THE SUM OF DIGITS OF PRIMES

Transcription

1 THE SUM OF DIGITS OF PRIMES Michael Drmota joint work with Christian Mauduit and Joël Rivat Institute of Discrete Mathematics and Geometry Vienna University of Technology supported by the Austrian Science Foundation FWF, grant S9604. ÖMG-DMV Kongress 2009, Graz,

2 Binary Representation of Primes 2 extremal cases p = 2 k + 1 Fermat prime (k = 2 m ) p = 2 k + 2 k Mersenne prime (k + 1 P) Question attributed to Ben Green Given k, does there exist a prime p and 0 = j 1 < j 2 < < j k with p = 2 j j j k??

3 Answer for large k s 2 (n)... number of powers of 2 in the binary expansion of n #{primes p < 2 2k : s 2 (p) = k} 2 2k 2π log 2 k 3 2 BIG open problem (which seems to be completely out of reach!!!) #{primes p : s 2 (p) = k} is finite???

4 Summary Thue-Morse sequence Gelfond s theorem on linear subsequences Gelfond s problems Exponential sum estimates A global central limit theorem A local central limit theorem Proof methods

5 q-ary Digital Expansion q 2... integer basis of digital expansion in N N = {0, 1,..., q 1}... set of digits n N = n = j 0 ε j (n) q j with ε j (n) N. Sum-of-digits function s q (n) = j 0 ε j (n)

6 Thue-Morse Sequence t n = 1 ( 1)s 2(n) 2 t n = { 0 if s2 (n) is even, 1 if s 2 (n) is odd.

7 Thue-Morse Sequence t n = 1 ( 1)s 2(n) 2 0

8 Thue-Morse Sequence t n = 1 ( 1)s 2(n) 2 01

9 Thue-Morse Sequence t n = 1 ( 1)s 2(n)

10 Thue-Morse Sequence t n = 1 ( 1)s 2(n)

11 Thue-Morse Sequence t n = 1 ( 1)s 2(n)

12 Thue-Morse Sequence t n = 1 ( 1)s 2(n)

13 Thue-Morse Sequence t n = 1 ( 1)s 2(n)

14 Thue-Morse Sequence t n = 1 ( 1)s 2(n) t 2 k +n = 1 t n (0 n < 2 k )

15 Thue-Morse Sequence t n = 1 ( 1)s 2(n) t 2 k +n = 1 t n (0 n < 2 k ) or t 2k = t k, t 2k+1 = 1 t k

16 Frequency of Letters #{n < N : t n = 0} = #{n < N : t n = 1} + O(1) = N 2 + O(1) equivalently #{n < N : s 2 (n) 0 mod 2} = #{n < N : s 2 (n) 1 mod 2} + O(1) = N 2 + O(1)

17 Subsequences of the Thue-Morse Sequence (n k ) k 0 increasing (sub)sequence of natural numbers Problem: #{k < K : t nk = 0} =???? Equivalently #{k < K : s 2 (n k ) 0 mod 2} =???? Examples: n k = ak + b n k = k-th prime p k n k = k 2 etc.

18 Linear Subsequences Gelfond 1967/1968 m, s... positive integers with (s, q 1) = 1. = #{n < N : n l mod m, s q (n) t mod s} = N ms + O(N λ ) with 0 < λ < 1. In particular: #{k < K : s 2 (ak + b) 0 mod 2} = #{k < K : t ak+b = 0} = K 2 + O(Kλ )

19 Linear Subsequences q = 2 Lemma n<2 L x s 2(n) y n = l<l (1 + xy 2l) Corollary 1 e(x) := e 2πix #{n < 2 L : n l mod m, s q (n) t mod s} = 1 ms m 1 i=0 s 1 j=0 e = 2L ms + O(2λL ) ( il m jt s ) l<l ( 1 + e ( )) i s + 2l j m

20 Uniform Distribution modulo 1 Corollary 2 α irrational, h 0 integer = e(h α s 2 (n)) = (1 + e(hα)) L = o(2 L ). n<2 L With a little bit more effort: n<n e(h α s 2 (n)) = o(n) Weyl s criterion = αs 2 (n) uniformly distributed modulo 1. By assuming certain Diophantine approximation properties for α these bounds also imply estimates for the discrepancy D N (α s 2 (n)).

