 ) =
2
|b
g (h0 − h1 ) gb(h0 )| .
0≤h0 <q µ2 −µ0
By periodicity modulo q µ2 −µ1 and (21) we have
X
−1
1 X
ν
πhs
min q , sin qµ2 −µ1
S 1≤s<S
µ
−µ
0≤h<q 2 0
µ1 −µ0 X
X
q
=
min q ν , sin qµπhs
2 −µ1
S 1≤s<S
0≤h<q µ2 −µ1
q µ1 −µ0 q ν τ q µ2 −µ1 + q µ2 −µ1 log q µ2 −µ1 .
−1
Hence
(74)
1 X
S5 (r, s) q ν+µ1 −µ0 τ q µ2 −µ1 + q µ2 −µ1 −ν log q µ2 −µ1 S8 (r),
S 1≤s<S
with
S8 (r) = q
2(µ2 −µ0 )
X
|h1 |≤H
ah1 (q
µ0 −µ2
q µ2
, H) min q ,
r |h1 |
2
µ
S7 (h1 ).
Taking (66) into account we split the summation S8 (r) in three parts
S8 (r) = S80 (r) + S800 (r) + S8000 (r)
(75)
depending on the size of |h1 |: |h1 | ≤ q 2ρ , q 2ρ < |h1 | ≤ q µ2 −µ0 , and q µ2 −µ0 < |h1 | ≤ H. Using
(14) in S80 (r) we have |ah1 (q µ0 −µ2 , H)| ≤ q µ0 −µ2 , thus
X
X
q µ2
2
0
2(µ2 −µ0 )
µ0 −µ2
µ
S8 (r) = q
ah1 (q
, H) min q ,
S7 (h1 ) ≤ q µ
S7 (h1 ).
r |h1 |
2ρ
2ρ
|h1 |≤q
|h1 |≤q
In order to prove that this short sum over h1 is small, we will assume in this section that the
following Lemma holds:
Lemma 10. If
(76)
µ ≤ 2 + 34 c ρ
20
CHRISTIAN MAUDUIT AND JOËL RIVAT
then uniformly for λ ∈ N with 31 (µ2 − µ0 ) ≤ λ ≤ 45 (µ2 − µ0 ) we have
X
X
|b
g (h + k) gb(h)|2 q −γ1 (λ,µ1 −µ0 ) (log q µ2 −µ1 )2
(77)
0≤h<q µ2 −µ0 0≤k<q µ2 −µ0 −λ
where
γ1 (λ, µ1 − µ0 ) =
(78)
γ(λ) − µ1 + µ0
.
2
Proof. Lemma 10 will be proved in section 7.
By (77), (50), (56), (59), (58) we have µ2 − µ0 ≤ 6ρ so that µ2 − µ0 − 2ρ ≤ 32 (µ2 − µ0 ) ≤
4
(µ2 − µ0 ) and
5
X
S7 (h1 ) q −γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) ,
|h1 |≤q 2ρ
hence
S80 (r) q µ−γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) .
(79)
Using (14) in S800 (r) we have |ah1 (q µ0 −µ2 , H)| ≤ q µ0 −µ2 , thus
X
q µ2
2
00
2(µ2 −µ0 )
µ0 −µ2
µ
ah1 (q
S8 (r) = q
, H) min q ,
S7 (h1 )
r
|h
|
1
µ −µ
2ρ
q
µ2
q
≤
r
<|h1 |≤q
X
q 2ρ <|h1 |≤q µ2 −µ0
2
0
q µ2 −2ρ
S7 (h1 )
≤
|h1 |
r
X
S7 (h1 ).
|h1 |≤q µ2 −µ0
As by (24) we have
X
(80)
|b
g (h)|2 = 1,
0≤h<q µ2 −µ0
we obtain
S800 (r) ≤
q µ2 −2ρ
,
r
hence using (50) and (48),
1 X 00
log R
S8 (r) q µ2 −2ρ
= q µ−ρ log q ρ .
R 1≤r<R
R
(81)
Using (14) in S8000 (r) we have |ah1 (q µ0 −µ2 , H)| ≤
S8000 (r)
= q
X
2(µ2 −µ0 )
ah1 (q
q µ2 −µ0 <|h1 |≤H
q 2(µ2 −µ0 )
q µ2
r
1
,
π|h1 |
X
q µ2 −µ0 <|h1 |≤H
thus
µ0 −µ2
q µ2
, H) min q ,
r |h1 |
2
µ
S7 (h1 )
S7 (h1 )
.
|h1 |3
Observing that S7 (h1 ) is q µ2 −µ0 periodic, we split the summation into jq µ2 −µ0 < |h1 | ≤ (j +
1)q µ2 −µ0 where 1 ≤ j < H/q µ2 −µ0 and bound |h1 |−3 by j −3 q −3(µ2 −µ0 ) :
X
q µ2 X
1
(82)
S8000 (r) q 2(µ2 −µ0 )
S7 (h1 ),
r j≥1 j 3 q 3(µ2 −µ0 )
µ −µ
0≤h1 <q
2
thus by (80) and (56)
S8000 (r) q −(µ2 −µ0 )
q µ2
q µ0
q µ−2ρ
=
≤
.
r
r
r
0
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
It follows from (75), (79), (81), (82) that
1 X
S8 (r) q µ−γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) + q µ−ρ log q ρ ,
R 1≤r<R
and using (74) we obtain
1 X X
S5 (r, s)
RS 1≤r<R 1≤s<S
q ν+µ1 −µ0 τ q µ2 −µ1 + q µ2 −µ1 −ν log q µ2 −µ1 q µ−γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) + q µ−ρ log q ρ .
