In mathematics, the prime-counting function is the function counting the number of prime numbers less than or equal to some real number x.^[1][2] It is denoted by π(x) (unrelated to the number π).

Growth rate

Of great interest in number theory is the growth rate of the prime-counting function.^[3][4] It was conjectured in the end of the 18th century by Gauss and by Legendre to be approximately

$${\displaystyle {\frac {x}{\log(x)))}$$

$${\displaystyle \lim _{x\rightarrow \infty }{\frac {\pi (x)}{x/\log(x)))=1.}$$

$${\displaystyle \lim _{x\rightarrow \infty }{\frac {\pi (x)}{\operatorname {li} (x)))=1}$$

More precise estimates

In 1899, de la Vallée Poussin proved that ^[6]

$${\displaystyle \pi (x)=\operatorname {li} (x)+O\left(xe^{-a{\sqrt {\log x))}\right)\quad {\text{as ))x\to \infty }$$

More precise estimates of π(x) are now known. For example, in 2002, Kevin Ford proved that^[7]

$${\displaystyle \pi (x)=\operatorname {li} (x)+O\left(x\exp \left(-0.2098(\log x)^{3/5}(\log \log x)^{-1/5}\right)\right).}$$

Mossinghoff and Trudgian proved^[8] an explicit upper bound for the difference between π(x) and li(x):

$${\displaystyle {\bigl |}\pi (x)-\operatorname {li} (x){\bigr |}\leq 0.2593{\frac {x}{(\log x)^{3/4))}\exp \left(-{\sqrt {\frac {\log x}{6.315))}\right)\quad {\text{for ))x\geq 229.}$$

For values of x that are not unreasonably large, li(x) is greater than π(x). However, π(x) − li(x) is known to change sign infinitely many times. For a discussion of this, see Skewes' number.

Exact form

For x > 1 let π₀(x) = π(x) − 1/2 when x is a prime number, and π₀(x) = π(x) otherwise. Bernhard Riemann, in his work On the Number of Primes Less Than a Given Magnitude, proved that π₀(x) is equal to^[9]

$${\displaystyle \pi _{0}(x)=\operatorname {R} (x)-\sum _{\rho }\operatorname {R} (x^{\rho }),}$$

$${\displaystyle \operatorname {R} (x)=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n))\operatorname {li} \left(x^{1/n}\right),}$$

$${\displaystyle \pi _{0}(x)\approx \operatorname {R} (x)-\sum _{\rho }\operatorname {R} \left(x^{\rho }\right)-{\frac {1}{\log(x)))+{\frac {1}{\pi ))\arctan {\frac {\pi }{\log(x))).}$$

The Riemann hypothesis suggests that every such non-trivial zero lies along Re(s) = 1/2.

Table of π(x), x/log(x), and li(x)

The table shows how the three functions π(x), x/log(x), and li(x) compared at powers of 10. See also,^[3][11] and^[12]

