The distribution of consecutive prime biases

Transcription

1 The distribution of consecutive prime biases Robert J. Lemke Oliver 1 Tufts University (Joint with K. Soundararajan) 1 Partially supported by NSF grant DMS

2 Chebyshev s Bias Let π(x; q, a) := #{p < x : p a(mod q)}.

3 Chebyshev s Bias Let π(x; q, a) := #{p < x : p a(mod q)}. Under GRH, we have π(x; q, a) = li(x) φ(q) + O(x 1/2+ɛ ) if (a, q) = 1.

4 Chebyshev s Bias Let π(x; q, a) := #{p < x : p a(mod q)}. Under GRH, we have π(x; q, a) = li(x) φ(q) + O(x 1/2+ɛ ) if (a, q) = 1. Observation (Chebyshev) Quadratic residues seem slightly less frequent than non-residues.

5 Chebyshev s Bias Let π(x; q, a) := #{p < x : p a(mod q)}. Under GRH, we have π(x; q, a) = li(x) φ(q) + O(x 1/2+ɛ ) if (a, q) = 1. Observation (Chebyshev) Quadratic residues seem slightly less frequent than non-residues. Theorem (Rubinstein-Sarnak) Under GRH(+ɛ), π(x; 3, 2) > π(x; 3, 1) for 99.9% of x,

6 Chebyshev s Bias Let π(x; q, a) := #{p < x : p a(mod q)}. Under GRH, we have π(x; q, a) = li(x) φ(q) + O(x 1/2+ɛ ) if (a, q) = 1. Observation (Chebyshev) Quadratic residues seem slightly less frequent than non-residues. Theorem (Rubinstein-Sarnak) Under GRH(+ɛ), π(x; 3, 2) > π(x; 3, 1) for 99.9% of x, and analogous results hold for any q.

7 The primes (mod 3) π(x) π(x; 3, 1) π(x; 3, 2)

8 The primes (mod 3) π(x) π(x; 3, 1) π(x; 3, 2)

9 The primes (mod 3) π(x) π(x; 3, 1) π(x; 3, 2)

10 The primes (mod 3) π(x) π(x; 3, 1) π(x; 3, 2)

11 The primes (mod 3) π(x) π(x; 3, 1) π(x; 3, 2)

12 The primes (mod 4) π(x) π(x; 4, 1) π(x; 4, 3)

13 The primes (mod 4) π(x) π(x; 4, 1) π(x; 4, 3)

14 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

15 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

16 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

17 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

18 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

19 The primes (mod 5) π(x) π(x; 5, 1) π(x; 5, 2) π(x; 5, 3) π(x; 5, 4)

20 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ),

21 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}.

22 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r,

23 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r, but little is known:

24 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r, but little is known: If r = 2 and φ(q) = 2, then each a occurs infinitely often

25 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r, but little is known: If r = 2 and φ(q) = 2, then each a occurs infinitely often Shiu: The pattern (a, a,..., a) occurs infinitely often

26 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r, but little is known: If r = 2 and φ(q) = 2, then each a occurs infinitely often Shiu: The pattern (a, a,..., a) occurs infinitely often Maynard: π(x; q, (a, a,..., a)) π(x)

27 Patterns of consecutive primes Let r 1 and a = (a 1,..., a r ), and set π(x; q, a) := #{p n < x : p n+i a i+1 (mod q) for 0 i < r}. We expect that π(x; q, a) li(x)/φ(q) r, but little is known: If r = 2 and φ(q) = 2, then each a occurs infinitely often Shiu: The pattern (a, a,..., a) occurs infinitely often Maynard: π(x; q, (a, a,..., a)) π(x) Question Are there biases between the different patterns a(mod q)?

