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Gaps and Differences Between Primes

Dan Goldston

San Jose State University

April 13, 2018

Dan Goldston Gaps and Differences Between Primes

A Recent Breakthrough on Primes

Twin Prime Conjecture: Does p − p′ = 2 have infinitely manysolutions?

Prime Pair Conjecture: For each k , Does p − p′ = 2k haveinfinitely many solutions?

Zhang 2013: There exists an even integer 2k < 70, 000, 000 wherethis is true!

Maynard-Tao-Polymath 2014: There exist an even integer2k ≤ 246 where this is true.

Distribution of Primes.

The 25 prime numbers less than 100:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,

71, 73, 79, 83, 89, 97

The first 10 primes starting at a billion:

1000000007, 1000000009, 1000000021, 1000000033, 1000000087,

1000000093, 1000000097, 1000000103, 1000000123, 1000000181

You never run out because if p1, p2, p3, . . . pk are k primes, then let

N = P + 1, P = p1p2p3 · · · pk

and N will have all new prime factors since

p1= p2p3 · · · pk +

p1is not an integer.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,

71, 73, 79, 83, 89, 97

1000000007, 1000000009, 1000000021, 1000000033, 1000000087,

1000000093, 1000000097, 1000000103, 1000000123, 1000000181

N = P + 1, P = p1p2p3 · · · pk

p1= p2p3 · · · pk +

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,

71, 73, 79, 83, 89, 97

1000000007, 1000000009, 1000000021, 1000000033, 1000000087,

1000000093, 1000000097, 1000000103, 1000000123, 1000000181

N = P + 1, P = p1p2p3 · · · pk

p1= p2p3 · · · pk +

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,

71, 73, 79, 83, 89, 97

1000000007, 1000000009, 1000000021, 1000000033, 1000000087,

1000000093, 1000000097, 1000000103, 1000000123, 1000000181

N = P + 1, P = p1p2p3 · · · pk

p1= p2p3 · · · pk +

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,

71, 73, 79, 83, 89, 97

1000000007, 1000000009, 1000000021, 1000000033, 1000000087,

1000000093, 1000000097, 1000000103, 1000000123, 1000000181

N = P + 1, P = p1p2p3 · · · pk

p1= p2p3 · · · pk +

Letπ(x) = the number of primes ≤ x =

∑p≤x

where p always denotes a prime.

20 40 60 80 100

Figure: π(x) = the number of primes ≤ x for 0 ≤ x ≤ 100

Letπ(x) = the number of primes ≤ x =

∑p≤x

where p always denotes a prime.

20 40 60 80 100

Figure: π(x) = the number of primes ≤ x for 0 ≤ x ≤ 100Dan Goldston Gaps and Differences Between Primes

200 400 600 800 1000

Figure: π(x) for 0 ≤ x ≤ 1000

200 000 400 000 600 000 800 000 1 ´ 106

20 000

40 000

60 000

80 000

Figure: π(x) for 0 ≤ x ≤ 106

This suggests the Prime Number Theorem:

π(x) ∼ x

log x, i. e. lim

x→∞

π(x)x

= 1, log x = loge x = ln x .

200 000 400 000 600 000 800 000 1 ´ 106

20 000

40 000

60 000

80 000

Figure: π(x) and xlog x for 1 ≤ x ≤ 106

This suggests the Prime Number Theorem:

π(x) ∼ x

log x, i. e. lim

x→∞

π(x)x

= 1, log x = loge x = ln x .

200 000 400 000 600 000 800 000 1 ´ 106

20 000

40 000

60 000

80 000

Figure: π(x) and xlog x for 1 ≤ x ≤ 106

A better approximation for π(x) is the Cramer Model:

Probability( n is a prime) = 1log n

π(x) ∼∑n≤x

log n∼ li(x),

li(x) =

log tdt

is called the logarithmic integral.

A better approximation for π(x) is the Cramer Model:

Probability( n is a prime) = 1log n

π(x) ∼∑n≤x

log n∼ li(x),

li(x) =

log tdt

is called the logarithmic integral.

1.02 ´ 106 1.04 ´ 106 1.06 ´ 106 1.08 ´ 106 1.10 ´ 106

80 000

81 000

82 000

83 000

84 000

85 000

Figure: π(x) and li(x) for 106 ≤ x ≤ 106 + 105

It took 100 years from when Gauss first conjectured this form ofthe PNT till it was proved in 1896 independently by Hadamard andde la Vallee Poussin. That li(x) is better than x

log x was proved byde la Vallee Poussin in 1899.