x	π(x)	π(x) − x/log(x)	li(x) − π(x)	x/π(x)	x/log(x)  % error
10	4	0	2	2.500	−8.57%
102	25	3	5	4.000	+13.14%
103	168	23	10	5.952	+13.83%
104	1,229	143	17	8.137	+11.66%
105	9,592	906	38	10.425	+9.45%
106	78,498	6,116	130	12.739	+7.79%
107	664,579	44,158	339	15.047	+6.64%
108	5,761,455	332,774	754	17.357	+5.78%
109	50,847,534	2,592,592	1,701	19.667	+5.10%
1010	455,052,511	20,758,029	3,104	21.975	+4.56%
1011	4,118,054,813	169,923,159	11,588	24.283	+4.13%
1012	37,607,912,018	1,416,705,193	38,263	26.590	+3.77%
1013	346,065,536,839	11,992,858,452	108,971	28.896	+3.47%
1014	3,204,941,750,802	102,838,308,636	314,890	31.202	+3.21%
1015	29,844,570,422,669	891,604,962,452	1,052,619	33.507	+2.99%
1016	279,238,341,033,925	7,804,289,844,393	3,214,632	35.812	+2.79%
1017	2,623,557,157,654,233	68,883,734,693,928	7,956,589	38.116	+2.63%
1018	24,739,954,287,740,860	612,483,070,893,536	21,949,555	40.420	+2.48%
1019	234,057,667,276,344,607	5,481,624,169,369,961	99,877,775	42.725	+2.34%
1020	2,220,819,602,560,918,840	49,347,193,044,659,702	222,744,644	45.028	+2.22%
1021	21,127,269,486,018,731,928	446,579,871,578,168,707	597,394,254	47.332	+2.11%
1022	201,467,286,689,315,906,290	4,060,704,006,019,620,994	1,932,355,208	49.636	+2.02%
1023	1,925,320,391,606,803,968,923	37,083,513,766,578,631,309	7,250,186,216	51.939	+1.93%
1024	18,435,599,767,349,200,867,866	339,996,354,713,708,049,069	17,146,907,278	54.243	+1.84%
1025	176,846,309,399,143,769,411,680	3,128,516,637,843,038,351,228	55,160,980,939	56.546	+1.77%
1026	1,699,246,750,872,437,141,327,603	28,883,358,936,853,188,823,261	155,891,678,121	58.850	+1.70%
1027	16,352,460,426,841,680,446,427,399	267,479,615,610,131,274,163,365	508,666,658,006	61.153	+1.64%
1028	157,589,269,275,973,410,412,739,598	2,484,097,167,669,186,251,622,127	1,427,745,660,374	63.456	+1.58%
1029	1,520,698,109,714,272,166,094,258,063	23,130,930,737,541,725,917,951,446	4,551,193,622,464	65.759	+1.52%

In the On-Line Encyclopedia of Integer Sequences, the π(x) column is sequence OEIS: A006880, π(x) − x/log x is sequence OEIS: A057835, and li(x) − π(x) is sequence OEIS: A057752.

The value for π(10²⁴) was originally computed by J. Buethe, J. Franke, A. Jost, and T. Kleinjung assuming the Riemann hypothesis.^[13] It was later verified unconditionally in a computation by D. J. Platt.^[14] The value for π(10²⁵) is due to J. Buethe, J. Franke, A. Jost, and T. Kleinjung.^[15] The value for π(10²⁶) was computed by D. B. Staple.^[16] All other prior entries in this table were also verified as part of that work.

The values for 10²⁷, 10²⁸, and 10²⁹ were announced by David Baugh and Kim Walisch in 2015,^[17], 2020,^[18] and 2022,^[19] respectively.

Algorithms for evaluating π(x)

A simple way to find π(x), if x is not too large, is to use the sieve of Eratosthenes to produce the primes less than or equal to x and then to count them.

A more elaborate way of finding π(x) is due to Legendre (using the inclusion–exclusion principle): given x, if p₁, p₂,…, p_n are distinct prime numbers, then the number of integers less than or equal to x which are divisible by no p_i is

${\displaystyle \lfloor x\rfloor -\sum _{i}\left\lfloor {\frac {x}{p_{i))}\right\rfloor +\sum _{i<j}\left\lfloor {\frac {x}{p_{i}p_{j))}\right\rfloor -\sum _{i<j<k}\left\lfloor {\frac {x}{p_{i}p_{j}p_{k))}\right\rfloor +\cdots }$

(where ⌊x⌋ denotes the floor function). This number is therefore equal to

${\displaystyle \pi (x)-\pi \left({\sqrt {x))\right)+1}$

when the numbers p₁, p₂,…, p_n are the prime numbers less than or equal to the square root of x.

The Meissel–Lehmer algorithm

In a series of articles published between 1870 and 1885, Ernst Meissel described (and used) a practical combinatorial way of evaluating π(x): Let p₁, p₂,…, p_n be the first n primes and denote by Φ(m,n) the number of natural numbers not greater than m which are divisible by none of the p_i for any i ≤ n. Then

${\displaystyle \Phi (m,n)=\Phi (m,n-1)-\Phi \left({\frac {m}{p_{n))},n-1\right).}$

Given a natural number m, if n = π(³√m) and if μ = π(√m) − n, then

${\displaystyle \pi (m)=\Phi (m,n)+n(\mu +1)+{\frac {\mu ^{2}-\mu }{2))-1-\sum _{k=1}^{\mu }\pi \left({\frac {m}{p_{n+k))}\right).}$

Using this approach, Meissel computed π(x), for x equal to 5×10⁵, 10⁶, 10⁷, and 10⁸.