28 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

29 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

30 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

31 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

32 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

33 Consecutive primes (mod 3) π(x) π(x; 3, (1, 1)) π(x; 3, (1, 2)) π(x; 3, (2, 1)) π(x; 3, (2, 2))

34 Consecutive primes (mod 4) π(x) π(x; 4, (1, 1)) π(x; 4, (1, 3)) π(x; 4, (3, 1)) π(x; 4, (3, 3))

35 Consecutive primes (mod 5) Let π(x 0 ) = 10 7.

36 Consecutive primes (mod 5) Let π(x 0 ) = We find: a π(x 0 ; 5, a) 1 2,499,755 a π(x 0 ; 5, a) 3 2,500, ,500, ,499,751

37 Consecutive primes (mod 5) Let π(x 0 ) = We find: a b π(x 0 ; 5, (a, b)) 1 1 2,499, ,500,283 a π(x 0 ; 5, a) 3 2,500, ,499,751

38 Consecutive primes (mod 5) Let π(x 0 ) = We find: a b π(x 0 ; 5, (a, b)) , , , , ,500,283 a π(x 0 ; 5, a) 3 2,500, ,499,751

39 Consecutive primes (mod 5) Let π(x 0 ) = We find: a b π(x 0 ; 5, (a, b)) , , , , , , , ,851 a b π(x 0 ; 5, (a, b)) , , , , , , , ,032

40 Consecutive primes (mod 5) Let π(x 1 ) = We find: a b π(x 1 ; 5, (a, b)) 1 1 4,623, ,504, ,429, ,442, ,373, ,439, ,755, ,431,870 a b π(x 1 ; 5, (a, b)) 3 1 6,010, ,043, ,442, ,502, ,991, ,012, ,372, ,622,916

41 What s the deal?

42 What s the deal? We conjecture that: There are large secondary terms in the asymptotic for π(x; q, a)

43 What s the deal? We conjecture that: There are large secondary terms in the asymptotic for π(x; q, a) The dominant factor is the number of a i a i+1 (mod q)

44 What s the deal? We conjecture that: There are large secondary terms in the asymptotic for π(x; q, a) The dominant factor is the number of a i a i+1 (mod q) There are smaller, somewhat erratic factors that affect non-diagonal a

45 The conjecture: explicit version Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2.

46 The conjecture: explicit version Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x + c 2(q; a) log x + O ( log 7/4 x) ],

47 The conjecture: explicit version Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x where c 1 (q; a) = φ(q) 2 + c 2(q; a) log x + O ( log 7/4 x) ], ( ) r 1 φ(q) #{1 i < r : a i a i+1 (mod q)},

48 The conjecture: explicit version Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x where c 1 (q; a) = φ(q) 2 + c 2(q; a) log x + O ( log 7/4 x) ], ( ) r 1 φ(q) #{1 i < r : a i a i+1 (mod q)}, and c 2 (q; a) is complicated but explicit.

49 An example Example Let q = 3 or 4.

50 An example Example Let q = 3 or 4. Then π(x; q, (a, b)) = li(x) 4 [ ( log log x 1 ± 2 log x )] ( log 2π/q + +O 2 log x x log 11/4 x ).

51 An example Example Let q = 3 or 4. Then π(x; q, (a, b)) = li(x) 4 [ ( log log x 1 ± 2 log x )] ( log 2π/q + +O 2 log x x log 11/4 x ). Conjecture (LO & Soundararajan) Let q = 3 or 4. If a b(mod q), then for all x > 5, π(x; q, (a, b)) > π(x; q, (a, a)).

52 Comparison with numerics (mod 3)

53 Comparison with numerics (mod 3) x π(x; 3, (1, 1)) π(x; 3, (1, 2)) 10 9 Actual

54 Comparison with numerics (mod 3) x π(x; 3, (1, 1)) π(x; 3, (1, 2)) 10 9 Actual Conj

55 Comparison with numerics (mod 3) x π(x; 3, (1, 1)) π(x; 3, (1, 2)) 10 9 Actual Conj

56 The conjecture when q = 5 When q = 5,

57 The conjecture when q = 5 When q = 5, we predict that c 2 (5; (a, a)) = 3 log(2π/5), 2

58 The conjecture when q = 5 When q = 5, we predict that c 2 (5; (a, a)) = and if a b, then c 2 (5; (a, b)) = log(2π/5) 2 3 log(2π/5), R ( L(0, χ)l(1, χ)a 5,χ [ χ(b a) + ]) χ(b) χ(a), 4

59 The conjecture when q = 5 When q = 5, we predict that c 2 (5; (a, a)) = and if a b, then c 2 (5; (a, b)) = log(2π/5) 2 3 log(2π/5), R ( L(0, χ)l(1, χ)a 5,χ [ χ(b a) + ]) χ(b) χ(a), 4 where A 5,χ = p 5 (1 ) (χ(p) 1)2 (p 1) i.

