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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.
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.
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.
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.
Dan Goldston Gaps and Differences Between Primes
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
N
p1= p2p3 · · · pk +
1
p1is not an integer.
Dan Goldston Gaps and Differences Between Primes
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
N
p1= p2p3 · · · pk +
1
p1is not an integer.
Dan Goldston Gaps and Differences Between Primes
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
N
p1= p2p3 · · · pk +
1
p1is not an integer.
Dan Goldston Gaps and Differences Between Primes
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
N
p1= p2p3 · · · pk +
1
p1is not an integer.
Dan Goldston Gaps and Differences Between Primes
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
N
p1= p2p3 · · · pk +
1
p1is not an integer.
Dan Goldston Gaps and Differences Between Primes
Letπ(x) = the number of primes ≤ x =
∑p≤x
1,
where p always denotes a prime.
20 40 60 80 100
5
10
15
20
25
Figure: π(x) = the number of primes ≤ x for 0 ≤ x ≤ 100
Dan Goldston Gaps and Differences Between Primes
Letπ(x) = the number of primes ≤ x =
∑p≤x
1,
where p always denotes a prime.
20 40 60 80 100
5
10
15
20
25
Figure: π(x) = the number of primes ≤ x for 0 ≤ x ≤ 100Dan Goldston Gaps and Differences Between Primes
200 400 600 800 1000
50
100
150
Figure: π(x) for 0 ≤ x ≤ 1000
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
This suggests the Prime Number Theorem:
π(x) ∼ x
log x, i. e. lim
x→∞
π(x)x
log 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
Dan Goldston Gaps and Differences Between Primes
This suggests the Prime Number Theorem:
π(x) ∼ x
log x, i. e. lim
x→∞
π(x)x
log 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
Dan Goldston Gaps and Differences Between Primes
A better approximation for π(x) is the Cramer Model:
Probability( n is a prime) = 1log n
Thus
π(x) ∼∑n≤x
1
log n∼ li(x),
where
li(x) =
∫ x
2
1
log tdt
is called the logarithmic integral.
Dan Goldston Gaps and Differences Between Primes
A better approximation for π(x) is the Cramer Model:
Probability( n is a prime) = 1log n
Thus
π(x) ∼∑n≤x
1
log n∼ li(x),
where
li(x) =
∫ x
2
1
log tdt
is called the logarithmic integral.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
The Riemann Zeta-function
Define
ζ(s) =∞∑n=1
1
ns= 1 +
1
2s+
1
3s+
1
4s+ · · · ,
where s = σ + it, and series conversed absolutely when σ > 1.Also for σ > 1
ζ(s) =∏
p prime
(1 +
1
ps+
1
p2s+
1
p3s+ · · ·
)=
∏p prime
(1− 1
ps
)−1This is called the Euler product.
Dan Goldston Gaps and Differences Between Primes
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
1
2+ i14.134,
1
2+ i21.022,
1
2+ i25.010,
1
2+ i30.424,
1
2+ i32.935,
Dan Goldston Gaps and Differences Between Primes
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
1
2+ i14.134,
1
2+ i21.022,
1
2+ i25.010,
1
2+ i30.424,
1
2+ i32.935,
Dan Goldston Gaps and Differences Between Primes
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
x iγ
has spikes when x equals a prime or prime power. Take absolutevalue of this we get
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
2 4 6 8 10
2
4
6
8
10
Figure: Using 10 zeros for 0 ≤ x ≤ 10
Theorem: 6 is not a prime.
Dan Goldston Gaps and Differences Between Primes
2 4 6 8 10
2
4
6
8
10
Figure: Using 10 zeros for 0 ≤ x ≤ 10
Theorem: 6 is not a prime.
Dan Goldston Gaps and Differences Between Primes
2 4 6 8 10
2
4
6
8
10
Figure: Using 20 zeros for 0 ≤ x ≤ 10
Dan Goldston Gaps and Differences Between Primes
5 10 15 20 25
5
10
15
20
Figure: Using 100 zeros for 0 ≤ x ≤ 25
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
Letπ2(x) = the number of pairs of twin primes ≤ x .
20 40 60 80 100
2
4
6
8
Figure: π2(x) for 0 ≤ x ≤ 100
Dan Goldston Gaps and Differences Between Primes
200 400 600 800 1000
5
10
15
20
25
30
35
Figure: π2(x) for 0 ≤ x ≤ 1000
Dan Goldston Gaps and Differences Between Primes
20 000 40 000 60 000 80 000 100 000
200
400
600
800
1000
1200
Figure: π2(x) for 0 ≤ x ≤ 105
Dan Goldston Gaps and Differences Between Primes
As for the Prime Number Theorem, we conjecture
π2(x) ∼ (1.3203 . . .)
∫ x
2
dt
(log t)2.