Finally by (72) we get
1 X X
(83)
|S 0 (r, s)|
RS 1≤r<R 1≤s<S 4
q µ+ν+µ1 −µ0 q −γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) + q −ρ log q ρ τ q µ2 −µ1 + q µ2 −µ1 −ν log q µ2 −µ1 .
6.4.2. Contribution of S400 (r, s).
Since h0 + h1 6= 0, after a summation over m, we get
X X
S400 (r, s) q 2(µ2 −µ0 )
ah0 (q µ0 −µ2 , H)ah1 (q µ0 −µ2 , H)
|h0 |≤H h1 6=−h0
X
|b
g (h3 − h1 ) gb(h3 )|
0≤h3 <q µ2 −µ0
0≤h2 <q µ2 −µ0
X
X
|b
g (h0 − h2 ) gb(−h2 ) |
1r
sin π (h0 +hq1µ)n+h
2
µ
min q ,
−1
.
n
Using (20) we have
X
1r
min q µ , sin π (h0 +hq1µ)n+h
2
−1
q ν−µ2 ((h0 + h1 , q µ2 )q µ + q µ2 log q µ2 ) ,
n
and observing that |h0 + h1 | ≤ 2H we get
X
1r
min q µ , sin π (h0 +hq1µ)n+h
2
−1
q ν−µ2 (Hq µ + q µ2 log q µ2 ) .
n
By (66) we have Hq µ ≥ q µ+µ2 −µ0 ≥ q µ2 ,
X
1r
min q µ , sin π (h0 +hq1µ)n+h
2
−1
q ν−µ2 H q µ log q µ2 .
n
Moreover we have by the Cauchy-Schwarz inequality and (80)
!1/2
!1/2
X
X
X
|b
g (−h2 )|2
= 1,
|b
g (h0 − h2 ) gb(−h2 ) | ≤
|b
g (h0 − h2 )|2
0≤h2 <q µ2 −µ0
h2
h2
and similarly
X
|b
g (h3 − h1 ) gb(h3 )| ≤ 1.
0≤h3 <q µ2 −µ0
Furthermore
X
ah (q µ0 −µ2 , H) ≤
|h|≤H
X
|h|≤q µ2 −µ0
1
q µ2 −µ0
+
X
q µ2 −µ0 <|h|≤H
1
log(H/q µ2 −µ0 ) log q ρ .
π |h|
Finally we obtain
|S400 (r, s)| (log q)2 ρ2 q 2(µ2 −µ0 ) q ν−µ2 H q µ log q µ2 ,
21
22
CHRISTIAN MAUDUIT AND JOËL RIVAT
which gives with the choice of H defined by (66):
|S400 (r, s)| (log q)3 (µ + ν)3 q µ+ν+3(µ2 −µ0 )+2ρ (q −µ2 + q −ν ).
(84)
6.5. Conclusion: estimate of SII (ϑ).
By (54), (62), (69), (71), (83) and (84), for all functions f satisfying Definition 1, f ∈ Fγ,c
in Definition 2 with c ≥ 10, (76) and (77) we obtain uniformly for ϑ ∈ R:
(85)
|SII (ϑ)|4 q 4µ+4ν+µ1 −µ0 q −γ1 (µ2 −µ0 −2ρ,µ1 −µ0 ) + q −ρ log q ρ
τ q µ2 −µ1 + q µ2 −µ1 −ν log q µ2 −µ1
+(log q)3 (µ + ν)3 q 4µ+4ν+3(µ2 −µ0 )+2ρ (q −µ2 + q −ν )
+ max(log q µ0 , τ (q µ0 )) q 4µ+4ν−2ρ
0
+ max(τ (q), log q) (µ + ν)ω(q) q 4µ+4ν−2ρ .
It remains to prove Lemma 10.
7. Distribution of the Discrete Fourier Transform
Let γ : R → R be a non decreasing function satisfying limλ→+∞ γ(λ) = +∞, f : N → U be
a function satisfying Definition 1 and f ∈ Fγ,c in Definition 2 for some c ≥ 10. The discrete
Fourier transform of the q λ -periodic function u 7→ fµ1 ,µ2 (rλ (u)q µ0 ) is a q λ -periodic function
Gµ0 ,λ defined for t ∈ R by
1 X
ut
µ0
(86)
Gµ0 ,λ (t) = λ
fµ1 ,µ2 (uq ) e − λ .
q
q
λ
0≤u<q
By (24) we have for t ∈ R and λ ∈ N,
X
(87)
|Gµ0 ,λ (h + t)|2 = 1,
0≤h<q λ
and by (61) and (70), for h ∈ Z we have
(88)
gb(h) = Gµ0 ,µ2 −µ0 (h).
In order to prove Lemma 10 we will need the following:
Lemma 11. If µ and ρ satisfy (76) then uniformly for λ ∈ N with
1
4
(µ2 − µ0 ) ≤ λ ≤ (µ2 − µ0 )
3
5
(89)
and t ∈ R we have
X
1
|Gµ0 ,µ2 −µ0 (k + t)|2 q 2 (µ1 −µ0 −γ(λ)) (log q µ2 −µ1 )2 .
0≤k<q µ2 −µ0 −λ
Proof. For 0 ≤ λ ≤ µ2 − µ0 and t ∈ R we can write
X
X
1
(u + vq λ )t
λ µ0
Gµ0 ,µ2 −µ0 (t) = µ2 −µ0
.
fµ1 ,µ2 ((u + vq )q ) e − µ2 −µ0