60 Comparison with numerics: q = 5 x π(x; 5, (1, 1)) π(x; 5, (1, 2)) π(x; 5, (1, 3)) π(x; 5, (1, 4))

61 Comparison with numerics: q = 5 x π(x; 5, (1, 1)) π(x; 5, (1, 2)) π(x; 5, (1, 3)) π(x; 5, (1, 4))

62 Comparison with numerics: q = 5 x π(x; 5, (1, 1)) π(x; 5, (1, 2)) π(x; 5, (1, 3)) π(x; 5, (1, 4))

63 Comparison with numerics: q = 5 x π(x; 5, (1, 1)) π(x; 5, (1, 2)) π(x; 5, (1, 3)) π(x; 5, (1, 4))

64 The heuristic when r = 2 π(x; q, (a, b)) =

65 The heuristic when r = 2 π(x; q, (a, b)) = n<x: n a(mod q) 1 P (n)

66 The heuristic when r = 2 π(x; q, (a, b)) = n<x: n a(mod q) 1 P (n) h>0: h b a(mod q) 1 P (n + h)

67 The heuristic when r = 2 π(x; q, (a, b)) = n<x: n a(mod q) 1 P (n) t<h: (t+a,q)=1 h>0: h b a(mod q) ( 1 1 P (n + t) 1 P (n + h) )

68 The heuristic when r = 2 π(x; q, (a, b)) n<x: n a(mod q) 1 P (n) t<h: (t+a,q)=1 h>0: h b a(mod q) ( 1 1 P (n + h) 1 log x ) Please do not try this at home!

69 The heuristic when r = 2 π(x; q, (a, b)) n<x: n a(mod q) 1 P (n) t<h: (t+a,q)=1 h>0: h b a(mod q) ( 1 q 1 ) φ(q) log x 1 P (n + h)

70 The heuristic when r = 2 π(x; q, (a, b)) n<x: n a(mod q) n<x: n a(mod q) 1 P (n) t<h: (t+a,q)=1 1 P (n) h>0: h b a(mod q) ( 1 q 1 ) φ(q) log x h>0: h b a(mod q) 1 P (n + h) h/ log x 1 P (n + h)e

71 The heuristic when r = 2 π(x; q, (a, b)) = n<x: n a(mod q) n<x: n a(mod q) 1 P (n) t<h: (t+a,q)=1 h b a(mod q) 1 P (n) h>0: h b a(mod q) ( 1 q 1 ) φ(q) log x h>0: h b a(mod q) h/ log x e n<x: n a(mod q) 1 P (n + h) h/ log x 1 P (n + h)e 1 P (n)1 P (n + h)

72 The Hardy-Littlewood conjecture We need to understand h/ log x e h b a(mod q) n<x n a(mod q) 1 P (n)1 P (n + h).

73 The Hardy-Littlewood conjecture We need to understand h/ log x e h b a(mod q) n<x n a(mod q) 1 P (n)1 P (n + h). Conjecture (Hardy-Littlewood) For any h 0, we have n<x x 1 P (n)1 P (n + h) S(h) log 2 x,

74 The Hardy-Littlewood conjecture We need to understand h/ log x e h b a(mod q) n<x n a(mod q) 1 P (n)1 P (n + h). Conjecture (Hardy-Littlewood) For any h 0, we have n<x x 1 P (n)1 P (n + h) S(h) log 2 x, where S(h) = p ( 1 ) ( #({0, h}mod p) 1 1 ) 2. p p