In 1959, Derrick Henry Lehmer extended and simplified Meissel's method. Define, for real m and for natural numbers n and k, P_k(m,n) as the number of numbers not greater than m with exactly k prime factors, all greater than p_n. Furthermore, set P₀(m,n) = 1. Then

${\displaystyle \Phi (m,n)=\sum _{k=0}^{+\infty }P_{k}(m,n)}$

where the sum actually has only finitely many nonzero terms. Let y denote an integer such that ³√m ≤ y ≤ √m, and set n = π(y). Then P₁(m,n) = π(m) − n and P_k(m,n) = 0 when k ≥ 3. Therefore,

${\displaystyle \pi (m)=\Phi (m,n)+n-1-P_{2}(m,n)}$

The computation of P₂(m,n) can be obtained this way:

${\displaystyle P_{2}(m,n)=\sum _{y<p\leq {\sqrt {m))}\left(\pi \left({\frac {m}{p))\right)-\pi (p)+1\right)}$

where the sum is over prime numbers.

On the other hand, the computation of Φ(m,n) can be done using the following rules:

1. ${\displaystyle \Phi (m,0)=\lfloor m\rfloor }$
2. ${\displaystyle \Phi (m,b)=\Phi (m,b-1)-\Phi \left({\frac {m}{p_{b))},b-1\right)}$

Using his method and an IBM 701, Lehmer was able to compute the correct value of π(10⁹) and missed the correct value of π(10¹⁰) by 1.^[20]

Further improvements to this method were made by Lagarias, Miller, Odlyzko, Deléglise, and Rivat.^[21]

Other prime-counting functions

Other prime-counting functions are also used because they are more convenient to work with.

Riemann's prime-power counting function

Riemann's prime-power counting function is usually denoted as Π₀(x) or J₀(x). It has jumps of 1/n at prime powers pⁿ and it takes a value halfway between the two sides at the discontinuities of π(x). That added detail is used because the function may then be defined by an inverse Mellin transform.

Formally, we may define Π₀(x) by

${\displaystyle \Pi _{0}(x)={\frac {1}{2))\left(\sum _{p^{n}<x}{\frac {1}{n))+\sum _{p^{n}\leq x}{\frac {1}{n))\right)\ }$

where the variable p in each sum ranges over all primes within the specified limits.

We may also write

${\displaystyle \ \Pi _{0}(x)=\sum _{n=2}^{x}{\frac {\Lambda (n)}{\log n))-{\frac {\Lambda (x)}{2\log x))=\sum _{n=1}^{\infty }{\frac {1}{n))\pi _{0}\left(x^{1/n}\right)}$

where Λ is the von Mangoldt function and

${\displaystyle \pi _{0}(x)=\lim _{\varepsilon \to 0}{\frac {\pi (x-\varepsilon )+\pi (x+\varepsilon )}{2)).}$

The Möbius inversion formula then gives

${\displaystyle \pi _{0}(x)=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n))\ \Pi _{0}\left(x^{1/n}\right),}$

where μ(n) is the Möbius function.

Knowing the relationship between the logarithm of the Riemann zeta function and the von Mangoldt function Λ, and using the Perron formula we have

${\displaystyle \log \zeta (s)=s\int _{0}^{\infty }\Pi _{0}(x)x^{-s-1}\,\mathrm {d} x}$

Chebyshev's function

The Chebyshev function weights primes or prime powers pⁿ by log(p):

${\displaystyle {\begin{aligned}\theta (x)&=\sum _{p\leq x}\log p\\\psi (x)&=\sum _{p^{n}\leq x}\log p=\sum _{n=1}^{\infty }\theta \left(x^{1/n}\right)=\sum _{n\leq x}\Lambda (n).\end{aligned))}$