q
q
λ
µ −µ −λ
0≤u<q 0≤v<q
2
0
Hence for µ1 − µ0 ≤ λ ≤ µ2 − µ0 , observing that 0 ≤ u + vq λ < q µ2 −µ0 and (u + vq λ )q µ0 ≡
uq µ0 mod q µ1 , using (55) we get for 0 ≤ u < q λ and 0 ≤ v < q µ2 −µ0 −λ
fµ1 ,µ2 ((u + vq λ )q µ0 ) = fµ2 ((u + vq λ )q µ0 )fµ1 ((u + vq λ )q µ0 )
= f (uq µ0 + vq µ0 +λ )fµ1 (uq µ0 ),
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
23
and this yields
Gµ0 ,µ2 −µ0 (t)
1
= µ2 −µ0 −λ
q
X
f (vq
µ0 +λ
)e −
0≤v<q µ2 −µ0 −λ
vt
q µ2 −µ0 −λ
1 X
ut
µ0
µ0 +λ
µ
+λ
µ
0
0
f (uq + vq
)f (vq
)fµ1 (uq ) e − µ2 −µ0 .
qλ
q
λ
0≤u<q
Given ρ3 ∈ N such that
1 ≤ ρ3 ≤ µ2 − µ0 − λ,
(90)
by Definition 1 the number of v ∈ {0, . . . , q µ2 −µ0 −λ −1} such that there exists u ∈ {0, . . . , q λ −1}
for which
f (uq µ0 + vq µ0 +λ )f (vq µ0 +λ ) 6= fµ0 +λ+ρ3 (uq µ0 + vq µ0 +λ )fµ0 +λ+ρ3 (vq µ0 +λ )
fλ of pairs (u, v) with this property satisfies
is O(q µ2 −µ0 −λ−ρ3 ). Hence the set W
fλ q µ2 −µ0 −ρ3 .
card W
(91)
This leads for all t ∈ R to write
Gµ0 ,µ2 −µ0 (t) = Gµ0 ,µ2 −µ0 ,λ,1 (t) + Gµ0 ,µ2 −µ0 ,λ,2 (t),
with
Gµ0 ,µ2 −µ0 ,λ,1 (t)
1
= µ2 −µ0 −λ
q
X
f (vq
µ0 +λ
)e −
0≤v<q µ2 −µ0 −λ
vt
q µ2 −µ0 −λ
1 X
ut
µ0
µ0 +λ
µ
+λ
µ
fµ0 +λ+ρ3 (uq + vq
)fµ0 +λ+ρ3 (vq 0 )fµ1 (uq 0 ) e − µ2 −µ0
qλ
q
λ
0≤u<q
and
Gµ0 ,µ2 −µ0 ,λ,2 (t)
X
1
(u + vq λ )t
µ0 +λ
µ
0
= µ2 −µ0
f (vq
)fµ1 (uq ) e − µ2 −µ0
q
q
f
(u,v)∈Wλ
f (uq µ0 + vq µ0 +λ )f (vq µ0 +λ ) − fµ0 +λ+ρ3 (uq µ0 + vq µ0 +λ )fµ0 +λ+ρ3 (vq µ0 +λ ) .
Let us introduce in Gµ0 ,µ2 −µ0 ,λ,1 (t) the residue w of v modulo q ρ3 in order to make the variables
u and v independent:
Gµ0 ,µ2 −µ0 ,λ,1 (t)
X
1
=
µ
−µ
2
0 −λ
q
0≤w<q ρ3
1
qλ
v−w
f (vq
e ` ρ3
q µ2 −µ0 −λ
q
ρ3
µ
−µ
−λ
0≤`<q
2
0
0≤v<q
X
ut
µ0
µ0 +λ
fµ0 +λ+ρ3 (uq + wq
)fµ0 +λ+ρ3 (wq µ0 +λ )fµ1 (uq µ0 ) e − µ2 −µ0 .
q
λ
X
µ0 +λ
)e −
vt
1
q ρ3
X
)e −
vt
0≤u<q
This gives
Gµ0 ,µ2 −µ0 ,λ,1 (t) =
X
0≤`<q ρ3
ce` (t)
q µ2 −µ0 −λ
X
0≤v<q µ2 −µ0 −λ
f (vq
µ0 +λ
q µ2 −µ0 −λ
v`
+ ρ3
q
,
24
CHRISTIAN MAUDUIT AND JOËL RIVAT
with
1
ce` (t) = ρ3
q
w`
cλ (w, t) e − ρ3
q
0≤w<q ρ3
X
and
ut
1 X
µ0 +λ
µ0
µ
+λ
µ
)fµ0 +λ+ρ3 (wq 0 )fµ1 (uq 0 ) e − µ2 −µ0 .
fµ0 +λ+ρ3 (uq + wq
cλ (w, t) = λ
q
q
λ
0≤u<q
By the Cauchy-Schwarz inequality we get
|Gµ0 ,µ2 −µ0 ,λ,1 (t)|2
2
!
X
≤
|e
c` (t)|2
0≤`<q ρ3
1
X
q µ2 −µ0 −λ
0≤`<q ρ3
µ0 +λ
f (vq
)e −
X
0≤v<q µ2 −µ0 −λ
vt
q µ2 −µ0 −λ
v`
+ ρ3
q
.
But
X
2
|e
c` (t)|
0≤`<q ρ3
(w − w0 )`
= 2ρ3
cλ (w, t)cλ (w0 , t)
e −
q 0≤w<qρ3 0≤w0 <qρ3
q ρ3
0≤`<q ρ3
1 X
|cλ (w, t)|2 .
= ρ3
q 0≤w<qρ3
1
X
X
X
Since fµ1 (u0 q µ0 + u1 q µ1 ) = fµ1 (u0 q µ0 ) for 0 ≤ u0 < q µ1 −µ0 and 0 ≤ u1 < q λ−µ1 +µ0 , we can write
cλ (w, t)
= fµ0 +λ+ρ3
(wq µ0 +λ ) e
1
q λ−µ1 +µ0
wq λ t
q µ2 −µ0
X
1
q µ1 −µ0
fµ0 +λ+ρ3 u0 q
X
fµ1 (u0 q µ0 )
0≤u0 <q µ1 −µ0
µ0
0≤u1 <q λ−µ1 +µ0
+ u1 q
µ1
+ wq
µ0 +λ
(u0 + u1 q µ1 −µ0 + wq λ )t
e −
q µ2 −µ0
.
The sum over u1 may be written as a sum over u0 such that 0 ≤ u0 < q λ+ρ3 and u0 = u0 +
u1 q µ1 −µ0 + wq λ for some u1 with 0 ≤ u1 < q λ−µ1 +µ0 . Hence this last line is