1.02 ´ 106 1.04 ´ 106 1.06 ´ 106 1.08 ´ 106 1.10 ´ 106

80 000

81 000

82 000

83 000

84 000

85 000

1.02 ´ 106 1.04 ´ 106 1.06 ´ 106 1.08 ´ 106 1.10 ´ 106

80 000

81 000

82 000

83 000

84 000

85 000

The Riemann Zeta-function

Define

ζ(s) =∞∑n=1

ns= 1 +

4s+ · · · ,

where s = σ + it, and series conversed absolutely when σ > 1.Also for σ > 1

ζ(s) =∏

p prime

p3s+ · · ·

∏p prime

(1− 1

)−1This is called the Euler product.

Zeros of the Riemann Zeta-function

We can analytically continue ζ(s), and find

ζ(s)− 1

s − 1

is analytic everywhere in the complex plane. We can then provethat ζ(s) has zeros at s = −2,−4,−6, . . . ”trivial zeros” andcomplex zeros ρ = β + iγ, with 0 ≤ β ≤ 1. Key step in proof ofPNT is β < 1. RH is conjecture β = 1/2.

First 5 zeros are

2+ i14.134,

2+ i21.022,

2+ i25.010,

2+ i30.424,

2+ i32.935,

Zeros of the Riemann Zeta-function

We can analytically continue ζ(s), and find

ζ(s)− 1

s − 1

is analytic everywhere in the complex plane. We can then provethat ζ(s) has zeros at s = −2,−4,−6, . . . ”trivial zeros” andcomplex zeros ρ = β + iγ, with 0 ≤ β ≤ 1. Key step in proof ofPNT is β < 1. RH is conjecture β = 1/2.First 5 zeros are

2+ i14.134,

2+ i21.022,

2+ i25.010,

2+ i30.424,

2+ i32.935,

The sequence of imaginary parts of the zeros

{14.134, 21.022, 25.01, 30.424, 32.935, 37.586, · · · }

and the sequence of primes

{2, 3, 5, 7, 11, 13, 17, . . .}

encode each other.

Landau formula: ∑0<γ≤T

has spikes when x equals a prime or prime power. Take absolutevalue of this we get

The sequence of imaginary parts of the zeros

{14.134, 21.022, 25.01, 30.424, 32.935, 37.586, · · · }

and the sequence of primes

{2, 3, 5, 7, 11, 13, 17, . . .}

encode each other.Landau formula: ∑

0<γ≤Tx iγ

has spikes when x equals a prime or prime power. Take absolutevalue of this we get

2 4 6 8 10

Figure: Using 10 zeros for 0 ≤ x ≤ 10

Theorem: 6 is not a prime.

2 4 6 8 10

Theorem: 6 is not a prime.

2 4 6 8 10

5 10 15 20 25

Twin Primes

Now consider the sequence

3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73, 101, 103, . . . .

These are pairs of primes that are two apart: the sequence of twinprimes.Here are the twin primes starting at a billion:

1000000007, 1000000009, 1000000409, 1000000411, 1000000931,

1000000933, 1000001447, 1000001449, 1000001789, 1000001791,

1000001801, 1000001803.

Twin Primes

3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73, 101, 103, . . . .

These are pairs of primes that are two apart: the sequence of twinprimes.

Here are the twin primes starting at a billion:

1000000007, 1000000009, 1000000409, 1000000411, 1000000931,

1000000933, 1000001447, 1000001449, 1000001789, 1000001791,

1000001801, 1000001803.

Twin Primes

3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73, 101, 103, . . . .

These are pairs of primes that are two apart: the sequence of twinprimes.Here are the twin primes starting at a billion:

1000000007, 1000000009, 1000000409, 1000000411, 1000000931,

1000000933, 1000001447, 1000001449, 1000001789, 1000001791,

1000001801, 1000001803.

Letπ2(x) = the number of pairs of twin primes ≤ x .

20 40 60 80 100

Figure: π2(x) for 0 ≤ x ≤ 100

200 400 600 800 1000

20 000 40 000 60 000 80 000 100 000

As for the Prime Number Theorem, we conjecture

π2(x) ∼ (1.3203 . . .)

(log t)2.

200000 400000 600000 800000 1·106

Figure: For 0 ≤ x ≤ 106

Why 1.3203 . . .?

