Formula for primes: Information from Answers.com

Wikipedia: Formula for primes

In mathematics, a formula for primes is a formula generating the prime numbers, exactly and without exception. No easily-computable such formula is known. A great deal is known about what, more precisely, such a "formula" can and cannot be.

1 Prime formulas and polynomial functions
2 Formula based on a system of Diophantine equations
3 Formulas using the floor function
4 Recurrence relation
5 Recursive formula
6 See also
7 References
8 External links

Prime formulas and polynomial functions

There is no general equation for finding prime numbers. It is known that no non-constant polynomial function P(n) exists that evaluates to a prime number for all integers n. The proof is simple: Suppose such a polynomial existed. Then P(1) would evaluate to a prime p, so $P(1) \equiv 0 \pmod p$ . But for any k, $P(1+kp) \equiv 0 \pmod p$ also, so $P (1 + k p)$ cannot also be prime (as it would be divisible by p) unless it were p itself, but the only way $P (1 + k p) = P (1)$ for all k is if the polynomial function is constant.

Using more algebraic number theory, one can show an even stronger result: no non-constant polynomial function P(n) exists that evaluates to a prime number for almost all integers n.

Euler first noticed (in 1772) that the quadratic polynomial

P(n) = n² + n + 41

is prime for all non-negative integers less than 40. The primes for n = 0, 1, 2, 3... are 41, 43, 47, 53, 61, 71... The differences between the terms are 2, 4, 6, 8, 10... For n = 40, it produces a square number, 1681, which is equal to 41×41, the smallest composite number for this formula. In fact if 41 divides n it divides P(n) too. The phenomenon is related to the Ulam spiral, which is also implicitly quadratic, and the class number; this polynomial is related to the Heegner number $163=4\cdot 41-1$ , and there are analogous polynomials for $p = 2,3,5,11, and 17$ , corresponding to other Heegner numbers.

It is known, based on Dirichlet's theorem on arithmetic progressions, that linear polynomial functions $L (n) = a n + b$ produce infinitely many primes as long as a and b are relatively prime (though no such function will assume prime values for all values of n). Moreover, the Green-Tao theorem says that for any k there exists a pair of a and b with the property that $L (n) = a n + b$ is prime for any n from 0 to k−1. However, the best known result of such type is for k = 25:

6171054912832631 + 81737658082080n is prime for all n from 0 to 24 (Andersen 2008).

It is not known whether there exists a univariate polynomial of degree at least 2 that assumes an infinite number of values that are prime.

Formula based on a system of Diophantine equations

A system of 14 Diophantine equations in 26 variables can be used to obtain a Diophantine representation of the set of all primes. Jones et al. (1976) proved that a given number k + 2 is prime if and only if the following system of 14 Diophantine equations has a solution in the natural numbers:

α₀ =

w z + h + j - q

= 0

α₁ =

(g k + 2 g + k + 1)(h + j) + h - z

= 0

α₂ =

16(k + 1) 3 (k + 2)(n + 1) 2 + 1 - f 2

= 0

α₃ =

2 n + p + q + z - e

= 0

α₄ =

e 3 (e + 2)(a + 1) 2 + 1 - o 2

= 0

α₅ =

(a 2 - 1) y 2 + 1 - x 2

= 0

α₆ =

16 r 2 y 4 (a 2 - 1) + 1 - u 2

= 0

α₇ =

n + l + v - y

= 0

α₈ =

(a 2 - 1) l 2 + 1 - m 2

= 0

α₉ =

a i + k + 1 - l - i

= 0

α₁₀ =

((a + u 2 (u 2 - a)) 2 - 1)(n + 4 d y) 2 + 1 - (x + c u) 2

= 0

α₁₁ =

p + l (a - n - 1) + b (2 a n + 2 a - n 2 - 2 n - 2) - m

= 0

α₁₂ =

q + y (a - p - 1) + s (2 a p + 2 a - p 2 - 2 p - 2) - x

= 0

α₁₃ =

z + p l (a - p) + t (2 a p - p 2 - 1) - p m

= 0

The 14 equations α₀, …, α₁₃ can be used to produce a prime-generating polynomial inequality in 26 variables:

$(k+2)(1-\alpha_0^2-\alpha_1^2-\cdots-\alpha_{13}^2) > 0$

ie:

(k + 2)(1 -

[w z + h + j - q] 2 -

[(g k + 2 g + k + 1)(h + j) + h - z] 2 -

[16(k + 1) 3 (k + 2)(n + 1) 2 + 1 - f 2] 2 -

[2 n + p + q + z - e] 2 -

[e 3 (e + 2)(a + 1) 2 + 1 - o 2] 2 -

[(a 2 - 1) y 2 + 1 - x 2] 2 -

[16 r 2 y 4 (a 2 - 1) + 1 - u 2] 2 -

[n + l + v - y] 2 -

[(a 2 - 1) l 2 + 1 - m 2] 2 -

[a i + k + 1 - l - i] 2 -

[((a + u 2 (u 2 - a)) 2 - 1)(n + 4 d y) 2 + 1 - (x + c u) 2] 2 -

[p + l (a - n - 1) + b (2 a n + 2 a - n 2 - 2 n - 2) - m] 2 -

[q + y (a - p - 1) + s (2 a p + 2 a - p 2 - 2 p - 2) - x] 2 -

[z + p l (a - p) + t (2 a p - p 2 - 1) - p m] 2)

> 0

is a polynomial inequality in 26 variables, and the set of prime numbers is identical to the set of positive values taken on by this polynomial inequality as the variables a, b, …, z range over the nonnegative integers.

A general theorem of Matiyasevich says that if a set is defined by a system of Diophantine equations, it can also be defined by a system of Diophantine equations in only 9 variables. Hence, there is a prime-generating polynomial as above with only 10 variables. However, its degree is large (in the order of 10⁴⁵). On the other hand, there also exists such a set of equations of degree only 4, but in 58 variables (Jones 1982).

Formulas using the floor function

Using the floor function $\lfloor x\rfloor$ (defined to be the largest integer less than or equal to the real number x), one can construct several formulas that take only prime numbers as values for all positive integers n.

Mills's formula

The first such formula known was established in 1947 by W. H. Mills, who proved that there exists a real number A such that

$\lfloor A^{3^{n}}\;\rfloor$

is a prime number for all positive integers n. If the Riemann hypothesis is true, then the smallest such A has a value of around 1.3063... and is known as Mills' constant. This formula has no practical value, because very little is known about the constant (not even whether it is rational), and there is no known way of calculating the constant without finding primes in the first place.

Converting Sieve of Eratosthenes to prime number formulas

There is another, quite different formula by Sebastián Martín-Ruiz (Rivera n.d.):

$\pi(k) = k - 1 + \sum_{j=1}^k \left\lfloor {2 \over j} \left(1 + \sum_{s=1}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j-1 \over s}\right\rfloor - \left\lfloor{j \over s}\right\rfloor\right) \right)\right\rfloor$

To understand the behavior of this function note the following equalities:

$\left(\left\lfloor{ j \over s}\right\rfloor - \left\lfloor{j-1 \over s}\right\rfloor\right) = \begin{cases} 1 & \text{s divides j} \\ 0 & \text{s does not divide j} \end{cases}$

$\sum_{s=2}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j \over s}\right\rfloor - \left\lfloor{j-1 \over s}\right\rfloor\right) = \text{number of divisors of j}$

$\left\lfloor {-1 \over j} \left(\sum_{s=2}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j \over s}\right\rfloor - \left\lfloor{j-1 \over s}\right\rfloor\right) \right)\right\rfloor=\begin{cases} 0 & \text{j is prime} \\ -1 & \text{j is composite} \end{cases}$

$IsPrime(x)=1+\left\lfloor {-1 \over j} \left(\sum_{s=2}^{\left\lfloor\sqrt{j}\right\rfloor} \left(\left\lfloor{ j \over s}\right\rfloor - \left\lfloor{j-1 \over s}\right\rfloor\right) \right)\right\rfloor=\begin{cases} 1 & \text{j is prime} \\ 0 & \text{j is composite} \end{cases}$

$\pi(k) =\sum_{j=2}^k IsPrime(j)$

Converting primality tests to prime number formulas

Any primality test can be used as the basis for a prime number formula. In effect, a test for the primality of $n$ is a computation of the function IsPrime( $n$ ), defined by:

$\operatorname{IsPrime}(x) = \begin{cases} 1 & n \text{ prime} \\ 0 & n \text{ composite} \end{cases}$

If the primality test is given by a condition on some formula involving $n,$ then that formula gives a formula for IsPrime( $n$ ). For example, Wilson's theorem says that $n$ is prime if and only if it divides $(n - 1)! + 1.$ To express this by an explicit formula, one may introduce two intermediate functions:

$r(n) = \frac{(n - 1)! + 1}{n},$

$\operatorname{IsInteger}(x) = \begin{cases} 1 & x \text{ is an integer} \\ 0 & x \text{ is not an integer} \end{cases}$

Then Wilson's theorem says that

$\operatorname{IsPrime}(n) = \operatorname{IsInteger}\bigl(r(n)\bigr).$

This can be further specified by an explicit formula for IsInteger( $x$ ). Some options are:

$\operatorname{IsInteger}(x) = \lfloor x\rfloor + \lfloor -x \rfloor + 1$

$\operatorname{IsInteger}(x) = \left\lfloor \frac{\lfloor x \rfloor}{x} \right\rfloor$

$\operatorname{IsInteger}(x) = \lfloor \cos^2(\pi x) \rfloor$ : formula by C. P. Willans (Bowyer n.d.)