21 Gelfond s Problems Gelfond 1967/ q 1, q 2,..., q d 2, (q i, q j ) = 1 for i j, (m j, q j 1) = 1: #{n < N : s qj (n) l j mod m j, 1 j d} = N m 1 m d + O(N 1 η ) 2. (m, q 1) = 1: #{primes p < N : s q (p) l mod m} = π(n) m + O(N 1 η ) 3. (m, q 1) = 1, P (x) N[x]: #{n < N : s q (P (n)) l mod m} = N m + O(N 1 η )

22 Gelfond s Problems Gelfond 1967/ q 1, q 2,..., q d 2, (q i, q j ) = 1 for i j, (m j, q j 1) = 1: Kim 1999 #{n < N : s qj (n) l j mod m j, 1 j d} = N m 1 m d + O(N 1 η ) 2. (m, q 1) = 1: Mauduit, Rivat #{primes p < N : s q (p) l mod m} = π(n) m + O(N 1 η ) 3. (m, q 1) = 1, P (x) N[x]: Mauduit, Rivat for P (x) = x 2 #{n < N : s q (n 2 ) l mod m} = N m + O(N 1 η )

23 Gelfond s 1 st Problem Besineau 1972: solution without error terms Kim 1999: bounds on exponential sums: (e(x) = e 2πix, x = min x k... distance to the nearest integer) 1 N n<n k Z e(α 1 s q1 (n) + α 2 s q2 (n) + + α d s qd (n)) exp η log N d j=1 (α j Q: Kim, α j R: Drmota, Larcher) (q j 1)α j 2,

24 Gelfond s 1 st Problem Applications of Kim s method Drmota, Larcher 2001: q 1, q 2,..., q d 2, (q i, q j ) = 1 for i j, α 1,..., α d irrational: ( α1 s q1 (n),..., α d s qd (n) ) n 0 Rd uniformly distributed mod 1. Thuswaldner, Tichy 2005: q 1, q 2,..., q d 2, (q i, q j ) = 1 for i j. For d > 2 k the number of representations of N = x k xk d with s qj (x j ) l j mod m j, 1 j d is asymptotically given by ( ) S(N) Γ d k m 1 m d Γ ( ) N d ( k 1 + O N k d 1 η). d k

25 Gelfond s 2 nd Problem Mauduit, Rivat 2005+: α real number 1 π(n) p<n e(αs q (p)) exp ( η log N (q 1)α 2 ) Applications α irrational = (αs q (p)) p prime is uniformly distributed mod 1.

26 Gelfond s 2 nd Problem Applications (cont.) Set α = j/m + discrete Fourier analysis = #{primes p < N : s q (p) l mod m} = π(n) m + O(N 1 η ) t n... Thue-Morse sequence = #{primes p < N : t p = 0} = π(n) 2 + O(N 1 η )

27 Gelfond s 2 nd Problem Gaussian primes q = a + i... basis for digital expansion in Z[i] (a {1, 2,...}) N = {0, 1,..., a 2 }... digit set z Z[i] = z = j 0 ε j (z) q j with ε j (z) N s q (z) = j 0 ε j (z)... sum-of-digits function Drmota, Rivat, Stoll 2008 (and generalized by Morgenbesser) Suppose that a 28 such that q = a + i is prime, i.e. a {36, 40, 54, 56, 66, 74, 84, 90, 94,...}. Then 1 N/ log N z 2 N, z prime e(αs q (z)) exp ( η log N (a 2 + 2a + 2)α 2).

28 Gelfond s 3 rd Problem Mauduit, Rivat 1995, c 7 5 : #{n < N : s q ([n c ]) l mod m} N m Dartyge, Tenenbaum 200?: There exists C > 0 with #{n < N : s q (n 2 ) l mod m} C N Drmota, Rivat 2005: s [<λ] 2 (n) = ɛ j (n), s [ λ] 2 (n) = ɛ j (n): j<λ j λ #{n < 2 L : s [<L] 2 (n 2 ) 0 mod 2} 2L 2, #{n < 2 L : s [ L] 2 (n 2 ) 0 mod 2} 2L 2.

