Skewes's number explained

x

for which

\pi(x)>\operatorname{li}(x),

where is the prime-counting function and is the logarithmic integral function. Skewes's number is much larger, but it is now known that there is a crossing between

\pi(x)<\operatorname{li}(x)

and

\pi(x)>\operatorname{li}(x)

near

e727.95133<1.397 x 10316.

It is not known whether it is the smallest crossing.

Skewes's numbers

J.E. Littlewood, who was Skewes's research supervisor, had proved in that there is such a number (and so, a first such number); and indeed found that the sign of the difference

\pi(x)-\operatorname{li}(x)

changes infinitely many times. All numerical evidence then available seemed to suggest that

\pi(x)

was always less than

\operatorname{li}(x).

Littlewood's proof did not, however, exhibit a concrete such number

x

.

proved that, assuming that the Riemann hypothesis is true, there exists a number

x

violating

\pi(x)<\operatorname{li}(x),

below
e79
e
e
1034
10
<10

.

Without assuming the Riemann hypothesis, proved that there exists a value of

x

below
e7.705
e
e
e
10964
10
<10

.

Skewes's task was to make Littlewood's existence proof effective: exhibiting some concrete upper bound for the first sign change. According to Georg Kreisel, this was at the time not considered obvious even in principle.

More recent estimates

These upper bounds have since been reduced considerably by using large-scale computer calculations of zeros of the Riemann zeta function. The first estimate for the actual value of a crossover point was given by, who showed that somewhere between

1.53 x 101165

and

1.65 x 101165

there are more than

10500

consecutive integers

x

with

\pi(x)>\operatorname{li}(x)

.Without assuming the Riemann hypothesis, proved an upper bound of

7 x 10370

. A better estimate was

1.39822 x 10316

discovered by, who showed there are at least

10153

consecutive integers somewhere near this value where

\pi(x)>\operatorname{li}(x)

. Bays and Hudson found a few much smaller values of

x

where

\pi(x)

gets close to

\operatorname{li}(x)

; the possibility that there are crossover points near these values does not seem to have been definitely ruled out yet, though computer calculations suggest they are unlikely to exist. gave a small improvement and correction to the result of Bays and Hudson. found a smaller interval for a crossing, which was slightly improved by . The same source shows that there exists a number

x

violating

\pi(x)<\operatorname{li}(x),

below

e727.9513468<1.39718 x 10316

. This can be reduced to

e727.9513386<1.39717 x 10316

assuming the Riemann hypothesis. gave

1.39716 x 10316

.
Yearnear x
  1. of complex
    zeros used
by
20001.39822 1 Bays and Hudson
20101.39801 1 Chao and Plymen
20101.397166 2.2 Saouter and Demichel
20111.397162 2.0 Stoll and Demichel

Rigorously, proved that there are no crossover points below

x=108

, improved by to

8 x 1010

, by to

1014

, by to

1.39 x 1017

, and by to

1019

.

There is no explicit value

x

known for certain to have the property

\pi(x)>\operatorname{li}(x),

though computer calculations suggest some explicit numbers that are quite likely to satisfy this.

Even though the natural density of the positive integers for which

\pi(x)>\operatorname{li}(x)

does not exist, showed that the logarithmic density of these positive integers does exist and is positive. showed that this proportion is about 0.00000026, which is surprisingly large given how far one has to go to find the first example.

Riemann's formula

Riemann gave an explicit formula for

\pi(x)

, whose leading terms are (ignoring some subtle convergence questions)

\pi(x)=\operatorname{li}(x)-\tfrac{1}{2}\operatorname{li}(\sqrt{x})-\sum\rho\operatorname{li}(x\rho)+smallerterms

where the sum is over all

\rho

in the set of non-trivial zeros of the Riemann zeta function.

The largest error term in the approximation

\pi(x)\operatorname{li}(x)

(if the Riemann hypothesis is true) is negative

\tfrac{1}{2}\operatorname{li}(\sqrt{x})

, showing that

\operatorname{li}(x)

is usually larger than

\pi(x)

. The other terms above are somewhat smaller, and moreover tend to have different, seemingly random complex arguments, so mostly cancel out. Occasionally however, several of the larger ones might happen to have roughly the same complex argument, in which case they will reinforce each other instead of cancelling and will overwhelm the term

\tfrac{1}{2}\operatorname{li}(\sqrt{x})

.