What is the probability that n and n + 2 are both

prime? The Cramer model would suggest that:

The chance that n is prime is 1/ log n.

The chance that n + 2 is prime is 1/ log(n + 2) ∼ 1/ log n.

Therefore by independence the probability of both being prime is1/(log n)2. Thus

π2(x) ∼∫ x

(log t)2dt

Why 1.3203 . . .? What is the probability that n and n + 2 are both

π2(x) ∼∫ x

(log t)2dt

π2(x) ∼∫ x

(log t)2dt

π2(x) ∼∫ x

(log t)2dt

π2(x) ∼∫ x

(log t)2dt

The problem with the Cramer’s model is that it fails to take intoaccount divisibility. Thus, for the primes p > 2, the probabilitythat p + 1 is prime is not 1/ log(p + 1) as suggested by the Cramermodel but rather 0 since p + 1 is even.

Further p + 2 is necessarily odd; therefore it is twice as likely to beprime as a random number. The conclusion is that n and n + 2being primes are not independent events.

The problem with the Cramer’s model is that it fails to take intoaccount divisibility. Thus, for the primes p > 2, the probabilitythat p + 1 is prime is not 1/ log(p + 1) as suggested by the Cramermodel but rather 0 since p + 1 is even.

Further p + 2 is necessarily odd; therefore it is twice as likely to beprime as a random number. The conclusion is that n and n + 2being primes are not independent events.

Correction: We need neither n and n + 2 to be divisible by2, 3, 5, 7, 11, · · · .

The chance that two random numbers are both odd is(1/2)(1/2) = 1/4, but, since n being odd forces n + 2 to be odd,the chance that n and n + 2 are both odd is 1/2, and thus twice aslarge as random.

The chance that neither of two random numbers is divisible by 3 is(2/3)(2/3) = 4/9, but the chance that neither n and n + 2 aredivisible by 3 is 1/3 since this occurs if and only if n is congruentto 2 modulo 3.

In general, the probability that neither of two random numbers aredivisible by p > 2 is (1− 1/p)2, while the probability that neitherof n and n + 2 is divisible by p is the slightly smaller 1− 2/p sincen must miss the two residue classes 0 and −2 modulo p.

Therefore, the correction factor to the Cramer model for lack ofindependence is 2 if p = 2, and for p ≥ 3 is(

1− 2

)(1− 1

)−2=

((1− 1

)(1− 1

)−2=

(1− 1

(p − 1)2

We conclude that the correct approximation for twin primes shouldbe

π2(x) ∼ 2∏p>2

(1− 1

(p − 1)2

(log x)2.

1− 2

)(1− 1

)−2=

((1− 1

)(1− 1

)−2=

(1− 1

(p − 1)2

π2(x) ∼ 2∏p>2

(1− 1

(p − 1)2

(log x)2.

1− 2

)(1− 1

)−2=

((1− 1

)(1− 1

)−2=

(1− 1

(p − 1)2

π2(x) ∼ 2∏p>2

(1− 1

(p − 1)2

(log x)2.

Hardy-Littlewood Conjecture

Let H = {h1, h2, . . . , hk} be a set of k distinct integers, anddenote by π(x ;H) the number of positive integers n ≤ x for whichn + h1, n + h2, . . . , n + hk are all primes.

In 1923, Hardy and Littlewood conjectured for x →∞

π(x ;H) ∼ S(H)lik(x),

lik(x) =

(log t)k,

and S(H) is the singular series defined by the product over allprimes p

S(H) =∏p

(1− 1

)−k (1− νH(p)

where νH(p) is the number of distinct residue classes occupied bythe elements of H. We assume S(H) 6= 0 here.

lik(x) =

(log t)k,

S(H) =∏p

(1− 1

)−k (1− νH(p)

lik(x) =

(log t)k,

S(H) =∏p

(1− 1

)−k (1− νH(p)

where νH(p) is the number of distinct residue classes occupied bythe elements of H.

We assume S(H) 6= 0 here.

lik(x) =

(log t)k,

S(H) =∏p

(1− 1

)−k (1− νH(p)

The Hardy-Littlewood prime pair conjecture is that

π2(x , d) :=∑p≤x

p−p′=d

1 ∼ S(d)x

(log x)2, as x →∞,

Here p′ is a previous prime before p but not necessarily adjacent top. The singular series S(d) is defined by

S(d) =

∏p|dp>2

(p − 1

p − 2

), if d is even;

0, if d is odd;

C2 =∏p>2

(1− 1

(p − 1)2

)= 0.66016 . . .

with the product extending over all primes p > 2.