µ1 −µ0
λ
X
X
1
u0 + u1 q
+ wq 

e `0
λ−µ
+µ
λ+ρ
1
0
3
q
q
0≤u1 <q λ−µ1 +µ0
0≤`0 <q λ+ρ3
X
1
u0 t
`0 u0
0 µ0
f (u q ) e − µ2 −µ0 − λ+ρ3
q λ+ρ3
q
q
λ+ρ
0
0≤u <q
3
In order to use (6) with κ = µ0 and λ replaced by λ + ρ3 we need to check that µ0 ≤ c(λ + ρ3 ).
By (59) we have µ0 ≤ µ1 , so that by (89) a sufficient condition would be that µ1 ≤ 3c (µ2 − µ1 ).
By (56) and (50) this is equivalent to (76). We are now ready to use (6) with κ = µ0 and λ
replaced by λ + ρ3 . Uniformly for t ∈ R and 0 ≤ w < q ρ3 we obtain:
0 µ1 −µ0 −1 !
X
q −γ(λ+ρ3 )
`q
min q λ−µ1 +µ0 , sin π
|cλ (w, t)| λ−µ1 +µ0
.
λ+ρ3
q
q
λ+ρ
0
0≤` <q
3
The sum over `0 runs over q µ1 −µ0 periods modulo q λ+ρ3 −µ1 +µ0 , thus
q −γ(λ+ρ3 ) λ+ρ3
q
log q λ+ρ3 −µ1 +µ0 = q ρ3 +µ1 −µ0 −γ(λ+ρ3 ) log q λ+ρ3 −µ1 +µ0 .
q λ−µ1 +µ0
Since the function γ is non decreasing and by (90) this gives uniformly for t ∈ R and 0 ≤ w <
q ρ3 :
|cλ (w, t)| q ρ3 +µ1 −µ0 −γ(λ) log q µ2 −µ1 .
|cλ (w, t)| PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
25
We deduce that
X
|e
c` (t)|2 q 2ρ3 +2(µ1 −µ0 )−2γ(λ) (log q µ2 −µ1 )2
0≤`<q ρ3
and
X
|Gµ0 ,µ2 −µ0 ,λ,1 (k + t)|2
0≤k<q µ2 −µ0 −λ
q 2ρ3 +2(µ1 −µ0 )−2γ(λ) (log q µ2 −µ1 )2
X
X
X
1
f (vq µ0 +λ )f (v 0 q µ0 +λ )
2(µ2 −µ0 −λ)
q
0≤`<q ρ3
0≤v<q µ2 −µ0 −λ 0≤v 0 <q µ2 −µ0 −λ
X
(v − v 0 )t (v − v 0 )`
(v − v 0 )k
e − µ2 −µ0 −λ +
e − µ2 −µ0 −λ
q
q ρ3
q
µ −µ −λ
0≤k<q
2
q 2ρ3 +2(µ1 −µ0 )−2γ(λ) (log q µ2 −µ1 )2
0
1
X
0≤`<q ρ3
X
q µ2 −µ0 −λ
1
0≤v<q µ2 −µ0 −λ
q 3ρ3 +2(µ1 −µ0 )−2γ(λ) (log q µ2 −µ1 )2 .
fλ we observe that
Denoting by Wλ the set of integers w = u + q λ v such that (u, v) ∈ W
X
1
wt
0
Gµ0 ,µ2 −µ0 ,λ,2 (t) = µ2 −µ0
cλ (w) e − µ2 −µ0 ,
q
q
µ −µ
w<q
2
0
where |c0λ (w)| ≤ 2 for 0 ≤ w < q µ2 −µ0 and c0λ (w) = 0 for w 6∈ Wλ . Therefore for all t ∈ R
X
|Gµ0 ,µ2 −µ0 ,λ,2 (k + t)|2
0≤k≤q µ2 −µ0
=
1
X
q 2(µ2 −µ0 )
0≤k≤q µ2 −µ0
1
q µ2 −µ0
c0λ (w)c0λ (w0 ) e
w<q µ2 −µ0 w0 <q µ2 −µ0
X
=
X
(w − w0 )t
− µ2 −µ0
q
(w − w0 )k
e − µ2 −µ0
q
X
1
2
|c0λ (w)| =
w<q µ2 −µ0
q µ2 −µ0
X
2
|c0λ (w)| ≤
w∈Wλ
1
q µ2 −µ0
X
22
w∈Wλ
q −ρ3 ,
where we take
ρ3 = max 1,
1
(γ(λ) − µ1 + µ0 )
2
.
≤ µ2 − µ0 . By (25) a sufficient condition to ensure
By (90) this is admissible at least if λ + γ(λ)
2
4
this inequality is that λ ≤ 5 (µ2 − µ0 ). Then we get
X
|Gµ0 ,µ2 −µ0 (k + t)|2
0≤k<q µ2 −µ0 −λ
≤ 2
X
|Gµ0 ,µ2 −µ0 ,λ,1 (k + t)|2 + 2
0≤k<q µ2 −µ0 −λ
q
1
(µ1 −µ0 −γ(λ))
2
X
|Gµ0 ,µ2 −µ0 ,λ,2 (k + t)|2
0≤k<q µ2 −µ0 −λ
(log q µ2 −µ1 )2 ,
which completes the proof of Lemma 11.
It follows from Lemma 11 and (80) that (77) holds with γ1 defined by (78), which completes
the proof of Lemma 10.
26
CHRISTIAN MAUDUIT AND JOËL RIVAT
8. End of the estimate of the sums of type II
It follows from (85) that for all functions f satisfying Definition 1 and f ∈ Fγ,c in Definition
2 for some c ≥ 10 that we have, under the condition (76), uniformly for any ϑ ∈ R:
1
(92)
|SII (ϑ)|4 q 4µ+4ν+µ1 −µ0 q 2 (µ1 −µ0 −γ(λ)) + q −ρ log q ρ
τ q µ2 −µ1 + q µ2 −µ1 −ν log q µ2 −µ1
+(log q)3 (µ + ν)3 q 4µ+4ν+3(µ2 −µ0 )+2ρ (q −µ2 + q −ν )
+ max(log q µ0 , τ (q µ0 )) q 4µ+4ν−2ρ
0
+ max(τ (q), log q) (µ + ν)ω(q) q 4µ+4ν−2ρ .
By (50), (56), (59), (58), (25) and since the function γ is non decreasing, γ(µ2 − µ0 − 2ρ) ≥
γ(µ2 −µ1 −2ρ) = γ(2ρ) and ρ ≥ γ(2ρ). By multiplicativity of the function τ we have τ (q µ2 −µ1 ) ≤
(µ2 − µ1 )ω(q) τ (q). By (50), (56), (45) and (47) we have µ2 − µ1 = 4ρ ≤ µ − 2ρ ≤ ν − 2ρ so that
q µ2 −µ1 −ν log q µ2 −µ1 1. This implies
3
1
|SII (ϑ)|4 τ (q)(µ2 − µ1 )ω(q) q 4µ+4ν+ 2 (µ1 −µ0 )− 2 γ(2ρ) log q ρ
+(log q)3 (µ + ν)3 q 4µ+4ν+3(µ1 −µ0 )+14ρ−µ
+ max(log q µ0 , τ (q µ0 )) q 4µ+4ν−2ρ
0
+ max(τ (q), log q) (µ + ν)ω(q) q 4µ+4ν−2ρ .
Taking
ρ0 = bγ(2ρ)/10c ,
(93)
we have (58) by (25) and by (59) µ1 − µ0 ≤ 2ρ0 ≤ γ(2ρ)/5 ≤ ρ/5 so that
3
γ(2ρ)
3γ(2ρ) γ(2ρ)
−γ(2ρ)
(µ1 − µ0 ) −
≤
−
≤
.
2
2
10
2
5
Choosing
ρ = bµ/15c
(94)
we get for µ ≥ 15 × 75 = 1125,
3(µ1 − µ0 ) + 14ρ − µ ≤
3ρ
−ρ
+ 14ρ − 15ρ + 15 ≤
.
5
5
In order to ensure (76) it is sufficient to check that
4 µ
µ≤ 2+ c
−1 ,
3
15
which is true for µ large enough provided c > 39/4. For convenience and in order to avoid that
the implied constants depend on c we take
(95)
c ≥ 10,
so that the inequality above is valid for µ ≥ 46 × 15 = 690. Finally we obtain
(96)
|SII (ϑ)|4 max(τ (q) log q, log3 q) (µ + ν)1+max(ω(q),2) q 4µ+4ν−γ(2bµ/15c)/5 ,
which completes the proof of (46) and Proposition 2.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
27
9. Proof of Theorems 1 and 2
Applying Lemma 1 of [20], or its analogue in the case of µ obtained using (13.40) instead
of (13.39) of [13], our estimate of the sums of type I in Proposition 1 and our estimate of the
µ
≤ 15
lead to
sums of type II in Proposition 2 with the observation that (45) implies µ+ν
60
X
9
1
9
1
Λ(n)f (n) e (ϑn) c1 (q)(log x) 4 + 4 max(ω(q),2) x q −γ(2b(log x)/60 log qc)/20 .
x/q<n≤x
and
X
µ(n)f (n) e (ϑn) c1 (q)(log x) 4 + 4 max(ω(q),2) x q −γ(2b(log x)/60 log qc)/20 .
x/q<n≤x
1
with c1 (q) = max(τ (q), log2 q)1/4 (log q)−2− 4 max(ω(q),2) . Now we will replace x by x/q k and sum
