Les chiffres des nombres premiers. (Digits of prime numbers)

Transcription

1 Les chiffres des nombres premiers (Digits of prime numbers) Joël RIVAT Institut de Mathématiques de Marseille, UMR 7373, Université d Aix-Marseille, France. joel.rivat@univ-amu.fr soutenu par le projet ANR MUNUM. Marseille, 22 septembre

2 Works with Christian MAUDUIT (Marseille) 2

3 (Pseudo-)Randomness and primes Are primes random? What type of results to expect? 3

4 Prime Number Theorem p is always a prime number. Von Mangoldt function: Λ(n) = log p if n = p k, Λ(n) = 0 otherwise. Prime Number Theorem (Hadamard, de la Vallée Poussin, 1896, independently): n x Λ(n) = x + o(x). We say that a reasonable function f satisfies a Prime Number Theorem (PNT) if we can get an assymptotic formula for n x Λ(n)f(n). Example: Vinogradov(-Helfgott) theorem with f(n) = exp(2iπαn). 4

5 Möbius Randomness Principle Möbius function: µ(n) = { ( 1) r if n = p 1 p r (distinct), 0 if p such that p 2 n. Exercise: n x Λ(n) = x + o(x) is equivalent to n x µ(n) = o(x). We say that a reasonable function f satisfies the Möbius Randomness Principle (MRP) if µ(n)f(n) is small. n x For general f the MRP might be (slightly) less difficult to show than the PNT. 5

6 What is a reasonable function f? Dynamical systems point of view: f is produced by (X, T ) if there exist x 0 X and ϕ continuous on X such that for any integer n, we have f(n) = ϕ(t n (x 0 )). Sarnak s conjecture: f satifies the MRP if f is bounded and produced by a dynamical system of zero topological entropy. (i.e. with a sub-exponential rate of increase of the number of ɛ-different orbits of length n, when n + and ɛ 0 + ). Complexity point of view: Bourgain, Green for MRP: f is of small computational complexity (boolean functions). 6

7 Are the digits of primes random? Does ε j (p) look like a random variable? 7

8 Prescribing the binary digits of primes Bourgain has given an asymptotic for the number of prime numbers with a positive proportion of preassigned digits, improving previous results due to himself (2013), Harman-Kátai (2008), Harman (2006), Wolke (2005) and Kátai (1986): Bourgain, 2014: There exists c > 0 such that given integers k and l with 1 l ck, for any 1 j 1 < < j l < k and for any (b 1,..., b l ) {0, 1} l and b l 0 when j l = k 1, then we have card{p < 2 k, ε j1 (p) = b 1,..., ε jl (p)) = b l } 1 2 l 2 k log 2 k (k ). 8

9 Gelfond s paper In base q 2 any n N can be written n = j 0 ε j (n) q j where ε j (n) {0,..., q 1}. Gelfond, 1968: The sum of digits s(n) = ε j (n) is well distributed in arithmetic progressions. Gelfond s problems, 1968: j 0 1. Given two bases q 1 and q 2 such that (q 1, q 2 ) = 1, evaluate the number of integers n x such that s q1 (n) a 1 mod m 1 and s q2 (n) a 2 mod m Evaluate the number of prime numbers p x such that s(p) a mod m. 3. Evaluate the number of integers n x such that s(p (n)) a mod m, where P is a suitable polynomial [for example P (n) = n 2 ]. 9

10 Proof of Gelfond s conjecture for primes Mauduit-Rivat, 2010: If (q 1)α R \ Z, there exists C q (α) > 0 and σ q (α) > 0, n x Λ(n) exp(2iπα s(n)) C q(α) x 1 σq(α). This the PNT with f(n) = exp(2iπα s(n)). Hence - for q 2 the sequence (α s(p n )) n 1 is equidistributed modulo 1 if and only if α R \ Q (here (p n ) n 1 denotes the sequence of prime numbers). - for q 2, m 2 such that (m, q 1) = 1 and a Z, p x s(p) a mod m 1 1 m p x 1 (x + ). 10