75 The Hardy-Littlewood conjecture We need to understand h/ log x e h b a(mod q) n<x n a(mod q) 1 P (n)1 P (n + h). Conjecture (Hardy-Littlewood) For any h 0, we have n<x n a(mod q) x 1 P (n)1 P (n + h) S(h) log 2 x, where S(h) = p ( 1 ) ( #({0, h}mod p) 1 1 ) 2. p p

76 The Hardy-Littlewood conjecture We need to understand h/ log x e h b a(mod q) n<x n a(mod q) 1 P (n)1 P (n + h). Conjecture (Hardy-Littlewood) For any h 0 with (h + a, q) = 1, we have n<x n a(mod q) 1 P (n)1 P (n + h) q φ(q) 2 S x q(h) log 2 x, where S q (h) = p q ( 1 ) ( #({0, h}mod p) 1 1 ) 2. p p

77 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x.

78 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x. Gallagher: S q (h) = 1 on average

79 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x. Gallagher: S q (h) = 1 on average Blindly plugging this in, we have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) h/ log x e

80 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x. Gallagher: S q (h) = 1 on average Blindly plugging this in, we have π(x; q, (a, b)) q φ(q) 2 q φ(q) 2 x log 2 x x log 2 x h b a(mod q) log x q h/ log x e

81 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x. Gallagher: S q (h) = 1 on average Blindly plugging this in, we have π(x; q, (a, b)) q φ(q) 2 q φ(q) 2 x log 2 x x log 2 x h b a(mod q) log x q h/ log x e = 1 x φ(q) 2 log x,

82 The main term We now have π(x; q, (a, b)) q φ(q) 2 x log 2 x h b a(mod q) S q (h)e h/ log x. Gallagher: S q (h) = 1 on average Blindly plugging this in, we have π(x; q, (a, b)) which is the main term. q φ(q) 2 q φ(q) 2 x log 2 x x log 2 x h b a(mod q) log x q h/ log x e = 1 x φ(q) 2 log x,

83 The source of the bias Consider h b a(mod q) S q (h)e h/ log x.

84 The source of the bias Consider Proposition We have h 0(mod q) h b a(mod q) S q (h)e h/ log x = log x q S q (h)e h/ log x. φ(q) 2q log log x + O(1),

85 The source of the bias Consider Proposition We have h 0(mod q) while for v 0(mod q), h b a(mod q) S q (h)e h/ log x = log x q h v(mod q) S q (h)e h/ log x. φ(q) 2q S q (h)e h/ log x = log x q log log x + O(1), + O(1).

86 The source of the bias Consider Proposition We have h 0(mod q) while for v 0(mod q), h b a(mod q) S q (h)e h/ log x = log x q h v(mod q) S q (h)e h/ log x. φ(q) 2q S q (h)e h/ log x = log x q Idea Only the first Dirichlet series has a pole at s = 0. log log x + O(1), + O(1).

87 Much needed rigor To do this properly, we need to be more careful with (1 1 P (n + t)) ( 1 t<h: (t+a,q)=1 t<h: (t+a,q)=1 q φ(q) log x ).

88 Much needed rigor To do this properly, we need to be more careful with (1 1 P (n + t)) ( 1 t<h: (t+a,q)=1 t<h: (t+a,q)=1 q φ(q) log x ). Could expand out and invoke Hardy-Littlewood,

89 Much needed rigor To do this properly, we need to be more careful with (1 1 P (n + t)) ( 1 t<h: (t+a,q)=1 t<h: (t+a,q)=1 q φ(q) log x ). Could expand out and invoke Hardy-Littlewood, but this is ugly!

90 Much needed rigor To do this properly, we need to be more careful with (1 1 P (n + t)) ( 1 t<h: (t+a,q)=1 t<h: (t+a,q)=1 q φ(q) log x ). Could expand out and invoke Hardy-Littlewood, but this is ugly! Better idea: Incorporate inclusion-exclusion directly into H-L.