The History of the Jumping Champion Problem

In the 1977-1978 volume of Journal of Recreational Mathematics,H. Nelson proposed Problem 654:

“Find the most probable difference between consecutiveprimes.”

Editor’s Comment in the 1978-1979 volume:

“No solution has been received, though there hasbeen a good deal of evidence presented pointing to thereasonable conjecture that there is no most probabledifference between consecutive primes. On the otherhand, there is also some evidence that 6 is the mostprobable such difference . . . However, there seems to begood reason to expect that 30 will eventually replace 6 asthe most probable difference and still later 210, 2310,30030, etc. will have their day.”

N(x , d) and N∗(x)

Let pn denote the nth prime. Let

N(x , d) :=∑pn≤x

pn−pn−1=d

N∗(x) := maxd

N(x , d).

The set D∗(x) of jumping champions for primes ≤ x is given by

D∗(x) := {d∗ : N(x , d∗) = N∗(x)}.

N(x , d) :=∑pn≤x

pn−pn−1=d

N∗(x) := maxd

N(x , d).

D∗(x) := {d∗ : N(x , d∗) = N∗(x)}.

N(x , d) :=∑pn≤x

pn−pn−1=d

N∗(x) := maxd

N(x , d).

D∗(x) := {d∗ : N(x , d∗) = N∗(x)}.

Example

x = 100

Primes ≤ x

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,

61, 67, 71, 73, 79, 83, 89, 97

Difference:

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8

N(100, 1) = 1,N(100, 2) = 8,N(100, 4) = 7,

N(100, 6) = 7,N(100, 8) = 1.

N∗(100) = 8, d∗ = 2

Example

x = 100

Primes ≤ x

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,

61, 67, 71, 73, 79, 83, 89, 97

Difference:

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8

N(100, 1) = 1,N(100, 2) = 8,N(100, 4) = 7,

N(100, 6) = 7,N(100, 8) = 1.

N∗(100) = 8, d∗ = 2

Example

x = 100

Primes ≤ x

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,

61, 67, 71, 73, 79, 83, 89, 97

Difference:

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8

N(100, 1) = 1,N(100, 2) = 8,N(100, 4) = 7,

N(100, 6) = 7,N(100, 8) = 1.

N∗(100) = 8, d∗ = 2

Example

x = 100

Primes ≤ x

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,

61, 67, 71, 73, 79, 83, 89, 97

Difference:

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8

N(100, 1) = 1,N(100, 2) = 8,N(100, 4) = 7,

N(100, 6) = 7,N(100, 8) = 1.

N∗(100) = 8, d∗ = 2

Table 1 below summarizes everything we presently know aboutjumping champions less than or equal to x .

Table: Known jumping champions for small x .

D∗(x) Smallest Prime Largest Known PrimeOccurrence of x Occurrence of x

{1} 3 3{1, 2} 5 5{2} 7 433{2, 4} 101 173{4} 131 541{2, 4, 6} 179 487{2, 6} 379 463{6} 389 > 1015

{4, 6} 547 941

This graph shows the PJC’s for all prime x ≤ 1800.

Figure: Prime jumping champions for all odd, prime x ≤ 1800

In 1999 A. Odlyzko, M. Rubinstein, and M. Wolf conjectured

Conjecture The jumping champions greater than 1 are 4 and theprimorials 2, 6, 30, 210, 2310, . . ..

Their main concern was finding the transition between 6 and 30,which they estimated to be at 1.7427 · 1035. (30 transitions to 210at about 10425 ?)

Theorem(Goldston-Ledoan 2015) Assume a form of theHardy-Littlewood prime pair and prime triple conjecture. Thensufficiently large PJC’s increases through the sequence ofprimorials.

S(d) = 2C2

∏p|dp>2

p − 2

Thus S(d) increases most rapidly on the sequence of primorials

For example, primes differing by 6 should occur twice as frequentlyas primes differing by 2.

(Aside: Primes differing by 6 are called Sexy Primes.)

