Unexpected biases in the distribution of consecutive primes. Robert J. Lemke Oliver, Kannan Soundararajan Stanford University

Transcription

1 Unexpected biases in the distribution of consecutive primes Robert J. Lemke Oliver, Kannan Soundararajan Stanford University

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

8 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}.

9 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,

10 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:

11 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

12 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

13 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)

14 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)?

15 The primes (mod 10) Let π(x 0 ) = 10 7.

16 The primes (mod 10) Let π(x 0 ) = We find: a π(x 0 ; 10, a) 1 2,499,755 a π(x 0 ; 10, a) 7 2,500, ,500, ,499,751

17 The primes (mod 10) Let π(x 0 ) = We find: a b π(x 0 ; 10, (a, b)) 1 1 2,499, ,500,209 a π(x 0 ; 10, a) 7 2,500, ,499,751

18 The primes (mod 10) Let π(x 0 ) = We find: a b π(x 0 ; 10, (a, b)) , , , , ,500,209 a π(x 0 ; 10, a) 7 2,500, ,499,751

19 The primes (mod 10) Let π(x 0 ) = We find: a b π(x 0 ; 10, (a, b)) , , , , , , , ,915 a b π(x 0 ; 10, (a, b)) , , , , , , , ,032

20 The primes (mod 10) Let π(x 1 ) = We find: a b π(x 1 ; 10, (a, b)) 1 1 4,623, ,429, ,504, ,442, ,010, ,442, ,043, ,502,896 a b π(x 1 ; 10, (a, b)) 7 1 6,373, ,755, ,439, ,431, ,991, ,372, ,012, ,622,916

21 The primes (mod 3) Let π(x 0 ) = 10 7.

22 The primes (mod 3) Let π(x 0 ) = We have π(x 0 ; 3, 1) = 4,999,505 and π(x 0 ; 3, 2) = 5,000,494

23 The primes (mod 3) Let π(x 0 ) = We have π(x 0 ; 3, 1) = 4,999,505 and π(x 0 ; 3, 2) = 5,000,494 while for r 2, we have a π(x 0 ; 3, a) 1,1 1,2 2,1 2,2

24 The primes (mod 3) Let π(x 0 ) = We have π(x 0 ; 3, 1) = 4,999,505 and π(x 0 ; 3, 2) = 5,000,494 while for r 2, we have a π(x 0 ; 3, a) 1,1 2,203,294 1,2 2,796,209 2,1 2,796,210 2,2 2,204,284

25 The primes (mod 3) Let π(x 0 ) = We have π(x 0 ; 3, 1) = 4,999,505 and π(x 0 ; 3, 2) = 5,000,494 while for r 2, we have a π(x 0 ; 3, a) 1,1 2,203,294 1,2 2,796,209 2,1 2,796,210 2,2 2,204,284 and a π(x 0 ; 3, a) 1,1,1 928,276 1,1,2 1,275,018 1,2,1 1,521,062 1,2,2 1,275,147 2,1,1 1,275,018 2,1,2 1,521,191 2,2,1 1,275,147 2,2,2 929,137

26 The primes (mod 3) Let π(x 0 ) = We have π(x 0 ; 3, 1) = 4,999,505 and π(x 0 ; 3, 2) = 5,000,494 while for r 2, we have a π(x 0 ; 3, a) 1,1 2,203,294 1,2 2,796,209 2,1 2,796,210 2,2 2,204,284 and a π(x 0 ; 3, a) 1,1,1 928,276 1,1,2 1,275,018 1,2,1 1,521,062 1,2,2 1,275,147 2,1,1 1,275,018 2,1,2 1,521,191 2,2,1 1,275,147 2,2,2 929,137 Observation The primes dislike to repeat themselves (mod q).

27 What s the deal?

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

29 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)

30 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

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

32 The conjecture: explicit version Conjecture (LO & S) 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) ],

33 The conjecture: explicit version Conjecture (LO & S) 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)},

34 The conjecture: explicit version Conjecture (LO & S) 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.

35 An example Example Let q = 3 or 4.

36 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 ).

37 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 & S) 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)).

38 Comparison with numerics: q = 3 Actual Conj. x π(x; 3, (1, 1)) π(x; 3, (1, 2))

39 Comparison with numerics: q = 3 Actual Pred. Conj. x π(x; 3, (1, 1)) π(x; 3, (1, 2))

40 Comparison with numerics: q = 3 Actual Pred. Conj. x π(x; 3, (1, 1)) π(x; 3, (1, 2))

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

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

43 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

44 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.

45 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))

46 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))

47 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))

48 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))

49 More on the conjectures for r = 2 If a = b then π(x; q, (a, a)) li(x) φ(q) 2 ( + 1 φ(q) 1 log log x 2 log x ( φ(q) log q 2π + log 2π φ(q) p q log p ) 1 ). p 1 2 log x

50 More on the conjectures for r = 2 If a = b then π(x; q, (a, a)) li(x) φ(q) 2 ( If a b then + 1 φ(q) 1 log log x 2 log x ( φ(q) log q 2π + log 2π φ(q) p q π(x; q, (a, b)) li(x) (1 φ(q) log log x 2 log x log p ) 1 ). p 1 2 log x + c 2(q; (a, b)) ). log x

51 More on the conjectures for r = 2 If a = b then π(x; q, (a, a)) li(x) φ(q) 2 ( If a b then + 1 φ(q) 1 log log x 2 log x ( φ(q) log q 2π + log 2π φ(q) p q π(x; q, (a, b)) li(x) (1 φ(q) log log x 2 log x Here c 2 is complicated, but log p ) 1 ). p 1 2 log x + c 2(q; (a, b)) ). log x Λ(q/(q, b a)) c 2 (q; (a, b)) + c 2 (q; (b, a)) = log(2π) φ(q) φ(q/(q, b a)).

52 Other consequences Conjecture Let q be prime. For large x p n x ( pn p ) n+1 li(x) ( 2π log x ) q 2 log x log. q

53 Other consequences Conjecture Let q be prime. For large x p n x ( pn p ) n+1 li(x) ( 2π log x ) q 2 log x log. q Conjecture #{p n x : p n p n+2 (mod q)}

54 Other consequences Conjecture Let q be prime. For large x p n x ( pn p ) n+1 li(x) ( 2π log x ) q 2 log x log. q Conjecture #{p n x : p n p n+2 (mod q)} li(x) φ(q) ( 1 φ(q) log x ).

55 The heuristic when r = 2

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

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

58 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)

59 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) )

60 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!

61 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)

62 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

63 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)

64 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).

65 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,

66 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

67 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

68 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

69 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.

70 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

71 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

72 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

73 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,

74 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,

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

76 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),

77 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).

78 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).

79 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 ).

80 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,

81 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!

82 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.

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

84 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,

85 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

86 Sums of modified singular series Theorem (Montgomery, S) 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 ),

87 Sums of modified singular series Theorem (Montgomery, S) 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.

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

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

90 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

91 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

92 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.

93 The conjecture Conjecture (LO & S) 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.