over k. Let K ∈ N such that q K ≤ x1/4 < q K+1 . Since γ is non decreasing we have
X x
X x
−γ(2blog(xq −k )/60 log q c)/20
−γ(2blog x3/4 /60 log q c)/20
q
≤
q
qk
qk
k≤K
k≤K
xq −γ(2blog x/80 log qc)/20
while
X x
X x3/4
−γ(2blog(xq −k )/60 log q c)/20
q
≤
x3/4 xq −γ(2blog x/80 log qc)/20 .
k
k
q
q
k>K
k≥0
Then Theorem 1 and Theorem 2 follow.
10. Proof of Corollaries 1, 2 and 3
In order to prove Corollaries 1 and 2 we use a classical partial summation. Using (for example)
Lemma 11 of [20], if f : N → C is such that |f (n)| ≤ 1 for all n ∈ N then
X
(97)
p≤x
f (p) ≤
X
√
2
max
Λ(n)f (n) + O( x).
log x t≤x n≤t
To prove Corollary 1 we observe first that if α ∈ Q, then the sequence (αb(p))p∈P(a,m) takes a
finite number of values modulo 1, and therefore is not equidistributed modulo 1. If α ∈ R \ Q,
then for all h ∈ Z such that h 6= 0, we have hα ∈ R \ Q, so that the function n 7→ e(hαb(n))
satisfies Definition 1 and f ∈ Fγ,c in Definition 2 for some c ≥ 10. By Theorem 1, we have for
all 0 ≤ j < m
X
jn
Λ(n) e hαb(n) +
= o(x),
m
n≤x
hence
X
n≤x
n≡a mod m
1 X
ja X
jn
Λ(n) e (hαb(n)) =
e −
Λ(n) e hαb(n) +
= o(x).
m 0≤j<m
m n≤x
m
By Inequality (97) and the Prime Number Theorem in arithmetic progressions (see for example
[2, Theorem 9.12]), we can write for gcd(a, m) = 1
X
√
x
e(hαb(p)) = o
+ O( x) = o(π(x; a, m)),
log x
p≤x
p≡a mod m
which proves that the sequence (αb(p))p∈P(a,m) is equidistributed modulo 1 according to Weyl’s
criterion (see for example [22, Chapter 1,p. 1]).
28
CHRISTIAN MAUDUIT AND JOËL RIVAT
In order to prove Corollary 2, we write
X j
X
X 1
j0
0
e
(p − a) + 0 (b(p) − a )
1 =
mm0 0≤j<m
m
m
p≤x
p≤x
p≡a mod m
b(p)≡a0 mod m0
0≤j 0 <m0
π(x; a, m)
1
=
+
0
m
mm0
X
0≤j<m
1≤j 0 <m0
aj a0 j 0 X
j
j0
e − − 0
e
p + 0 b(p) .
m
m p≤x
m
m
0
Since for 1 ≤ j 0 < m0 the functions n 7→ e( mj 0 b(n)) satisfy Definition 1 and f ∈ Fγ,c in Definition
2 for some c ≥ 10, by Theorem 1 and using (97), for all 0 ≤ j < m and 1 ≤ j 0 < m0 we obtain
X j0
j
e
b(p) + p = o(π(x)).
m0
m
p≤x
As the integers m and m0 are fixed, it follows by using the Prime Number Theorem in arithmetic
progressions (see for example [2, Theorem 9.12]), that for gcd(a, m) = 1
X
π(x; a, m)
1 = (1 + o(1))
,
0
m
p≤x
p≡a mod m
b(p)≡a0 mod m0
which proves Corollary 2.
In order to prove Corollary 3 we observe first that if ϑ ∈ Q, then the sequence (ϑp)p∈B(a,m,a0 ,m0 )
takes a finite number of values modulo 1, and therefore is not equidistributed modulo 1. If
ϑ ∈ R \ Q, we write
X X
X 1
j0
j
0
e hϑp + (p − a) + 0 (b(p) − a )
e(hϑp) =
mm0 0≤j<m
m
m
p≤x
p≤x
p≡a mod m
b(p)≡a0 mod m0
0≤j 0 <m0
1 X
aj X
j
=
e −
e
hϑ +
p
mm0 0≤j<m
m p≤x
m
X aj a0 j 0 X 1
j0
j
+
e − − 0
p + 0 b(p) .
e
hϑ +
mm0 0≤j<m
m
m p≤x
m
m
1≤j 0 <m0
It follows from [8] that for all h ∈ Z \ {0}, ϑ ∈ R \ Q, 0 ≤ j < m we have
X j
e
hϑ +
p = o(π(x)).
m
p≤x
0
Since for 1 ≤ j 0 < m0 the functions n 7→ e( mj 0 b(n)) satisfy Definition 1 and f ∈ Fγ,c in Definition
2 for some c ≥ 10, by Theorem 1 and using (97), for all h ∈ Z \ {0}, ϑ ∈ R, 0 ≤ j < m and
1 ≤ j 0 < m0 we obtain
X j0
j
e
b(p) + hϑ +
p = o(π(x)).
m0
m
p≤x
It follows that
X
e(hϑp) = o(π(x)),
p≤x
p≡a mod m
b(p)≡a0 mod m0
which, as the integers m and m0 are fixed, proves that the sequence (ϑp)p∈B(a,m,a0 ,m0 ) is equidistributed modulo 1 according to Weyl’s criterion (see for example [22, Chapter 1,p. 1]).
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
29
11. Application to Rudin-Shapiro sequences
11.1. Rudin-Shapiro sequences of order δ. For any n ∈ N we denote by
X
n=
εk (n)
k≥0
its representation in base 2, where εk (n) denotes the k-th least significant digit of n in base
2. Let δ ∈ N and βδ (n) the number of occurencies of patterns 1w1 (where w ∈ {0, 1}δ ) in the
representation of n in base 2:
X
βδ (n) =
εk−δ−1 (n) εk (n).
k≥δ+1
For α ∈ R we consider in this section f (n) = e (βδ (n)α) . By (4) for all λ ≥ δ + 2 we have
!
X
fλ = e α
εi−δ−1 εi .
δ+1≤i<λ
Therefore considering fκ+ρ in (5), the inequality may occur only by carry propagation when
the digits of `q κ + k1 of indexes κ,. . . ,κ + ρ − 1 are equal to 1, i.e. for integers ` with 2ρ
least significant digits equal to 1. It follows that f satisfies Definition 1. For δ + 2 ≤ µ1 < µ2 ,
by (55) we have
!
X
εi−δ−1 εi .
fµ1 ,µ2 = fµ2 fµ1 = e α
µ1 ≤i<µ2
It follows that fµ1 ,µ2 (n) depends only on the digits of n of index µ1 − δ − 1, µ1 , . . . , µ2 − 1.
Therefore in (59) we can choose any value of µ0 at most equal to µ1 − δ − 1 and (60) will be
satisfied for any value of ρ0 at most equal to ρ, which makes the choice (93) admissible. The
aim of Proposition 3 is to show that for any α ∈ R, the function n 7→ e(αβδ (n)) belongs to
some Fγ,c in Definition 2 (observe that βδ (2κ n) = βδ (n) for any κ ∈ N).
Proposition 3. For any δ ∈ N, α ∈ R, ϑ ∈ R and λ ∈ N we have
λ/2
X
δ+1
1 + |cos πα|
−λ
2
2
e(αβδ (n) + ϑn) ≤ 2
.
2
λ
0≤n<2
Remark 3. This is Theorem 3.1 of [1], but we will present here a direct proof.
Proof. For 0 ≤ i < 2δ+1 we write
Γ[i] (n) =
δ
X
εk (n) εδ−k (i),
X
[i]
Sλ (α, ϑ) =
e αΓ[i] (n) + αβδ (n) + ϑn
0≤n<2λ
k=0
and