200000 400000 600000 800000 1·106
2000
4000
6000
8000
Figure: For 0 ≤ x ≤ 106
Dan Goldston Gaps and Differences Between Primes
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
2
1
(log t)2dt
Dan Goldston Gaps and Differences Between Primes
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
2
1
(log t)2dt
Dan Goldston Gaps and Differences Between Primes
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
2
1
(log t)2dt
Dan Goldston Gaps and Differences Between Primes
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
2
1
(log t)2dt
Dan Goldston Gaps and Differences Between Primes
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
2
1
(log t)2dt
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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
p
)(1− 1
p
)−2=
((1− 1
p
)2
− 1
p2
)(1− 1
p
)−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
)x
(log x)2.
Dan Goldston Gaps and Differences Between Primes
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
p
)(1− 1
p
)−2=
((1− 1
p
)2
− 1
p2
)(1− 1
p
)−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
)x
(log x)2.
Dan Goldston Gaps and Differences Between Primes
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
p
)(1− 1
p
)−2=
((1− 1
p
)2
− 1
p2
)(1− 1
p
)−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
)x
(log x)2.
Dan Goldston Gaps and Differences Between Primes
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),
where
lik(x) =
∫ x
2
dt
(log t)k,
and S(H) is the singular series defined by the product over allprimes p
S(H) =∏p
(1− 1
p
)−k (1− νH(p)
p
),
where νH(p) is the number of distinct residue classes occupied bythe elements of H. We assume S(H) 6= 0 here.
Dan Goldston Gaps and Differences Between Primes
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),
where
lik(x) =
∫ x
2
dt
(log t)k,
and S(H) is the singular series defined by the product over allprimes p
S(H) =∏p
(1− 1
p
)−k (1− νH(p)
p
),
where νH(p) is the number of distinct residue classes occupied bythe elements of H. We assume S(H) 6= 0 here.
Dan Goldston Gaps and Differences Between Primes
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),
where
lik(x) =
∫ x
2
dt
(log t)k,
and S(H) is the singular series defined by the product over allprimes p
S(H) =∏p
(1− 1
p
)−k (1− νH(p)
p
),
where νH(p) is the number of distinct residue classes occupied bythe elements of H.
We assume S(H) 6= 0 here.
Dan Goldston Gaps and Differences Between Primes
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),
where
lik(x) =
∫ x
2
dt
(log t)k,
and S(H) is the singular series defined by the product over allprimes p
S(H) =∏p
(1− 1
p
)−k (1− νH(p)
p
),
where νH(p) is the number of distinct residue classes occupied bythe elements of H. We assume S(H) 6= 0 here.
Dan Goldston Gaps and Differences Between Primes
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) =
2C2
∏p|dp>2
(p − 1
p − 2
), if d is even;
0, if d is odd;
where
C2 =∏p>2
(1− 1
(p − 1)2
)= 0.66016 . . .
with the product extending over all primes p > 2.
Dan Goldston Gaps and Differences Between Primes
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.”
Dan Goldston Gaps and Differences Between Primes
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.”
Dan Goldston Gaps and Differences Between Primes
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.”
Dan Goldston Gaps and Differences Between Primes
N(x , d) and N∗(x)
Let pn denote the nth prime. Let
N(x , d) :=∑pn≤x
pn−pn−1=d
1
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)}.
Dan Goldston Gaps and Differences Between Primes
N(x , d) and N∗(x)
Let pn denote the nth prime. Let
N(x , d) :=∑pn≤x
pn−pn−1=d
1
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)}.
Dan Goldston Gaps and Differences Between Primes
N(x , d) and N∗(x)
Let pn denote the nth prime. Let
N(x , d) :=∑pn≤x
pn−pn−1=d
1
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)}.
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
This graph shows the PJC’s for all prime x ≤ 1800.
Figure: Prime jumping champions for all odd, prime x ≤ 1800
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
Since
S(d) = 2C2
∏p|dp>2
(1 +
1
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.)
Dan Goldston Gaps and Differences Between Primes
Since
S(d) = 2C2
∏p|dp>2
(1 +
1
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.)
Dan Goldston Gaps and Differences Between Primes
Since
S(d) = 2C2
∏p|dp>2
(1 +
1
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.)
Dan Goldston Gaps and Differences Between Primes
Since
S(d) = 2C2
∏p|dp>2
(1 +
1
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.)
Dan Goldston Gaps and Differences Between Primes
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 ′
1
Dan Goldston Gaps and Differences Between Primes
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 ′
1
Dan Goldston Gaps and Differences Between Primes
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).
Dan Goldston Gaps and Differences Between Primes
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 .
Dan Goldston Gaps and Differences Between Primes
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 .
Dan Goldston Gaps and Differences Between Primes
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 .
Dan Goldston Gaps and Differences Between Primes
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 .
Dan Goldston Gaps and Differences Between Primes
Unfortunately we can’t compute anything
Problem: All computations show that 6 is the PJC when x > 1000.
Dan Goldston Gaps and Differences Between Primes
Unfortunately we can’t compute anything
Problem: All computations show that 6 is the PJC when x > 1000.