91 Modified Hardy-Littlewood Define 1 P (n) = 1 P (n) q φ(q) log n.

92 Modified Hardy-Littlewood Define 1 P (n) = 1 P (n) q φ(q) log n. Conjecture If H = k with (h + a, q) = 1 for all h H, then n x h H n a(mod q) 1P (n + h) qk 1 φ(q) k S x q,0(h) log k x,

93 Modified Hardy-Littlewood Define 1 P (n) = 1 P (n) q φ(q) log n. Conjecture If H = k with (h + a, q) = 1 for all h H, then n x h H n a(mod q) 1P (n + h) qk 1 φ(q) k S x q,0(h) log k x, where S q,0 (H) := ( 1) H\T S q (T ). T H

94 Sums of modified singular series Theorem (Montgomery, Soundararajan) H [1,h] H =k where µ k = 0 if k is odd. S 0 (H) = µ k k! ( h log h + Ah)k/2 + O k (h k/2 δ ),

95 Sums of modified singular series Theorem (Montgomery, Soundararajan) H [1,h] H =k where µ k = 0 if k is odd. S 0 (H) = µ k k! ( h log h + Ah)k/2 + O k (h k/2 δ ), Point We can discard H with H 3.

96 Notes on assembly Three flavors of two-term H:

97 Notes on assembly Three flavors of two-term H: H = {0, h}: Contributes main-term, bias against diagonal

98 Notes on assembly Three flavors of two-term H: H = {0, h}: Contributes main-term, bias against diagonal H = {0, t}, {t, h}: (One interloper) Contributes to c 2

99 Notes on assembly Three flavors of two-term H: H = {0, h}: Contributes main-term, bias against diagonal H = {0, t}, {t, h}: (One interloper) Contributes to c 2 H = {t 1, t 2 }: (Two interlopers) Unbiased

100 Notes on assembly Three flavors of two-term H: H = {0, h}: Contributes main-term, bias against diagonal H = {0, t}, {t, h}: (One interloper) Contributes to c 2 H = {t 1, t 2 }: (Two interlopers) Unbiased Remark We expect H with H 3 to contribute further lower-order terms.

101 The conjecture Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x where c 1 (q; a) = φ(q) 2 + c 2(q; a) log x + O ( log 7/4 x) ], ( ) r 1 φ(q) #{1 i < r : a i a i+1 (mod q)}, and c 2 (q; a) is complicated but explicit.

102 The constant c 2 (q; a)

103 The constant c 2 (q; a) If q is prime, then c 2 (q; (a, a)) = q 2 2 log(q/2π),

104 The constant c 2 (q; a) If q is prime, then while c 2 (q; (a, b)) q = 1 φ(q) c 2 (q; (a, a)) = q 2 2 χ χ 0 (mod q) log(q/2π), χ(b a)l(0, χ)l(1, χ)a q,χ + o(1),

105 The constant c 2 (q; a) If q is prime, then while c 2 (q; (a, b)) q where = 1 φ(q) c 2 (q; (a, a)) = q 2 2 χ χ 0 (mod q) A q,χ = p q (1 log(q/2π), χ(b a)l(0, χ)l(1, χ)a q,χ + o(1), ) (χ(p) 1)2 (p 1) 2.

106 How big is c 2 (q; a)? A heuristic

107 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ,

108 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ, so roughly 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ

109 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ, so roughly 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ q1/2 φ(q) χ χ 0 (mod q) e(θ χ )

110 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ, so roughly 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ q1/2 φ(q) χ χ 0 (mod q) q1/2 φ(q) q1/2 e(θ χ )

111 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ, so roughly 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ q1/2 φ(q) χ χ 0 (mod q) q1/2 φ(q) q1/2 1. e(θ χ )

112 How big is c 2 (q; a)? A heuristic Expect L(1, χ) 1 and L(0, χ) q 1/2 for typical χ, so roughly 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ q1/2 φ(q) χ χ 0 (mod q) q1/2 φ(q) q1/2 1. Thus, we should typically expect c 2 (q; a)/q 1. e(θ χ )

113 The distribution of c 2 (q; a) Theorem (LO & Soundararajan) As q, c 2 (q; (a, b))/q has a continuous limiting distribution,

114 The distribution of c 2 (q; a) Theorem (LO & Soundararajan) As q, c 2 (q; (a, b))/q has a continuous limiting distribution, e.g. #{a b(mod q) : c 2 (q; (a, b))/q τ} lim q φ(q)(φ(q) 1) exists and is non-zero.