For x ≥ 2,^[22]

${\displaystyle \vartheta (x)=\pi (x)\log x-\int _{2}^{x}{\frac {\pi (t)}{t))\,\mathrm {d} t}$

and

${\displaystyle \pi (x)={\frac {\vartheta (x)}{\log x))+\int _{2}^{x}{\frac {\vartheta (t)}{t\log ^{2}(t)))\mathrm {d} t.}$

Formulas for prime-counting functions

Formulas for prime-counting functions come in two kinds: arithmetic formulas and analytic formulas. Analytic formulas for prime-counting were the first used to prove the prime number theorem. They stem from the work of Riemann and von Mangoldt, and are generally known as explicit formulae.^[23]

We have the following expression for the second Chebyshev function ψ:

${\displaystyle \psi _{0}(x)=x-\sum _{\rho }{\frac {x^{\rho )){\rho ))-\log 2\pi -{\frac {1}{2))\log \left(1-x^{-2}\right),}$

where

${\displaystyle \psi _{0}(x)=\lim _{\varepsilon \to 0}{\frac {\psi (x-\varepsilon )+\psi (x+\varepsilon )}{2)).}$

Here ρ are the zeros of the Riemann zeta function in the critical strip, where the real part of ρ is between zero and one. The formula is valid for values of x greater than one, which is the region of interest. The sum over the roots is conditionally convergent, and should be taken in order of increasing absolute value of the imaginary part. Note that the same sum over the trivial roots gives the last subtrahend in the formula.

For Π₀(x) we have a more complicated formula

${\displaystyle \Pi _{0}(x)=\operatorname {li} (x)-\sum _{\rho }\operatorname {li} \left(x^{\rho }\right)-\log 2+\int _{x}^{\infty }{\frac {\mathrm {d} t}{t\left(t^{2}-1\right)\log t)).}$

Again, the formula is valid for x > 1, while ρ are the nontrivial zeros of the zeta function ordered according to their absolute value. The integral is equal to the series over the trivial zeros:

${\displaystyle \int _{x}^{\infty }{\frac {\mathrm {d} t}{t\left(t^{2}-1\right)\log t))=\int _{x}^{\infty }{\frac {1}{t\log t))\left(\sum _{m}t^{-2m}\right)\,\mathrm {d} t=\sum _{m}\int _{x}^{\infty }{\frac {t^{-2m)){t\log t))\,\mathrm {d} t\,\,{\overset {\left(u=t^{-2m}\right)}{=))-\sum _{m}\operatorname {li} \left(x^{-2m}\right)}$

The first term li(x) is the usual logarithmic integral function; the expression li(x^ρ) in the second term should be considered as Ei(ρ log x), where Ei is the analytic continuation of the exponential integral function from negative reals to the complex plane with branch cut along the positive reals.

Thus, Möbius inversion formula gives us^[10]

${\displaystyle \pi _{0}(x)=\operatorname {R} (x)-\sum _{\rho }\operatorname {R} \left(x^{\rho }\right)-\sum _{m}\operatorname {R} \left(x^{-2m}\right)}$

valid for x > 1, where

${\displaystyle \operatorname {R} (x)=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n))\operatorname {li} \left(x^{1/n}\right)=1+\sum _{k=1}^{\infty }{\frac {\left(\log x\right)^{k)){k!k\zeta (k+1)))}$

is Riemann's R-function^[24] and μ(n) is the Möbius function. The latter series for it is known as Gram series.^[25][26] Because log x < x for all x > 0, this series converges for all positive x by comparison with the series for e^x. The logarithm in the Gram series of the sum over the non-trivial zero contribution should be evaluated as ρ log x and not log x^ρ.

Folkmar Bornemann proved,^[27] when assuming the conjecture that all zeros of the Riemann zeta function are simple,^{[note 1]} that

${\displaystyle \operatorname {R} \left(e^{-2\pi t}\right)={\frac {1}{\pi ))\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}t^{-2k-1)){(2k+1)\zeta (2k+1)))+{\frac {1}{2))\sum _{\rho }{\frac {t^{-\rho )){\rho \cos {\frac {\pi \rho }{2))\zeta '(\rho )))}$

where ρ runs over the non-trivial zeros of the Riemann zeta function and t > 0.