Sλ (α, ϑ) = 
[0]
Sλ (α, ϑ)
..
.
[2δ+1 −1]
Sλ


.
(α, ϑ)
δ
If 0 ≤ i < 2 , we have εδ (i) = 0, so that for any n ∈ N and ε ∈ {0, 1}
[i]
Γ (2n + ε) =
δ
X
εk (2n + ε) εδ−k (i) =
k=0
=
δ−1
X
k=0
δ
X
εk (2n + ε) εδ−k (i) =
k=1
εk (n) εδ−k (2i) =
δ
X
k=0
εk (n) εδ−k (2i) = Γ[2i] (n)
δ
X
k=1
εk−1 (n) εδ−k+1 (2i)
30
CHRISTIAN MAUDUIT AND JOËL RIVAT
and εδ (2δ + i) = 1, so that
Γ
[2δ +i]
δ
X
(2n + ε) =
δ
εk (2n + ε) εδ−k (2 + i) = ε0 (2n + ε) · 1 +
k=0
= ε+
δ
X
εk (2n + ε) εδ−k (2δ + i)
k=1
δ
X
εk−1 (n) εδ−k (i) = ε +
k=1
δ
X
εk−1 (n) εδ−k+1 (2i) = ε + Γ[2i] (n).
k=1
Furthermore
[2i+1]
Γ
(n) =
δ
X
εk (n) εδ−k (2i + 1) = εδ (n)ε0 (2i + 1) +
k=0
δ−1
X
εk (n) εδ−k (2i + 1)
k=0
= εδ (n) +
δ−1
X
εk (n) εδ−k (2i) = εδ (n) + Γ[2i] (n).
k=0
It follows from the definition of βδ that
X
βδ (2n) =
X
εk−δ−1 (2n) εk (2n) = ε0 (2n) εδ+1 (2n) +
k≥δ+1
εk−δ−2 (n) εk−1 (n) = βδ (n)
k≥δ+2
and
βδ (2n + 1) =
X
εk−δ−1 (2n + 1) εk (2n + 1)
k≥δ+1
= ε0 (2n + 1) εδ+1 (2n + 1) +
X
εk−δ−2 (n) εk−1 (n) = εδ (n) + βδ (n),
k≥δ+2
so that for any i ∈ {0, . . . , 2δ − 1} and λ ∈ N we have
[i]
Sλ+1 (α, ϑ) =
X
e αΓ[i] (n) + αβδ (n) + ϑn
0≤n<2λ+1
=
X
e αΓ[i] (2n) + αβδ (2n) + 2nϑ
0≤n<2λ
X
+
e αΓ[i] (2n + 1) + αβδ (2n + 1) + (2n + 1)ϑ
0≤n<2λ
=
X
e αΓ[2i] (n) + αβδ (n) + 2nϑ
0≤n<2λ
+ e(ϑ)
X
e αΓ[2i] (n) + αεδ (n) + αβδ (n) + 2nϑ
0≤n<2λ
[2i]
[2i+1]
= Sλ (α, 2ϑ) + e(ϑ) Sλ
(α, 2ϑ)
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
31
and
[2δ +i]
Sλ+1 (α, ϑ) =
δ
e αΓ[2 +i] (n) + αβδ (n) + ϑn
X
0≤n<2λ+1
=
δ
e αΓ[2 +i] (2n) + αβδ (2n) + 2nϑ
X
0≤n<2λ
δ
e αΓ[2 +i] (2n + 1) + αβδ (2n + 1) + (2n + 1)ϑ
X
+
0≤n<2λ
=
X
e αΓ[2i] (n) + αβδ (n) + 2nϑ
0≤n<2λ
+ e(ϑ)
X
e α + αΓ[2i] (n) + αεδ (n) + αβδ (n) + 2nϑ
0≤n<2λ
=
[2i]
Sλ (α, 2ϑ)
[2i+1]
+ e(α + ϑ) Sλ
(α, 2ϑ).
This yields
(98)
Sλ+1 (α, ϑ) = M (α, ϑ)Sλ (α, 2ϑ),
δ+1
where M (α, ϑ) denotes the 2
× 2δ+1 matrix

1
e(ϑ)
0
···
0
0
1
e(ϑ)

 ..
.
···
···
···

···
···
···
0

···
···
···
0
M (α, ϑ) = 
1 e(α + ϑ) 0
0
0
1 e(α + ϑ)

.
.
.
···
···
···

0
···
···
···
0
···
···
···
···
0
···
···
···
···
···
···
0
0
..
.





···
···

e(ϑ)
0
0


0
1
e(ϑ) 
.
0


···
···
0


..

···
···
.


e(α + ϑ) 0
0
0
1 e(α + ϑ)
··· ···
0
1
··· ···
···
0 ···
··· ···
0
1
··· ···
Writing
A(α, ϑ) =
2
e(ϑ) + e(α + ϑ)
e(−ϑ) + e(−α − ϑ)
2
we observe that t M (α, ϑ)M (α, ϑ) is the block matrix