115 The distribution of c 2 (q; a) Theorem (LO & Soundararajan) As q, c 2 (q; (a, b))/q has a continuous limiting distribution, e.g. #{a b(mod q) : c 2 (q; (a, b))/q τ} lim q φ(q)(φ(q) 1) exists and is non-zero. Moreover, there is A > 0 such that [ ] c 2 (q; (a, b)) P q > τ exp( Ae τ /τ).

116 The distribution of c 2 (q; a) Theorem (LO & Soundararajan) As q, c 2 (q; (a, b))/q has a continuous limiting distribution, e.g. #{a b(mod q) : c 2 (q; (a, b))/q τ} lim q φ(q)(φ(q) 1) exists and is non-zero. Moreover, there is A > 0 such that [ ] c 2 (q; (a, b)) P q > τ exp( Ae τ /τ). Remark This is reminiscent of the distribution of L(1, χ).

117 A simplified version Recall that c 2 (q; (a, b)) q = 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ + o(1).

118 A simplified version Recall that c 2 (q; (a, b)) q = 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ + o(1). Consider instead the simplified version 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ).

119 A simplified version Recall that c 2 (q; (a, b)) q = 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ + o(1). Consider instead the simplified version 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ). It turns out this is related to the Dedekind sum s q (a),

120 A simplified version Recall that c 2 (q; (a, b)) q = 1 φ(q) χ χ 0 (mod q) χ(b a)l(0, χ)l(1, χ)a q,χ + o(1). Consider instead the simplified version 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ). It turns out this is related to the Dedekind sum s q (a), given by s q (a) = x(mod q) ψ(x/q)ψ(ax/q), ψ(x) = x x 1/2.

121 Dedekind sums The Dedekind sum s q (a) is large when a/q is close to a rational with small denominator,

122 Dedekind sums The Dedekind sum s q (a) is large when a/q is close to a rational with small denominator, e.g. s q (1) = q q,

123 Dedekind sums The Dedekind sum s q (a) is large when a/q is close to a rational with small denominator, e.g. s q (1) = q q, but it s typically much smaller:

124 Dedekind sums The Dedekind sum s q (a) is large when a/q is close to a rational with small denominator, e.g. s q (1) = q q, but it s typically much smaller: Theorem (Vardi) As q, s q(a) log q is dist. according to the Cauchy distribution.

125 Dedekind sums The Dedekind sum s q (a) is large when a/q is close to a rational with small denominator, e.g. s q (1) = q q, but it s typically much smaller: Theorem (Vardi) As q, s q(a) log q is dist. according to the Cauchy distribution. Remark dµ Cauchy = 2 1+4π 2 x 2 dx has infinite expectation.

126 The Fourier transform of Dedekind sums Consider for 0 t q 1 the Fourier transform ŝ q (t) := 1 q s q (a)e(at/q). a(mod q)

127 The Fourier transform of Dedekind sums Consider for 0 t q 1 the Fourier transform ŝ q (t) := 1 q s q (a)e(at/q). a(mod q) Then 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ)

128 The Fourier transform of Dedekind sums Consider for 0 t q 1 the Fourier transform ŝ q (t) := 1 q s q (a)e(at/q). a(mod q) Then 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ) = πi ŝ q (t).

129 The Fourier transform of Dedekind sums Consider for 0 t q 1 the Fourier transform ŝ q (t) := 1 q s q (a)e(at/q). a(mod q) Then 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ) = πi ŝ q (t). Theorem (LO & Soundararajan) As q, ŝ q (t) has a continuous limiting distribution.