29 Gelfond s 3 rd Problem Mauduit, Rivat N n<n e(αs q (n 2 )) (log N) A exp ( η log N (q 1)α 2) Applications #{n < N : s q (n 2 ) l mod m} = N m + O(N 1 η ) #{n < N : t n 2 = 0} = N 2 + O(N λ ) (α s q (n 2 )) n 0 is uniformly distributed modulo 1

30 Global Results Central limit theorem for s q (n): 1 ( ) N #{n N : s q(n) µ q log q N + y σq 2 1 log q N} = Φ(y) + O, log N where and Φ(y) = 1 2π y e 1 2 t2 dt µ q := q 1 2, σ2 q := q distribution function of the normal distribution Remark (q = 2) 1 2 L e its 2(n) = n<2 L ( ) 1 + e it L = CLT 2

31 Global Results Central limit theorem for s q (p): Bassily, Katai 1 π(n) #{p N : s q(p) µ q log q N + y σq 2 log q N} = Φ(y) + o(1). Central limit theorem for s q (n 2 ): Bassily, Katai 1 N #{n N : s q(n 2 ) 2µ q log q N + y 2σq 2 log q N} = Φ(y) + o(1) Remark. The results also hold for s q (P (n)) and s q (P (p)) if P (x) in a non-negative integer polynomial.

32 Local Results for all n #{n < N : s q (n) = k} = N 2πσq 2 log q N ( exp ( (k µ q log q N) 2 2σ 2 q log q N ) + O((log N) 1 2 +ε ) ), Remark: This asymptotic expansion is only significant if Note that 1 N k µ q log q N C(log N) 1 2 n<n s q (n) µ q log q N.

33 Local Results for all n More precise results (only stated for q = 2) Mauduit, Sarközy ( ) k ( #{n < N : s q (n) = k} = F log N, log [log 2 N 2 N] ) ( ( O k log N uniformly for ε log 2 N k (1 ε) log 2 N, where F (x, t) is analytic in x and periodic in t. )), Proof is based on an representation of the form n<n x s 2(n) = F (x, log 2 N)(1 + x) log 2 N. and a saddle point analysis applied to the integral in Cauchy s formula.

34 Local Results for primes Drmota, Mauduit, Rivat 2009: (q, k 1) = 1 #{primes p < N : s q (p) = k} = q 1 ϕ(q 1) where π(n) 2πσq 2 log q N ( exp ( (k µ q log q N) 2 2σ 2 q log q N µ q := q 1 2, σ2 q := q ) + O((log N) 1 2 +ε ) ), Remark: This asymptotic expansion is only significant if Note that 1 π(n) k µ q log q N C(log N) 2 1 p<n s q (p) µ q log q N.

35 Local Results for primes This result does NOT apply for k = 2 and k = [log 2 p] (for q = 2): p is Fermat prime s 2 (p) = 2. p is Mersenne prime s 2 (p) = [log 2 p].

36 Local Results for primes... but we have: #{primes p < 2 2k : s 2 (p) = k} 2 2k 2π log 2 k 3 2

37 Local Results for squares Drmota, Mauduit, Rivat 2009+, Q(a, d) = #{0 n < d : n 2 a mod d}: #{n < N : s q (n 2 ) = k} = N Q(k, q 1) 4πσ 2 q log q N ( exp ( (k 2µ q log q N) 2 4σ 2 q log q N ) + O((log N) 1 2 +ε ) ). Remark: Again this asymptotic expansion is only significant if which is the significant range. k 2µ q log q N C(log N) 1 2

38 Idea of the proof for primes Lemma 1 For every fixed integer q 2 there exist two constants c 1 > 0, c 2 > 0 such that for every k with (k, q 1) = 1 p N p k mod q 1 e(αs q (p)) (log N) 3 N 1 c 1 (q 1)α 2 uniformly for real α with (q 1)α c 2 (log N) 1 2. Remark. This is a refined version of the previous estimate by Mauduit and Rivat.