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
Prime Difference 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
1
Dan Goldston Gaps and Differences Between Primes
Figure: Difference count for x = 105
Dan Goldston Gaps and Differences Between Primes
Here is a zoom in:
Figure: Detail of difference count for x = 105; the PDC is 2310 and thecorresponding count is circled.
Dan Goldston Gaps and Differences Between Primes
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}.
Dan Goldston Gaps and Differences Between Primes
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.
Dan Goldston Gaps and Differences Between Primes
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 −
Dan Goldston Gaps and Differences Between Primes
Figure: Zoom-in view of D∗(x) over region where transition from 210 to2310 occurs.
Dan Goldston Gaps and Differences Between Primes
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
x
log xshould be replaced by li(x) =
∫ x
2
dt
log t,
so we say a prime p has a density or probability of occuring of 1log p
rather than constant density 1log x .
Dan Goldston Gaps and Differences Between Primes
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
x
log xshould be replaced by li(x) =
∫ x
2
dt
log t,
so we say a prime p has a density or probability of occuring of 1log p
rather than constant density 1log x .
Dan Goldston Gaps and Differences Between Primes
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
1,
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
2
dt
log t log(t + d)+ E (x , d), as x →∞,
where E (x , d) represents an error term.
Dan Goldston Gaps and Differences Between Primes
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
1,
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
2
dt
log t log(t + d)+ E (x , d), as x →∞,
where E (x , d) represents an error term.
Dan Goldston Gaps and Differences Between Primes
Let I (x , d) :=
∫ x−d
2
dt
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 .
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Figure: Asymptotic difference count, G (x , d), for x = 105.
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Figure: Difference count for x = 105
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Figure: Error term associated with asymptotic count for x = 105.
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Consider the variance
ν(x) :=∑d
(G (x , d)− G (x , d)
)2.
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.
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Consider the variance
ν(x) :=∑d
(G (x , d)− G (x , d)
)2.
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.
Dan Goldston Gaps and Differences Between Primes
Numerical evidence for the HL pair conjecture
Figure: Plot of variance, ν(x), divided by π(x)2 for certain x = pn.
Dan Goldston Gaps and Differences Between Primes
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 →∞.
Dan Goldston Gaps and Differences Between Primes
Unconditional Results on PDC
TheoremWe have ∑
p≤2 log xp-d∗(x)
1
p≤ log(2C)(1 + o(1)) < 2.08.
Corollary
We have ∑p|d∗(x)
1
p∼ max
1≤d≤x
∑p|d
1
p∼ log log log x .
Corollary
For sufficiently large x , every PDC is divisible by an odd prime≤ 25, 583.
Dan Goldston Gaps and Differences Between Primes
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,
Dan Goldston Gaps and Differences Between Primes
How much is left by the sieve?
bxc −⌊x
2
⌋−⌊x
3
⌋+⌊x
6
⌋. . .
The proportion of numbers left is(1− 1
2
)(1− 1
3
)(1− 1
5
)· · ·
This product converges to 0, as we expect
Dan Goldston Gaps and Differences Between Primes
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∑
d |n
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.
Dan Goldston Gaps and Differences Between Primes
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∑
d |n
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.
Dan Goldston Gaps and Differences Between Primes
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∑
d |n
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.
Dan Goldston Gaps and Differences Between Primes
∑n≤x
gcd(n,P(√x))=1
1 =∑n≤x
∑d |(n,P(
√x))
µ(d)
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)
µ(d)
d−
∑d |P(√x)
µ(d){ xd
}
= x∏
p≤√x
(1− 1
p
)−
∑d |P(√x)
µ(d){ xd
}.
Dan Goldston Gaps and Differences Between Primes
Recall Merten’s theorem:∏p≤y
(1− 1
p
)∼ e−γ
log y
and e−γ = 0.561459 . . ..Hence
π(x)− π(√x) + 1 ∼ 2e−γ
x
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.
Dan Goldston Gaps and Differences Between Primes
Recall Merten’s theorem:∏p≤y
(1− 1
p
)∼ e−γ
log y
and e−γ = 0.561459 . . ..Hence
π(x)− π(√x) + 1 ∼ 2e−γ
x
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.
Dan Goldston Gaps and Differences Between Primes
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
βd
Dan Goldston Gaps and Differences Between Primes
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
βd
Dan Goldston Gaps and Differences Between Primes
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
βd
Dan Goldston Gaps and Differences Between Primes
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 ∑
d |n
µ(d) ≤
∑d |n
λd
2
.
Now choose λd = 0 for d > z , and optimize.
Dan Goldston Gaps and Differences Between Primes
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
Dan Goldston Gaps and Differences Between Primes
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∑
n=N+1
(k∑
i=1
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.
Dan Goldston Gaps and Differences Between Primes
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∑
n=N+1
(k∑
i=1
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.
Dan Goldston Gaps and Differences Between Primes
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∑
n=N+1
(k∑
i=1
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.
Dan Goldston Gaps and Differences Between Primes