A(α, ϑ)
t
M (α, ϑ)M (α, ϑ) = 
0
..
.
.
0

A(α, ϑ)
Denoting by ρ(A(α, ϑ)) the spectral radius of A(α, ϑ), it follows that
p
p
p
kM (α, ϑ)k2 = ρ(A(α, ϑ)) = 2 + |e(ϑ) + e(α + ϑ)| = 2(1 + |cos πα|).
By (98) this permits to write
kSλ+1 (α, ϑ)k2 ≤ kM (α, ϑ)k2 kSλ (α, 2ϑ)k2 ≤
p
2(1 + |cos πα|) kSλ (α, 2ϑ)k2 .
By induction we get
X
[0]
e(αβδ (n) + ϑn) = Sλ (α, ϑ)
≤ kSλ (α, ϑ)k2
0≤n<2λ
≤ (2 + 2 |cos πα|)λ/2 S0 (α, 2λ ϑ)
=2
δ+1
2
λ/2
(2 + 2 |cos πα|)
,
2
32
CHRISTIAN MAUDUIT AND JOËL RIVAT
which gives Theorem 3.
Observing that βδ (u2κ ) = βδ (u), Proposition 3 shows that f (n) = e (βδ (n)α) belongs to Fγ,c
in Definition 2 for any c > 0 and
1 + |cos πα|
δ+1
λ
log
−
.
(99)
γ(λ) = −
2 log 2
2
2
Applying Theorem 1 and Theorem 2 we obtain
Theorem 3. For any δ ∈ N, α ∈ R, ϑ ∈ R and x ≥ 2 we have
(100)
X
11
Λ(n) e (βδ (n)α + ϑn) x (log x) 4 2−γ(2b(log x)/80 log 2c)/20
n≤x
and
(101)
X
11
µ(n) e (βδ (n)α + ϑn) x (log x) 4 2−γ(2b(log x)/80 log 2c)/20 ,
n≤x
where γ is defined by (99).
11.2. Rudin-Shapiro sequences of degree d.
Let d ∈ N, d ≥ 2 and bd (n) the number of occurencies of blocks of d consecutive 1’s in the
representation of n in base 2:
X
bd (n) =
εk−d+1 (n) · · · εk (n).
k≥d−1
For α ∈ R we consider in this section f (n) = e (bd (n)α) . By (4) for all λ ≥ d we have
!
X
εi−d+1 · · · εi−1 εi .
fλ = e α
d−1≤i<λ
Therefore considering fκ+ρ in (5), the inequality may occur only by carry propagation when
the digits of `2κ + k1 of indexes κ,. . . ,κ + ρ − 1 are equal to 1, i.e. for integers ` with 2ρ
least significant digits equal to 1. It follows that f satisfies Definition 1. For d ≤ µ1 < µ2 , by
(55) we have
!
X
fµ1 ,µ2 = fµ2 fµ1 = e α
εi−d+1 · · · εi−1 εi .
µ1 ≤i<µ2
It follows that fµ1 ,µ2 (n) depends only on the digits of n of index µ1 − d + 1, . . . , µ2 − 1. Given
any ρ0 satisfying (d − 1)/2 ≤ ρ0 ≤ ρ (which implies (58)), we can choose µ0 = µ1 − d + 1 so that
(59) and (60) are satisfied. This makes the choice (93) admissible.
The aim of Proposition 4 is to show that for any α ∈ R, the function n 7→ e(αbd (n)) belongs
to some Fγ,c in Definition 2 (observe that bd (2κ n) = bd (n) for any κ ∈ N).
Proposition 4. For any d ≥ 2, α ∈ R, ϑ ∈ R and λ ∈ N we have
2 bλ/dc
X
πkαk
−λ
3−d
.
2
e (bd (n)α + nϑ) ≤ 1 − 2
sin 4
0≤n<2λ
Proof. For k ∈ N we define χk : N → {0, 1} by χk (n) = 1 if the k least significant digits of n are
1’s and χk (n) = 0 otherwise. This means that χ0 = 1 and for k ≥ 1, χk (n) = εk−1 (n) · · · ε0 (n).
In particular χk (n) = 0 for k ≥ 1 and n < 2k−1 .
For n ∈ N we define
[1]
χd (n) = 0
and for i ∈ {2, . . . , d},
[i]
χd (n) = χd−i+1 (n) + · · · + χd−1 (n).
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
33
We define
[i]
Sλ (α, ϑ) =
[i]
e χd (n)α + bd (n)α + nϑ
X
0≤n<2λ
[1]
and notice that we are interested by Sλ (α, ϑ) . For n ∈ N we have
bd (2n) = bd (n), bd (2n + 1) = bd (n) + χd−1 (n)
and for k ≥ 1,
χk (2n) = 0, χk (2n + 1) = χk−1 (n),
so that for i ∈ {1, . . . , d},
[i]
[1]
[1]
[2]
χd (2n + 1) = 0 = χd (n) − χd−1 (n)
χd (2n) = 0 = χd (n),
and for i ∈ {2, . . . , d − 1}
[i]
[i+1]
[d]
[d]
χd (2n + 1) = χd−i (n) + · · · + χd−2 (n) = χd
(n) − χd−1 (n),
while
χd (2n + 1) = χ0 (n) + · · · + χd−2 (n) = 1 + χd (n) − χd−1 (n).
It follows that for i = 1, . . . , d − 1
[i]
[1]
[i+1]
Sλ+1 (α, ϑ) = Sλ (α, 2ϑ) + e(ϑ) Sλ
(α, 2ϑ)
and
[d]
[1]
[d]
Sλ+1 (α, ϑ) = Sλ (α, 2ϑ) + e(α + ϑ) Sλ (α, 2ϑ).
Let us introduce the d × d matrix M (α, ϑ) and the vector Sλ (α, ϑ) defined by


1 e(ϑ) 0 · · · · · ·
0

1 0 e(ϑ) · · · · · ·
 [1]

0

.
S
(α,
ϑ)
.
..

.
λ
..
.


.

..
,
S
(α,
ϑ)
=
M (α, ϑ) = 


.
λ
.
.

1 0
..
0
0
[d]


Sλ (α, ϑ)
1 0
0
e(ϑ) 
1 0
0 e(α + ϑ)
We have
Sλ+1 (α, ϑ) = M (α, ϑ) Sλ (α, 2ϑ),
and writing
M [λ] (α, ϑ) = M (α, ϑ)M (α, 2ϑ) · · · M (α, 2λ−1 ϑ),
we get
Sλ (α, ϑ) = M [λ] (α, ϑ) S0 (α, 2λ ϑ).
Therefore
[1]
Sλ (α, ϑ) ≤ kSλ (α, ϑ)k∞ ≤ M [λ] (α, ϑ)
so that Proposition 4 follows from the following Lemma.
Lemma 12. We have
M [d] (α, ϑ)
∞
2
.
≤ 2d − 8 sin πkαk
4
∞
,
34
CHRISTIAN MAUDUIT AND JOËL RIVAT
Proof. Let us first observe that (M [d] (0, 0))i,j = ((M (0, 0))d )i,j = max(2, 2d−j ): indeed it easy
to show by induction on k that for 1 ≤ k ≤ d − 1,
 k−1 k−2

2
2
. . . 20 1 0 . . . . . . 0
.. 
...
 k−1 k−2
2
20 0 1
.
2
 .
..
.. .. . . . . . . .. 
 ..
.
.
. .
.
. .


 ..

.
.
.
k
.
.
.
.
.
.
.
(M (0, 0)) =  .
.
. 0 .
.
.
.


..
.. ..
 ..