The sum over non-trivial zeta zeros in the formula for π₀(x) describes the fluctuations of π₀(x) while the remaining terms give the "smooth" part of prime-counting function,^[28] so one can use

${\displaystyle \operatorname {R} (x)-\sum _{m=1}^{\infty }\operatorname {R} \left(x^{-2m}\right)}$

as a good estimator of π(x) for x > 1. In fact, since the second term approaches 0 as x → ∞, while the amplitude of the "noisy" part is heuristically about √x/log x, estimating π(x) by R(x) alone is just as good, and fluctuations of the distribution of primes may be clearly represented with the function

${\displaystyle {\bigl (}\pi _{0}(x)-\operatorname {R} (x){\bigr )}{\frac {\log x}{\sqrt {x))}.}$

Inequalities

Here are some useful inequalities for π(x).

${\displaystyle {\frac {x}{\log x))<\pi (x)<1.25506{\frac {x}{\log x))}$

for x ≥ 17.

The left inequality holds for x ≥ 17 and the right inequality holds for x > 1. The constant 1.25506 is 30 log 113/113 to 5 decimal places, as π(x) log x/x has its maximum value at x = 113.^[29]

Pierre Dusart proved in 2010:^[30]

${\displaystyle {\frac {x}{\log x-1))<\pi (x)\quad {\text{for ))x\geq 5393,}$

and

${\displaystyle \pi (x)<{\frac {x}{\log x-1.1))\quad {\text{for ))x\geq 60184.}$

Here are some inequalities for the nth prime, p_n. The upper bound is due to Rosser (1941),^[31] the lower one to Dusart (1999):^[32]

${\displaystyle n(\log(n\log n)-1)<p_{n}<n{\log(n\log n)}\quad {\text{for ))n\geq 6.}$

The left inequality holds for n ≥ 2 and the right inequality holds for n ≥ 6.

An approximation for the nth prime number is

${\displaystyle p_{n}=n(\log(n\log n)-1)+{\frac {n(\log \log n-2)}{\log n))+O\left({\frac {n(\log \log n)^{2)){(\log n)^{2))}\right).}$

Ramanujan^[33] proved that the inequality

${\displaystyle \pi (x)^{2}<{\frac {ex}{\log x))\pi \left({\frac {x}{e))\right)}$

holds for all sufficiently large values of x.

In 2010^[30] Dusart proved (Proposition 6.6) that, for n ≥ 688383,

${\displaystyle p_{n}\leq n\left(\log n+\log \log n-1+{\frac {\log \log n-2}{\log n))\right),}$

and (Proposition 6.7) that, for n ≥ 3,

${\displaystyle p_{n}\geq n\left(\log n+\log \log n-1+{\frac {\log \log n-2.1}{\log n))\right).}$

More recently, Dusart^[34] has proved (Theorem 5.1) that, for x > 1,

${\displaystyle \pi (x)\leq {\frac {x}{\log x))\left(1+{\frac {1}{\log x))+{\frac {2}{\log ^{2}x))+{\frac {7.59}{\log ^{3}x))\right),}$

and that, for x ≥ 88789,

${\displaystyle \pi (x)>{\frac {x}{\log x))\left(1+{\frac {1}{\log x))+{\frac {2}{\log ^{2}x))\right).}$

The Riemann hypothesis

The Riemann hypothesis implies a much tighter bound on the error in the estimate for π(x), and hence to a more regular distribution of prime numbers,

${\displaystyle \pi (x)=\operatorname {li} (x)+O({\sqrt {x))\log {x}).}$

Specifically,^[35]

${\displaystyle |\pi (x)-\operatorname {li} (x)|<{\frac {\sqrt {x)){8\pi ))\,\log {x},\quad {\text{for all ))x\geq 2657.}$

Dudek (2015) proved that the Riemann hypothesis implies that for all x ≥ 2 there is a prime p satisfying

${\displaystyle x-{\frac {4}{\pi )){\sqrt {x))\log x<p\leq x.}$