11 Gaussian distribution around the mean value Drmota-Mauduit-Rivat, 2009: uniformly for all integers k 0 with (k, q 1) = 1 where card{p x : s(p) = k} = q 1 ϕ(q 1) π(x) 2πσq 2 log q x ( exp ( (k µq log q x) 2 2σ 2 q log q x µ q = q 1 2, σ2 q = q and ε > 0 is arbitrary but fixed. ) + O((log x) 1 2 +ε ) ), Such a local result was considered by Erdős as hopelessly difficult. 11

12 Discrete Fourier Transform Let f : N C. For λ N let f λ : Z C be 2 λ -periodic function such that f λ (0) = f(0),..., f λ (2 λ 1) = f(2 λ 1). The Discrete Fourier Transform of f λ is defined for t R by f λ (t) = 1 ( 2 λ f λ (u) e ut ) 0 u<2 λ 2 λ = 12 λ f(u) e 0 u<2 λ ( ut ) 2 λ. By orthogonality we always have f λ 2 = 0 h<2 λ f λ (h) 2 1/2 = 1. Giving a non-trivial upper bound for f λ or f λ 1 is a challenging problem. Getting f λ 1 = O(2 ηλ ) with η < 1/2 was crucial for solving Gelfond s conjecture. 12

13 The Rudin-Shapiro sequence f(n) = e 1 2 j 1 ε j 1 (n)ε j (n) = ( 1) j 1 ε j 1(n)ε j (n) This sequence is famous to satisfy As usual f λ 2 1 λ 2. f λ 2 = 1. By Cauchy-Schwarz it is easy to deduce that 2 λ 1 2 f λ 1 2 λ 2. Therefore f λ 1 is quite big and (so to say) η = 1 2. The proof for the sum of digits function cannot be adapted for the Rudin-Shapiro sequence. 13

14 General result Definitions Let U = {z C, z = 1}. Definition 1 A function f : N U has the carry property if, uniformly for (λ, κ, ρ) N 3 with ρ < λ, the number of integers 0 l < q λ such that there exists (k 1, k 2 ) {0,..., q κ 1} 2 with f(lq κ + k 1 + k 2 ) f(lq κ + k 1 ) f κ+ρ (lq κ + k 1 + k 2 ) f κ+ρ (lq κ + k 1 ) is at most O(q λ ρ ) where the implied constant may depend only on q and f. Definition 2 Given a non decreasing function γ : R R satisfying lim λ + γ(λ) = + and c > 0 we denote by F γ,c the set of functions f : N U such that for (κ, λ) N 2 with κ cλ and t R: q λ 0 u<q λ f(uq κ ) e ( ut) q γ(λ). 14

15 General result Mauduit-Rivat, 2015: Let γ : R R be a non decreasing function satisfying lim λ + γ(λ) = +, c 10 and f : N U be a function satisfying Definition 1 and f F γ,c in Definition 2. For any θ R we have n x Λ(n)f(n) e (θn) c 1(q)(log x) c2(q) x q γ(2 (log x)/80 log q )/20, with explicit c 1 (q) and c 2 (q). The same estimate holds if we replace the von Mangoldt function Λ by the Möbius function µ. Corollary: we get a quantitative PNT and MRP for the Rudin-Shapiro sequence and some of its generalizations. Note: Tao has sketched on Mathoverflow an approach which might have permitted to get (only) the MRP for the Rudin-Shapiro sequence. 15

16 Ideas and tools 16

17 Approach to digital problem 1. Reduce the problem to an exponential sum, 2. apply several times the van der Corput inequality to remove the upper and lower digits, 3. separate into a discrete Fourier transform part and an analytic part, 4. handle the analytic part to see which Fourier estimates are needed, 5. obtain the corresponding Fourier estimates. 17

18 Sum over prime numbers By partial summation f(p) Λ(n) f(n) where Λ(n) is von Mangoldt s function. p x n x Advantage: convolutions! Λ 1l = log, i.e. d n Λ(d) = log n. A classical process (Vinogradov, Vaughan, Heath-Brown) remains (using some more technical details), for some 0 < β 1 < 1/3 and 1/2 < β 2 < 1, to estimate uniformly the sums S I := m M n N f(mn) for M x β 1 (type I) where MN = x (which implies that the easy sum over n is long) and for all complex numbers a m, b n with a m 1, b n 1 the sums S II := m M n N (which implies that both sums have a significant length). a m b n f(mn) for x β 1 < M x β 2 (type II), 18