130 A closer connection to L(1, χ) Using L(1, χ) = b 1 χ(b) b and L(0, χ) = 1 q with orthogonality of characters, a(mod q) χ(a)ψ(a/q)

131 A closer connection to L(1, χ) Using L(1, χ) = b 1 χ(b) b and L(0, χ) = 1 q a(mod q) with orthogonality of characters, we find 1 φ(q) χ χ 0 (mod q) χ(t)l(0, χ)l(1, χ) χ(a)ψ(a/q)

132 A closer connection to L(1, χ) Using L(1, χ) = b 1 χ(b) b and L(0, χ) = 1 q a(mod q) with orthogonality of characters, we find 1 φ(q) χ χ 0 (mod q) χ(a)ψ(a/q) χ(t)l(0, χ)l(1, χ) = b 1 (b,q)=1 ψ(t b/q), b

133 A closer connection to L(1, χ) Using L(1, χ) = b 1 χ(b) b and L(0, χ) = 1 q a(mod q) with orthogonality of characters, we find 1 φ(q) χ χ 0 (mod q) χ(a)ψ(a/q) χ(t)l(0, χ)l(1, χ) = where b is the inverse of b(mod q). b 1 (b,q)=1 ψ(t b/q), b

134 The random model Thus, we ve found ŝ q (t) = 1 πi b 1 (b,q)=1 ψ(t b/q). b

135 The random model Thus, we ve found ŝ q (t) = 1 πi b 1 (b,q)=1 ψ(t b/q). b Idea: Compare this to the random model 1 πi b 1 ψ(x/b), x R, b

136 The random model Thus, we ve found ŝ q (t) = 1 πi b 1 (b,q)=1 ψ(t b/q). b Idea: Compare this to the random model 1 πi b 1 and use the method of moments. ψ(x/b), x R, b

137 The random model Thus, we ve found ŝ q (t) = 1 πi b 1 (b,q)=1 ψ(t b/q). b Idea: Compare this to the random model 1 πi b 1 and use the method of moments. ψ(x/b), x R, b Remark A similar expression holds for c 2 (q; a).

138 Another connection: A conjecture of Montgomery

139 Another connection: A conjecture of Montgomery Conjecture (Montgomery) Define R(x) by φ(n) = 3 π 2 x 2 + R(x). n x

140 Another connection: A conjecture of Montgomery Conjecture (Montgomery) Define R(x) by φ(n) = 3 π 2 x 2 + R(x). n x Then R(x) x log log x and R(x) = Ω ± (x log log x).

141 Another connection: A conjecture of Montgomery Conjecture (Montgomery) Define R(x) by φ(n) = 3 π 2 x 2 + R(x). n x Then R(x) x log log x and R(x) = Ω ± (x log log x). Theorem (LO & Soundararajan) R(x)/x has a limiting distribution with doubly exponential decay.

142 Wait, how is this related?

143 Wait, how is this related? Theorem (Montgomery) R(x) = Ω ± (x log log x).

144 Wait, how is this related? Theorem (Montgomery) R(x) = Ω ± (x log log x). Idea: Montgomery shows that R(x) = x b x µ(b)ψ(x/b) b + O(x/ log x),

145 Wait, how is this related? Theorem (Montgomery) R(x) = Ω ± (x log log x). Idea: Montgomery shows that R(x) = x b x and chooses x very cleverly. µ(b)ψ(x/b) b + O(x/ log x),

146 Wait, how is this related? Theorem (Montgomery) R(x) = Ω ± (x log log x). Idea: Montgomery shows that R(x) = x b x and chooses x very cleverly. µ(b)ψ(x/b) b But this is very close to our random model! + O(x/ log x),

147 The conjecture Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x where c 1 (q; a) = φ(q) 2 and c 2 (q; a) is complicated, explicit, + c 2(q; a) log x + O ( log 7/4 x) ], ( ) r 1 φ(q) #{1 i < r : a i a i+1 (mod q)},

148 The conjecture Conjecture (LO & Soundararajan) Let a = (a 1,..., a r ) with r 2. Then π(x; q, a) = li(x) [ log log x φ(q) r 1 + c 1 (q; a) log x where c 1 (q; a) = φ(q) 2 + c 2(q; a) log x + O ( log 7/4 x) ], ( ) r 1 φ(q) #{1 i < r : a i a i+1 (mod q)}, and c 2 (q; a) is complicated, explicit, and nicely distributed.