S(d) = 2C2

∏p|dp>2

p − 2

S(d) = 2C2

∏p|dp>2

p − 2

S(d) = 2C2

∏p|dp>2

p − 2

Idea for getting Jumping Champions

Erdos and Straus (1980) proved that asymptotically the number ofdifferences p − p′ = d that are not between consecutive primes areinsignificant as long as d grows smaller than the average spacing ofprimes log x for primes ≤ x .

For if p and p′ are not consecutive primes, then there must exist athird prime p′′ such that p′ < p′′ < p. Writing d ′ = p − p′′, wecount the number of such triplets of primes with

π3(x , d , d ′) :=∑p≤x

p−p′=dp−p′′=d ′

Idea for getting Jumping Champions

Erdos and Straus (1980) proved that asymptotically the number ofdifferences p − p′ = d that are not between consecutive primes areinsignificant as long as d grows smaller than the average spacing ofprimes log x for primes ≤ x .

For if p and p′ are not consecutive primes, then there must exist athird prime p′′ such that p′ < p′′ < p. Writing d ′ = p − p′′, wecount the number of such triplets of primes with

π3(x , d , d ′) :=∑p≤x

p−p′=dp−p′′=d ′

Inclusion-Exclusion gives

π2(x , d)−∑

1≤d ′<d

π3(x , d , d ′) ≤ N(x , d) ≤ π2(x , d).

We can use a sieve to bound the number of triples and get theresult. To prove the G-L result you use Hardy-Littlewoodconjectures for pairs and triples, plus sieve bounds for quadruplesto get N(x , d).

Unfortunately we can’t prove anything unconditionally

Problem: Prove the PJC increase to infinity.

Problem: Prove some even numbers are prime jumping LOSERS.

Problem: Prove some even number is or isn’t a PJC for sufficientlylarge x .

Unfortunately we can’t compute anything

Problem: All computations show that 6 is the PJC when x > 1000.

Unfortunately we can’t compute anything

Problem: All computations show that 6 is the PJC when x > 1000.

Prime Difference Champions

Joint work with Scott Funkhouser, Dipendra Sengupta, and JharnaSengupta.

In a 2012 SPAWAR technical report (unclassified) Cotwright,Funkhouser, Sengupta, and Williams looked at all the differencesbetween primes up to 106 and studied numerically patterns in thedifferences.

SPAWAR: Space and Naval Warfare Systems Command

Let D∗(x) be the most common difference between the primes≤ x . We call this the PDC.

Conjecture: The PDC are 4 and the primorials.

This looks very similar to earlier work on differences betweenconsecutive primes: Prime Jumping Champions

PDC are similar to PJC but we can prove and compute thingsabout them. Analogously to N(x , d) define

G (x , d) :=∑

p,p′≤xp′−p=d

Figure: Difference count for x = 105

Here is a zoom in:

Figure: Detail of difference count for x = 105; the PDC is 2310 and thecorresponding count is circled.

The PDC for a given x is any d for which G (x , d) attains itsmaximum,

G ∗(x) := maxd

G (x , d) ,

D∗(x) := {d : G (x , d) = G ∗(x)} .

For example: D∗(5) = {1, 2, 3}.

Figure: Prime difference champions for all odd, prime x no greater than2× 108. The dot-dashed line is a plot of log10(x/ log(x)) and the dashedline is a plot of log10(x/ log2(x)), in association with Theorem 1.

Transistion Ranges

Aside from 1, 3 and 4 all PDC’s in the numerically explored rangeof x are primorials. Additionally the PDC’s run through theprimorials in a broadly step-wise manner as x increases.

Table: Transition points for PDC’s

primorial smallest x for largest known prime x foroccurrence as PDC occurrence as PDC

6 p7 = 17 p41 = 17930 p32 = 131 p269 = 1723

210 p224 = 1423 p3523 = 328432310 p2718 = 24499 p35000 = 414977

30030 p34903 = 413863 p609928 = 9120277510510 p607867 = 9087131 p10657197 = 191945597

9699690 p10561154 = 190107653 −

Figure: Zoom-in view of D∗(x) over region where transition from 210 to2310 occurs.

The Hardy-Littlewood prime pair conjecture

The simplest conjecture

G (x , d) ∼ S(d)x

(log x)2, as x →∞

isn’t precise enough when d is large to handle PDC’s.

In the prime number theorem we know

log xshould be replaced by li(x) =

log t,

so we say a prime p has a density or probability of occuring of 1log p

rather than constant density 1log x .