39 Idea of the proof for primes Lemma 2 Suppose that 0 < ν < 2 1 and 0 < η < 2 ν. Then for every k with (k, q 1) = 1 we have p N, p k mod q 1 e(αs q (p)) = π(n) ϕ(q 1) e(αµ q log q N) (e 2π2 α 2 σq 2 ) log q N (1 + O ( α )) + O ( α (log N) ν ) uniformly for real α with α (log N) η 1 2.

40 Idea of the proof for primes Lemma imply the result Set S(α) := p N e(αs q (p)) and S k (α) := p N, p k mod q 1 e(αs q (p)). Then by using s q (p) p mod (q 1) we get #{p N : s q (p) = k} = 1 1 2(q 1) 1 2(q 1) S(α)e( αk) dα = (q 1) 1 2(q 1) 1 2(q 1) = (q 1) p N, p k mod q 1 1 2(q 1) 1 2(q 1) e(αs q (p)) e( αk) dα S k (α) e( αk) dα.

41 Idea of the proof for primes We split up the integral 1 2(q 1) 1 2(q 1) = α (log N) η 1/2 + (log N) η 1/2 < α 1/(2(q 1)) From Lemma 1 we get an upper bound for the second integral S k (α) e( αk) dα (log N) 2 N e c 1(q 1) 2 (log N) 2η (log N) η 1/2 < α 1/(2(q 1)) π(n) log N.

42 Idea of the proof for primes Set α := t/(2πσ q log q N) and k = k µ q log q N σ 2 q log q N. Then by Lemma 2 we an asymptotic expansion for the first integral: S k(α)e( αk) dα α (log N) η 1/2 = π(n) ϕ(q 1) ( α (log N) η 1/2 e(α(µ q log q N k)) e 2π2 α 2 σ 2 q log q N (1 + O ( +O π(n) α (log α (log N) η 1/2 N)ν dα 1 π(n) = ϕ(q 1) 2πσ q logq N eit k t 2 /2 dt + O (π(n)e 2π2 σq 2 ( ) ( ) π(n) π(n) +O + O log N (log N) 1 ν 2η ( ) 1 π(n) = ϕ(q 1) 2πσq 2 log e 2 k /2 + O((log N) 1 2 +ν+2η )). q N ) (log N)2η)

43 Idea of the proof for primes Proof idea of Lemma 1 Lemma 1 For every fixed integer q 2 there exist two constants c 1 > 0, c 2 > 0 such that for every k with (k, q 1) = 1 p N p k mod q 1 e(αs q (p)) (log N) 3 N 1 c 1 (q 1)α 2 uniformly for real α with (q 1)α c 2 (log N) 1 2.

44 Idea of the proof for primes Proof idea of Lemma 1: Vaughan s method Let q 2, x q 2, 0 < β 1 < 1/3, 1/2 < β 2 < 1. Let g be an arithmetic function. Suppose that uniformly for all real numbers M x and all complex numbers a m, b n such that a m 1, b n 1, we have max x qm <t xq M M q <m M Then M q <m M x qm <n x m g(mn) x qm <n t a m b n g(mn) x/q<p x, p prime U for M x β 1 (type I)., U for x β 1 M x β 2 (type II) g(p) U (log x) 2.

45 Idea of the proof for primes Proof idea of Lemma 1: Type I - sums For q 2, x 2, and for every α such that (q 1)α R \ Z we have max x qm <t xq M M/q<m M for 1 M x 1/3 and e(α s q (mn)) x qm <n t 0 < κ q (α) := min ( 1 6, 1 3 (1 γ q(α)) ) q x 1 κ q(α) log x where 0 γ q (α) < 1 is defined by ( q γq(α) = max 1, max ϕ q (α + t) ϕ q (α + qt) t R with ϕ q (t) = { sin πqt / sin πt if t R \ Z, q if t Z. )

46 Idea of the proof for primes Proof idea of Lemma 1: Type II - sums For q 2, there exist β 1, β 2 and δ verifying 0 < δ < β 1 < 1/3 and 1/2 < β 2 < 1 such that for all α with (q 1)α R \ Z, there exist ξ q (α) > 0 for which, uniformly for all complex numbers b n with b n 1, we have q µ 1 <m q µ whenever q ν 1 <n q ν b n e(αs q (mn)) β 1 δ µ µ + ν β 2 + δ. q (µ + ν)q (1 1 2 ξ q(α))(µ+ν),