The reason why the Skewes number is so large is that these smaller terms are quite a lot smaller than the leading error term, mainly because the first complex zero of the zeta function has quite a large imaginary part, so a large number (several hundred) of them need to have roughly the same argument in order to overwhelm the dominant term. The chance of

N

random complex numbers having roughly the same argument is about 1 in

2N

.This explains why

\pi(x)

is sometimes larger than

\operatorname{li}(x),

and also why it is rare for this to happen.It also shows why finding places where this happens depends on large scale calculations of millions of high precision zeros of the Riemann zeta function.

The argument above is not a proof, as it assumes the zeros of the Riemann zeta function are random, which is not true. Roughly speaking, Littlewood's proof consists of Dirichlet's approximation theorem to show that sometimes many terms have about the same argument.In the event that the Riemann hypothesis is false, the argument is much simpler, essentially because the terms

\operatorname{li}(x\rho)

for zeros violating the Riemann hypothesis (with real part greater than) are eventually larger than

\operatorname{li}(x1/2)

.

The reason for the term

\tfrac{1}{2}li(x1/2)

is that, roughly speaking,

li(x)

actually counts powers of primes, rather than the primes themselves, with

pn

weighted by
1
n
. The term

\tfrac{1}{2}li(x1/2)

is roughly analogous to a second-order correction accounting for squares of primes.

Equivalent for prime k-tuples

An equivalent definition of Skewes' number exists for prime k-tuples . Let

P=(p,p+i1,p+i2,...,p+ik)

denote a prime (k +&thinsp;1)-tuple,

\piP(x)

the number of primes

p

below

x

such that

p,p+i1,p+i2,...,p+ik

are all prime, let

\operatorname{liP}(x)=

x
\int
2
dt
(lnt)k+1
and let

CP

denote its Hardy–Littlewood constant (see First Hardy–Littlewood conjecture). Then the first prime

p

that violates the Hardy–Littlewood inequality for the (k +&thinsp;1)-tuple

P

, i.e., the first prime

p

such that

\piP(p)>CP\operatorname{li}P(p),

(if such a prime exists) is the Skewes number for

P.

The table below shows the currently known Skewes numbers for prime k-tuples:

Prime k-tupleSkewes numberFound by
(p, p&thinsp;+&thinsp;2)1369391
(p, p&thinsp;+&thinsp;4)5206837
(p, p&thinsp;+&thinsp;2, p&thinsp;+&thinsp;6)87613571 Tóth (2019)
(p, p&thinsp;+&thinsp;4, p&thinsp;+&thinsp;6)337867 Tóth (2019)
(p, p&thinsp;+&thinsp;2, p&thinsp;+&thinsp;6, p&thinsp;+&thinsp;8)1172531 Tóth (2019)
(p, p&thinsp;+&thinsp;4, p&thinsp;+6&thinsp;, p&thinsp;+&thinsp;10)827929093 Tóth (2019)
(p, p&thinsp;+&thinsp;2, p&thinsp;+&thinsp;6, p&thinsp;+&thinsp;8, p&thinsp;+&thinsp;12)21432401 Tóth (2019)
(p, p&thinsp;+4&thinsp;, p&thinsp;+6&thinsp;, p&thinsp;+&thinsp;10, p&thinsp;+&thinsp;12)216646267 Tóth (2019)
(p, p&thinsp;+&thinsp;4, p&thinsp;+&thinsp;6, p&thinsp;+&thinsp;10, p&thinsp;+&thinsp;12, p&thinsp;+&thinsp;16)251331775687 Tóth (2019)
(p, p+2, p+6, p+8, p+12, p+18, p+20)7572964186421 Pfoertner (2020)
(p, p+2, p+8, p+12, p+14, p+18, p+20)214159878489239 Pfoertner (2020)
(p, p+2, p+6, p+8, p+12, p+18, p+20, p+26)1203255673037261 Pfoertner / Luhn (2021)
(p, p+2, p+6, p+12, p+14, p+20, p+24, p+26)523250002674163757 Luhn / Pfoertner (2021)
(p, p+6, p+8, p+14, p+18, p+20, p+24, p+26)750247439134737983 Pfoertner / Luhn (2021)

The Skewes number (if it exists) for sexy primes

(p,p+6)

is still unknown.

It is also unknown whether all admissible k-tuples have a corresponding Skewes number.

References

External links