The Hardy-Littlewood prime pair conjecture

The simplest conjecture

G (x , d) ∼ S(d)x

(log x)2, as x →∞

isn’t precise enough when d is large to handle PDC’s.In the prime number theorem we know

log xshould be replaced by li(x) =

log t,

so we say a prime p has a density or probability of occuring of 1log p

rather than constant density 1log x .

For G (x , d) if d > 0 then the conditions p, p′ ≤ x and p′ = p + dimplies that p ≤ x − d . Hence we see for d > 0 that

G (x , d) =∑

p≤x−dp+d is prime

Thus we conjecture for d > 0 : p has density 1log p ; p + d has

density 1log(p+d) and hence

G (x , d) = S(d)

∫ x−d

log t log(t + d)+ E (x , d), as x →∞,

where E (x , d) represents an error term.

For G (x , d) if d > 0 then the conditions p, p′ ≤ x and p′ = p + dimplies that p ≤ x − d . Hence we see for d > 0 that

G (x , d) =∑

p≤x−dp+d is prime

Thus we conjecture for d > 0 : p has density 1log p ; p + d has

density 1log(p+d) and hence

G (x , d) = S(d)

∫ x−d

log t log(t + d)+ E (x , d), as x →∞,

where E (x , d) represents an error term.

Let I (x , d) :=

∫ x−d

log t log(t + d)and G (x , d) := S(d)I (x , d).

Then the conjecture is

G (x , d) = G (x , d) + E (x , d) ,

where G (x , d) represents presumably the asymptotic differencecount.A strong conjecture is that for 2 ≤ d ≤ x − xε

E (x , d)� (x − d)12+ε.

The actual conjecture we need is much weaker:Conjecture We have E (x , d) = o( x

(log x)4) uniformly for

2 ≤ d ≤ 89x .

Numerical evidence for the HL pair conjecture

Figure: Asymptotic difference count, G (x , d), for x = 105.

Figure: Difference count for x = 105

Figure: Error term associated with asymptotic count for x = 105.

Consider the variance

ν(x) :=∑d

(G (x , d)− G (x , d)

a measure of the total error in the HL pair conjecture over all pairs.

We might conjecture that

ν(x) ∼ cπ(x)2 ∼ cx2

(log x)2

Numerical evidence suggests this might be true with c about size0.16.

Consider the variance

ν(x) :=∑d

(G (x , d)− G (x , d)

a measure of the total error in the HL pair conjecture over all pairs.We might conjecture that

ν(x) ∼ cπ(x)2 ∼ cx2

(log x)2

Numerical evidence suggests this might be true with c about size0.16.

Figure: Plot of variance, ν(x), divided by π(x)2 for certain x = pn.

Unconditional Results on PDC

TheoremSuppose d∗ = d∗(x) is a prime difference champion for primes≤ x . Then d∗ →∞ as x →∞ and the number of distinct primefactors of d∗ also goes to infinity as x →∞.

Unconditional Results on PDC

TheoremWe have ∑

p≤2 log xp-d∗(x)

p≤ log(2C)(1 + o(1)) < 2.08.

Corollary

We have ∑p|d∗(x)

p∼ max

1≤d≤x

∑p|d

p∼ log log log x .

Corollary

For sufficiently large x , every PDC is divisible by an odd prime≤ 25, 583.

The Sieve of Eratosthenes

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,

25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, . . .

First prime is 2, so remove multiples of 2:

1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35

Next prime 3, remove multiples of 3

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35

Get prime 5, remove multiples

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 39, 41, 43, 47, 49, 53,

How much is left by the sieve?

bxc −⌊x

⌋−⌊x

⌋+⌊x

⌋. . .

The proportion of numbers left is(1− 1

)(1− 1

)· · ·

This product converges to 0, as we expect

The Sieve of Eratosthenes: It’s Bad at Proving Things

Here is how we do this in equations: Let µ(1) = 1,µ(p1p2p3 · · · pr ) = (−1)r , µ(n) = 0 if p2|n.

µ is a multiplicativefunction. For squarefree multiplicative functions∑

f (d) =∏p|n

(1 + f (p)) ,

Thus ∑d |n

µ(d) =

{1, if n = 1,

0, if n 6= 1.

Now, let P(y) =∏

p≤y p, Then∑n≤x

gcd(n,P(√x))=1

1 = π(x)− π(√x) + 1.