19 Sums of type I The knowledge of the function f should permit to deal with the sum f(mn) = f(k) = 1 ( kl f(k) e m m n N k MN k 0 mod m 0 l<m k MN ). In our case f(n) is some digital function. Some arguments from Fouvry and Mauduit (1996) can be generalized. We use the large sieve inequality: If x 1,..., x R R satisfy x r x s δ for all r s and a 1,..., a K C, then R K r=1 k=1 a k e(kx r ) 2 ( K ) K a k 2. δ k=1 19

20 Sums of type II By Cauchy-Schwarz: S II 2 M m M n N b n f(mn) Expanding the square and exchanging the summations leads to a smooth sum over m, but also two free variables n 1 and n 2 with no control. 2. Using Van der Corput s inequality we have n 1 = n + r and n 2 = n so that the size of n 1 n 2 = r is small. It remains to estimate n N b n+r b n m M If f(k) = e(αg(k)) we have the difference f(m(n + r))f(mn). g(mn + mr) g(mn) where the upper digits might be removed. 20

21 Removing the upper digits f is q-additive if for all k 0 and all (a 0,..., a k ) {0,..., q 1} k+1, f(a 0 + a 1 q + + a k q k ) = f(a 0 ) + f(a 1 q) + + f(a k q k ). Consider the difference f(a + b) f(a), with b q β much smaller that a q α. Example: a = α {}}{ , b = , } {{ } β In the sum a + b the digits of index β may change only by carry propagation. The proportion of pairs (a, b) for which the carry propagation exceeds β + ρ is likely to be O(q ρ ). If so we can ignore these exceptional pairs and replace f(a + b) f(a) by f β+ρ (a + b) f β+ρ (a) where f β+ρ is the truncated f function which considers only the digits of index < β + ρ: which is periodic of period q β+ρ. f β+ρ (n) := f(n mod q β+ρ ) 21

22 A variant of van der Corput s inequality For all complex numbers z 1,..., z L and integers k 1, R 1 we have 2 L z l l=1 L + kr k R L l=1 z l R 1 r=1 (for k = 1 this is the classical van der Corput s inequality.) ( 1 r ) L kr R l=1 R ( ) z l+kr z l. Taking k = 1, this permits to remove the upper digits. Taking k = q µ 1, this permits to remove the lower digits. 22

23 After two van der Corput s inequalities We have also removed the lower digits and are led to consider m n f µ1,µ 2 (mn + mr + q µ 1sn + q µ 1rs)f µ1,µ 2 (mn + mr) f µ1,µ 2 (mn + q µ 1sn)f µ1,µ 2 (mn), where f µ1,µ 2 depends only on the digits of index in {µ 1,..., µ 2 1}. so that for mn and mn + mr we need to detect their digits of indexes between µ 1 and µ 2 1 while for sn and rs it is sufficient to look the digits of index < µ 2 µ 1. This is done by Fourier analysis. 23

24 Sums of type II - Fourier analysis We are now working modulo q µ 2 µ 1. We consider Discrete Fourier Transform f µ1,µ 2 (t) = 1 ( ) q µ f µ1,µ 2 µ 1 2 (q µ ut 1u) e q µ. 2 µ 1 0 u<q µ 2 µ 1 By Fourier inversion formula and exchanges of summations we must show that the quantity h 0, h 1, h 2, h 3 <q µ 2 µ 1 e n N m M is estimated by O(q µ+ν ρ ). f µ1,µ 2 (h 0 ) f µ1,µ 2 (h 1 ) f µ1,µ 2 (h 2 ) f µ1,µ 2 (h 3 ) ( (h0 + h 1 + h 2 + h 3 )mn + (h 0 + h 1 )mr + (h 0 + h 2 )q µ 1sn q µ 2 ) The geometric sums over m and n give very good estimates except for diagonal terms like those for h 0 + h 1 + h 2 + h 3 = 0. Then the proof depends on the properties of f µ1,µ 2. 24

25 Open problems What is the numbers of prime numbers with missing digits? (e.g. without the digit 9 in base 10) What is the number of prime numbers with twice more 1 than 0 (in base 2)? 25