.
. .
0 1
 .
 .
..
.. ..
.. .. 
 ..
.
. .
. .
k−1
k−2
0
2
2
... 2 0 ... ... 0 1
We remark that for any (i, j) ∈ {1, . . . , d}2 the coefficient (M [d] (α, ϑ))i,j is a sum of complex
numbers of modulus 1. The number of them is precisely (M [d] (0, 0))i,j = max(2, 2d−j ) while the
argument of each of them being the coding of a path of length d going from the vertex i to
the vertex j in the following graph:
0
1
1
1
1
1
d−1
2
0
1∗
d
0
0
0
The coding (given by the rules of matrix product) is the following: for any path of length d
from i to j and for any t ∈ {1, . . . , d}:
• crossing an arc labeled by 0 at step t adds 0 to the argument;
• crossing an arc labeled by 1 at step t adds 2t−1 ϑ to the argument;
• crossing an arc labeled by 1∗ at step t adds 2t−1 ϑ + α to the argument.
For any i ∈ {1, . . . , d} there are exactly two paths of length d going from the vertex i to the
vertex d :
(1) the path
1
1
1
i+1
i
d
1∗ (i times)
that corresponds to the coding
ϑ + 2ϑ + · · · + 2d−i−1 ϑ + (2d−i ϑ + α) + · · · + (2d−1 ϑ + α) = (2d − 1)ϑ + iα;
(2) the path
1
1
1
1
i
d−1
0
that corresponds to the coding
ϑ + 2ϑ + · · · + 2d−1 ϑ = (2d − 2)ϑ.
It follows that for any i ∈ {1, . . . , d} we have
(102)
(M [d] (α, ϑ))i,d = e (2d − 2)ϑ + e (2d − 1)ϑ + iα .
d
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
35
For any i ∈ {1, . . . , d} there are exactly 2d−1 paths of length d going from the vertex i to the
vertex 1 among which we have the two following ones:
(1) the path
1
1
1
1
i+1
i
d
1∗ (i − 1 times)
0
that corresponds to the coding
ϑ + 2ϑ + · · · + 2d−i−1 ϑ + (2d−i ϑ + α) + · · · + (2d−2 ϑ + α) + 0 = (2d−1 − 1)ϑ + (i − 1)α;
(2) the path
1
1
1
1
2
1
1
d−1
i
0 (step 1)
0
that corresponds to the coding
0 + 2ϑ + · · · + 2d−2 ϑ + 0 = (2d−1 − 2)ϑ.
It follows that for any i ∈ {1, . . . , d} we have
(M [d] (α, ϑ))i,1 = e (2d−1 − 2)ϑ + e (2d−1 − 1)ϑ + (i − 1)α
(103)
+(2d−1 − 2) terms of modulus 1.
It follows from (102) and (103) that for any i ∈ {1, . . . , d} we have
d
X
[d]
(M (α, ϑ))i,j
≤
d−1
X
2d−j + (2d−1 − 2)
j=2
j=1
+ e (2d−1 − 2)ϑ + e (2d−1 − 1)ϑ + (i − 1)α
+ e (2d − 2)ϑ + e (2d − 1)ϑ + iα ,
which implies
d
X
(M [d] (α, ϑ))i,j ≤ 2d − 4 + 2 |cos π (ϑ + (i − 1)α)| + 2 |cos π (ϑ + iα)| .
j=1
For any x ∈ R we have
0
0
|cos πx| + |cos π(x + α)| = |cos πx + cos π(x + α0 )| ≤ eiπx + eiπ(x+α ) = 2 cos πα
2
with α0 = α if both cosinus have the same sign and α0 = α + 1 otherwise. Hence
πα
,
sin
|cos πx| + |cos π(x + α)| ≤ 2 max cos πα
2
2
and observing that α = n ± kαk with n ∈ Z, and 0 ≤ kαk ≤ 12 , we get
2
πkαk
πkαk
πkαk
πkαk
πα
,
sin
=
max
,
sin
=
cos
=
1
−
2
sin
.
max cos πα
cos
2
2
2
2
2
4
Taking x = ϑ + (i − 1)α we obtain
d
X
j=1
[d]
d
(M (α, ϑ))i,j ≤ 2 − 8 sin
πkαk
4
2
.
36
CHRISTIAN MAUDUIT AND JOËL RIVAT
which completes the proof of Lemma 12.
We will now prove Proposition 4. We have
X
2−λ
[1]
e (bd (n)α + nϑ) = 2−λ Sλ (α, ϑ) ≤ 2−λ M [λ] (α, ϑ)
∞
0≤n<2λ
and
M [λ] (α, ϑ) = M (α, ϑ)M (α, 2ϑ) · · · M (α, 2λ−1 ϑ)
= M [d] (α, ϑ)M [d] (α, 2d ϑ) · · · M [d] (α, 2dbλ/dc ϑ)
M (α, 2dbλ/dc ϑ) · · · M (α, 2λ−1 ϑ).
Using the sub-multiplicativity of the matrix norm, the bound kM (α, ϑ0 )k∞ = 2 and Lemma 12,
we get
2 bλ/dc
πkαk
d
[λ]
λ−dbλ/dc
2 − 8 sin 4
M (α, ϑ) ∞ ≤ 2
,
so that
−λ
2
X
e (bd (n)α + nϑ) ≤
3−d
1−2
sin
πkαk
4
2 bλ/dc
,
0≤n<2λ
which completes the proof of Proposition 4. Taking into account that bd (n2k ) = bd (n) for all
k ∈ N, d ≥ 2 and 0 ≤ kαk ≤ 1/2, it follows from Proposition 4 that for f (n) = e (bd (n)α) and
for all k ∈ N and all ϑ ∈ R,
−λ
2
X
k
f (u2 ) e (uϑ)
2 bλ/dc
πkαk
3−d
≤
1−2
sin 4
0≤u<2λ
≤
3−d
1−2
sin
−1
π 2
8
3−d
1−2
sin
πkαk
4
2 λ/d
,
2 λ/d
√
πkαk
3−d
≤
2 1−2
sin 4
.
It follows that f belongs to Fγ,c in Definition 2 for all c > 0 and
2 1
−λ
πkαk
3−d
log 1 − 2
sin 4
(104)
γ(λ) =
− .
d log 2
2
Applying Theorem 1 and Theorem 2 we obtain
Theorem 4. For d ≥ 2, α ∈ R, ϑ ∈ R and x ≥ 2 we have
(105)
X
11
Λ(n) e (bd (n)α + ϑn) x (log x) 4 2−γ(2b(log x)/80 log 2c)/20
n≤x
and
(106)
X
11
µ(n) e (bd (n)α + ϑn) x (log x) 4 2−γ(2b(log x)/80 log 2c)/20 ,
n≤x
where γ is defined by (104).
Acknowledgements. We thank Guy Barat for helpful comments concerning this work.
PRIME NUMBERS ALONG RUDIN–SHAPIRO SEQUENCES
37
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