47 Idea of the proof for primes Proof idea of Lemma 1: Methods Van-der-Corput Inequality Neglecting large digits periodic structure Discrete Fourier analysis with Fourier terms F λ (h, α) = q λ ( αsq (u) huq λ). 0 u<q λ e

48 Idea of the proof for primes Proof idea of Lemma 2: Lemma 2 Suppose that 0 < ν < 2 1 and 0 < η < 2 ν. Then for every k with (k, q 1) = 1 we have p N, p k mod q 1 e(αs q (p)) = π(n) ϕ(q 1) e(αµ q log q N) (e 2π2 α 2 σq 2 ) log q N (1 + O ( α )) + O ( α (log N) ν ) uniformly for real α with α (log N) η 1 2.

49 Idea of the proof for primes Proof idea of Lemma 2: Interpretation as sum of random variables Lemma 2 is equivalent to S N = S N (p) = s q (p) = ϕ 1 (t) := E e it(s N Lµ q )/(Lσ 2 q )1/2 = e t2 /2 that is uniform for t (log N) η. ( j log q N 1 + O ε j (p). ( )) t t + O log N (log N) 2 1 ν

50 Idea of the proof for primes Proof idea of Lemma 2: Truncation of digits L = log q N, L = #{j Z : L ν j L L ν } = L 2L ν + O(1) (0 < ν < 1 2 ), T N = T N (p) = L ν j L L ν ε j (p) = L ν j L L ν D j, ϕ 2 (t) := E e it(t N L µ q )/(L σ 2 q )1/2 We have, uniformly for all real t ϕ 1 (t) ϕ 2 (t) = O t (log N) 1 2 ν

51 Idea of the proof for primes Proof idea of Lemma 2: variables Approximation by sum of iid random Z j (j 0) iid random variables with range {0, 1,..., q 1} and P{Z j = l} = 1 q. T N := L ν j L L ν Z j. We have and E T N = L µ q and V T N = L σ 2 q ϕ 3 (t) := E e it(t N L µ q )/(L σq 2)1/2 = e t2 /2 1 + O ( ( )) t log N (This is the classical central limit theo- uniformly for t (log N) 1 2. rem.)

52 Idea of the proof for primes Proof idea of Lemma 2: quantification We have uniformly for real t with t L η ϕ 2 (t) ϕ 3 (t) = O ( t e c 1L κ), where 0 < 2η < κ < ν. and c 1 > 0 is a constant depending on η and κ. This estimate directly proves Lemma 2. Remark: Taylor theorem gives for random variables X, Y : Ee itx Ee ity = + O d<d 0 d<d ( t D D! (it) d d! ( E X d E Y d) E X D E Y D t D + E Y D D! ).

53 Idea of the proof for primes Proof idea of Lemma 2: comparision of moments We have uniformly for 1 d L E T N L µ q L σq 2 d = E T N L µ q L σq 2 d + O (( 4q σ q ) L (12 +ν)d e c 4L ν ), which can be reduced to compare frequencies Pr{D j1,m = l 1,..., D jd,n = l d } = Pr{Z j1 = l 1,..., Z jd = l d } + O ( (4L ν ) d e c 4L ν) Recall: T N := L ν j L L ν D j, T N := Z j L ν j L L ν

54 Idea of the proof for primes Proof idea of Lemma 2: comparision of moments Proof of this property uses exponential sum estimats over primes ( ) A e Q p (log x) 5 x q K/2 p x Erdős-Turán inequality Discrepancy of (x n ) 1 H + H h=1 1 h 1 N e(hx n), trivial observation: ɛ j (n) = d { n } q j+1 [ d q, d + 1 ). q

55 Open problem Polynomials: P (x)... integer polynomial of degree d 3: distribution properties of s q (P (n))?? There are only results for very large q q 0 (d). General problem: Let S be a set of natural numbers that are determined by congruence conditions and bounds on the exponents of the prime factorization. What is the distribution of (s q (n)) n S?? Examples: primes, squares, cubes, square-free numbers etc.

56 Thank You!