The sum of digits of primes in Z[i] Thomas Stoll (TU Wien) Journées de Numération, Graz 2007 April 19, 2007 (joint work with M Drmota and J Rivat) Research supported by the Austrian Science Foundation, no9604
Gaussian primes Complex sum-of-digits Pointillism: Gaussian primes in the first quadrant
Gaussian primes Introduction Gaussian primes Complex sum-of-digits Gaussian primes p: (1) 1 + i and its associates, (2) the rational primes 4k + 3 and their associates, (3) the factors in Z[i] of the rational primes 4k + 1 Hecke prime number theorem: π i (N) := {p Z[i] non-associated : p 2 N} N log N Complex Von Mangoldt function Λ i : { log p, n = εp Λ i (n) = ν, ε unit, ν Z + ; 0, otherwise
The complex sum-of-digits function Gaussian primes Complex sum-of-digits [Kátai/Kovács, Kátai/Szabó] Let q = a ± i (choose a sign) with a Z + Then every n Z[i] has a unique finite representation λ 1 n = ε j q j, j=0 where ε j {0, 1,, a 2 } are the digits and ε λ 1 0 Let s q (n) = λ 1 j=0 ε j be the sum-of-digits function in Z[i]
I Introduction Statements Tools and difference process Recall q = a ± i, a Z + and write e(x) = exp(2πix) Theorem For any α R with (a 2 + 2a + 2)α Z, a even, there is σ q (α) > 0 such that Λ i (n) e(αs q (n)) N 1 σq(α) n 2 N Theorem The sequence (αs q (p)), a even, running over Gaussian primes p is uniformly distributed modulo 1 if and only if α R \ Q
Statements Tools and difference process II Theorem Let a 2, a even, b Z 0, m Z +, m 2 and set d = (m, a 2 + 2a + 2) If (b, d) = 1 then there exists σ q,m > 0 such that # { p Z[i] : p 2 N, s q (p) b mod m} d = m ϕ(d) π i(n) + O q,m (N 1 σq,m ) If (b, d) 1 then the set has cardinality O q,m (N 1 σq,m )
Statements Tools and difference process Inequalities à la Vaughan and van der Corput Lemma (Vaughan) Let β 1 (0, 1 3 ), β 2 ( 1 2, 1) Further suppose that for all a n, b n with a n, b n 1, n Z[i] and all M x we uniformly have (put Q = a 2 + 1) max e(αs q (mn)) U for M x β 1, M Q < m 2 M x Q m 2 <t x m 2 M Q < m 2 x M Q m 2 < n 2 x Then x Q < n 2 x x Q m 2 < n 2 t a m b n e(αs q (mn)) U for x β 1 M x β 2 m 2 Λ i (n) e(αs q (n)) U(log x)2
Statements Tools and difference process Lemma (Van der Corput) Let z n C with n Z[i] and A, B 0 Then for all R 1 we have A< n <B z n 2 C 3 ( B A R ) + 2 B + A R r <2R ( 1 r ) 2R A< n <B A< n+r <B z n+r z n Now, start with S = Q µ 1 < m 2 Q µ b n e(αs q (mn)) Q ν 1 < n 2 Q ν
How to get the difference process: Statements Tools and difference process Denote f (n) = αs q (n) Then with Cauchy-Schwarz ineq, S 2 Q µ Q µ 1 < m 2 Q µ and with Van der Corput ineq, 2 b n e(f (mn)), Q ν 1 < n 2 Q ν S 2 Q 2(µ+ν) ρ + Q µ+ν max 1 r < q ρ Q ν 1 < n 2 Q ν e (f (m(n + r)) f (mn)) Q µ 1 < m 2 Q µ
The truncated sum-of-digits function Statements Tools and difference process We introduce the truncated sum-of-digits function of Z[i], defined by λ 1 λ 1 f λ (z) = f (ε j q j ) = α ε j, where λ Z and λ 0 j=0 Periodicity property: For any d Z[i], j=0 f λ (z + dq λ ) = f λ (z), z Z[i] Reason: Let d = x + iy Use the identities iq = aq + q 2, Q = (a 1) 2 q + (2a 1)q 2 + q 3