Here is how we do this in equations: Let µ(1) = 1,µ(p1p2p3 · · · pr ) = (−1)r , µ(n) = 0 if p2|n. µ is a multiplicativefunction. For squarefree multiplicative functions∑

f (d) =∏p|n

(1 + f (p)) ,

Thus ∑d |n

µ(d) =

{1, if n = 1,

0, if n 6= 1.

Now, let P(y) =∏

gcd(n,P(√x))=1

1 = π(x)− π(√x) + 1.

Here is how we do this in equations: Let µ(1) = 1,µ(p1p2p3 · · · pr ) = (−1)r , µ(n) = 0 if p2|n. µ is a multiplicativefunction. For squarefree multiplicative functions∑

f (d) =∏p|n

(1 + f (p)) ,

Thus ∑d |n

µ(d) =

{1, if n = 1,

0, if n 6= 1.

Now, let P(y) =∏

gcd(n,P(√x))=1

1 = π(x)− π(√x) + 1.

∑n≤x

gcd(n,P(√x))=1

1 =∑n≤x

∑d |(n,P(

√x))

Since d |(n,P(

√x)) is equivalent to d |n and d |P(

√x), this is

d |P(√x)

µ(d)∑n≤xd |n

1 =∑

d |P(√x)

µ(d)⌊ xd

= x∑

d |P(√x)

∑d |P(√x)

µ(d){ xd

= x∏

p≤√x

(1− 1

∑d |P(√x)

µ(d){ xd

Recall Merten’s theorem:∏p≤y

(1− 1

)∼ e−γ

and e−γ = 0.561459 . . ..Hence

π(x)− π(√x) + 1 ∼ 2e−γ

log x−

∑d |P(√x)

µ(d){ xd

Thus we get prime number theorem wrong!

Moral: ∑d |P(√x)

µ(d){ xd

}isn’t as uncorrelated as you would guess.

Recall Merten’s theorem:∏p≤y

(1− 1

)∼ e−γ

and e−γ = 0.561459 . . ..Hence

π(x)− π(√x) + 1 ∼ 2e−γ

log x−

∑d |P(√x)

µ(d){ xd

Thus we get prime number theorem wrong!Moral: ∑

d |P(√x)

µ(d){ xd

}isn’t as uncorrelated as you would guess.

Modern Sieve Methods

The number of d |P(√x) is ∼ e2

√x/ log x which is huge. To keep

this from swamping the main term we would need to replace√x

by ≤ c(log x) which is tiny.

Example P = 2× 3× 5× 7× 11× 13× 17× 19× 23× 29 will letus sieve up to 312 = 961 while

Number of divisors of n = d(n) = 210 = 1024

Solution: Try to get upper and lower bounds only:∑d |n

αd ≤∑d |n

µ(d) ≤∑d |n

αd ≤∑d |n

µ(d) ≤∑d |n

αd ≤∑d |n

µ(d) ≤∑d |n

Brun chose αd and βd to be µ(d) or 0 depending on if d has ≤ rdistinct prime factors, and whether r is odd or even. In 1947Selberg made a brilliant choose for βd : Let λ1 = 1, λd arbitraryreal, then ∑

µ(d) ≤

∑d |n

Now choose λd = 0 for d > z , and optimize.

Bounded gaps between primes

In late 2013 and 2014 James Maynard and independently TerryTao used the multi-dimension Selberg weight

ΛR(n,Hk ,F ) =∑

di |(n+hi ),1≤i≤kd1d2...dk≤R

µ(d1)µ(d2) · · ·µ(dk)Λd1,d2,...,dk

where the Λ’s are smooth weights

Detection of Primes

To detect primes among the numbers inHk = {n + h1, n + h2, . . . , n + hk} for N < n ≤ 2N, let 1P(n) = 1if n is a prime and zero otherwise. Let

S(Hk ,m) =2N∑

1p(n + hi )−m

)ΛR(n,Hk ,F )2.

If there is a Hk with S(Hk , 1) > 0 then there must exist some nfor which two of the n + hi are both prime.

If there is a Hk with S(Hk ,m) > 0 then there must exist some nfor which m + 1 of the n + hi are all prime.

Detection of Primes

S(Hk ,m) =2N∑

1p(n + hi )−m

)ΛR(n,Hk ,F )2.

Detection of Primes

S(Hk ,m) =2N∑

1p(n + hi )−m

)ΛR(n,Hk ,F )2.

Gaps and Differences Between Primes - math.berkeley.edumolsson/Goldston.pdf · Gaps and Di erences...

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