Statements Tools and difference process At only small cost: Carry propagation lemma Put λ = µ + 2ρ Lemma Let a 2 For all integers µ > 0, ν > 0, 0 ρ < ν/2 and r Z[i] with r < q ρ denote by E(r, µ, ν, ρ) the number of pairs (m, n) Z[i] Z[i] with Q µ 1 < m 2 Q µ, Q ν 1 < n 2 Q ν and f (m(n + r)) f (mn) f λ (m(n + r)) f λ (mn) Then we have E(r, µ, ν, ρ) Q µ+ν ρ
Statements Tools and difference process At only small cost II: The addition automaton 0 1 a 2 1 a 2 a 2 0 [ ] P R 0 2a 1 (a 1) 2 a 2 Q 0 (a 1) 2 2a 1 a 2 0 2a a 2 2a a 2 0 a 2 2a + 2 2a 2 a 2 a 2 2a + 2 0 a 2 2a 2 2a 0 a 2 a 2 2a (a 1) 2 0 2a 1 0 Q a 2 2a 1 a 2 (a 1) 2 1 0 a 2 a 2 1 R P [ ] 0 a 2 Performs addition by 1 (P), by a i (R) and by a 1 + i (Q) Example: Let a = 3 and z = 58 40i = (ε 0, ε 1, ε 2) = (8, 2, 7), and consider z + 2 + i The corresponding walk is Q 8 3 Q 2 7 Q 7 2 Q 0 5 Q 0 5 P 0 1 [ ], thus z + 2 + i = (3, 7, 2, 5, 5, 1)
Statements Tools and difference process At only small cost III: Proof Idea of proof of the carry propagation lemma : Assume mr = x + iy = y( a i) + (x + ay) with y < 0, x > ay Write f (mn + mr) f (mn) = f (mn + mr) f (mn + (a + i) + mr) + f (mn + (a + i) + mr) f (mn + 2(a + i) + mr) + + f (mn (y + 1)(a + i) + mr) f (mn y(a + i) + mr) + f (mn y(a + i) + mr) f (mn y(a + i) + mr 1) + f (mn y(a + i) + mr 1) f (mn y(a + i) + mr 2) + + f (mn y(a + i) + mr x ay + 1) f (mn)
Orthogonality relation Introduction Preliminaries Some estimates Let F λ = { λ 1 j=0 ε jq j : ε j N } be the fundamental region of the number system, which is a complete system of residues mod q λ with #F λ = Q λ Hence, z F λ e where tr (z) = 2R(z) Put ( tr (hzq λ ) ) = { Q λ, h 0 mod q λ ; 0, otherwise, λ F λ (h, α) = Q λ ( ϕ Q α tr (hq j ) ), j=1 where ϕ Q (t) = { sin(πqt) / sin(πt), t R \ Z; Q, t Z
Transformation Introduction Preliminaries Some estimates Let Then S 2(n) = 1 Q 2λ S 2(n) = u F λ Q µ 1 < m 2 Q µ e (f λ (m(n + r)) f λ (mn)) e(f λ (u) f λ (v)) v F λ e h F λ k F λ Q µ 1 < m 2 Q µ = h F λ k F λ F λ (h, α)f λ ( k, α) ( h(m(n + r) u) tr q λ + tr Q µ 1 < m 2 Q µ e ( tr ) k(mn v) q λ (h + k)mn + hmr q λ )
Preliminaries Some estimates of F λ Lemma For all α R, ξ C and a 3 we have λ 1 α tr (ξq j ) 2 λ 2 (a 2 2(a 2 + 1) 2 + 2a + 2)α 2 j=0 Corollary There exists a constant C a > 0 only depending on a such that F λ (h, α) exp( C a λ (a 2 + 2a + 2)α 2 ) uniformly for all h Z[i], α R and integers λ 0
Preliminaries Some estimates of G λ Define G λ (b, d, α) := h F λ h b mod d F λ (h, α), G λ (α) = G λ (0, 1, α) = Lemma For a 2, b Z[i], α R, 0 δ λ there is η Q < 1 2 and G λ (b, q δ, α) = h F λ h b mod q δ F λ (h, α) Q η Q(λ δ) F δ (b, α) h F λ F λ (h, α) In particular, G λ (α) = G λ (0, 1, α) Q η Qλ
Preliminaries Some estimates II of G λ Lemma For a 2, b Z[i], α R, 0 δ λ, k Z[i] and k q λ δ, q k we have G λ (b, kq δ, α) 2 k 2η 5 Q η 5(λ δ) F δ (b, α) Lemma For a even we have F λ (h, α) 2 = F δ (b, α) 2 h F λ h b mod q δ and thus, in particular, h F λ F λ (h, α) 2 = 1