Then, for example, taking the first option gives a formula for IsPrime( $n$ ) using Wilson's theorem:

$\operatorname{IsPrime}(n) = \left\lfloor \frac{(n - 1)! + 1}{n} \right\rfloor + \left\lfloor -\frac{(n - 1)! + 1}{n} \right\rfloor + 1$

Once IsPrime( $n$ ) can be computed, the prime counting function $π(n)$ can as well, since by definition

$\pi(n) = \text{number of primes less than or equal to } n = \sum_{k = 1}^n \operatorname{IsPrime}(k).$

$π(n)$ can then compute a function testing whether a given integer $n$ is the $m$ ^th prime:

$\operatorname{IsNthPrime}(n,m) = \operatorname{IsPrime}(n) * \operatorname{IsZero}(\pi(n) - m) = \begin{cases} 1 & n \text{ is the } m^\text{th} \text{ prime} \\ 0 & \text{otherwise} \end{cases}$

The function IsZero( $x$ ) can likewise be expressed by a formula:

$\operatorname{IsZero}(n) = \operatorname{IsNonNegative}(x) * \operatorname{IsNonNegative}(-x),$ where for example when

x

is an integer

n

(such as is the case in IsNthPrime(

n, m

)),

$\operatorname{IsNonNegative}(n) = \operatorname{IsInteger}(2^n).$

Finally, the IsNthPrime( $n, m$ ) function can be used to produce a formula for the $n$ ^th prime:

$p_n = \sum_{m = 1}^{2^n} \operatorname{IsNthPrime}(n,m).$

The upper bound 2^$n$ comes from Bertrand's postulate, which implies that there is a sequence of primes

$2 = p_1 < p_2 < \dots < p_n,$ where

p i + 1 < 2 p i .

Thus,

p n < 2 n .

Substituting the formulas above and applying Wilson's theorem gives a formula for $p n$ involving the arithmetic operations and the floor function. Other such formulas are:

$p_n = 1 + \sum_{k=1}^{2^n} \left\lfloor \left\lfloor { n \over 1 + \pi(k) } \right\rfloor^{1 \over n} \right\rfloor.$

in which the following equality is important:

$\left\lfloor \left\lfloor { n \over 1 + \pi(k) } \right\rfloor^{1 \over n} \right\rfloor={\begin{cases} 1, & \mbox{if } \pi(k) <= n-1 \\ 0, & \mbox{if } \pi(k) > n-1 \end{cases}}$

and

$p_n = 1 + \sum_{k=1}^{2(\lfloor n \ln(n)\rfloor+1)} \left(1 - \left\lfloor{\pi(k) \over n} \right\rfloor\right).$ by Sebastián Martín-Ruiz.

Recurrence relation

Another prime generator is defined by the recurrence relation

$a_n = a_{n-1} + \operatorname{gcd}(n,a_{n-1}), \quad a_1 = 7,$

where gcd(x, y) denotes the greatest common divisor of x and y. The sequence of differences a_{n + 1} − a_n starts with 1, 1, 1, 5, 3, 1, 1, 1, 1, 11, 3, 1, 1 (sequence A132199 in OEIS). Rowlands (2008) proved that this sequence contains only ones and prime numbers.

Recursive formula

This is a primitive recursive formula by Liu Fengsui. Let

$A= \langle a_{1},a_{2},\dots,a_{i},\dots,a_{n}\rangle,$

$B= \langle b_{1},b_{2},\dots,b_{j},\dots,b_{m}\rangle$

be the arbitrary finite sets of natural numbers, we define $A+B= \langle a_{1}+b_{1},a_{2}+b_{1},\dots,a_{i}+b_{j}\dots,a_{n-1}+b_{m},a_{n}+b_{m}\rangle$ $AB=\langle a_{1}b_{1},a_{2}b_{1},\dots,a_{i}b_{j}\dots,a_{n-1}b_{m},a_{n}b_{m}\rangle .$

Let $A\backslash B$ be the set subtraction. Let $m_{i+1}=\prod_{0}^{i}p_{j}$ , let $U(\langle t_{1},t_{2},\ldots,t_{n} \rangle )=t_{2}$ be the projective function. Then we have a formula

$\begin{array}{lcl} T_{1}&= & \langle 1 \rangle ,\\ p_{1}&= & 3 ,\\ T_{i+1}&= &(T_{i}+ \langle m_{i} \rangle \langle0,1,2,\dots,p_{i}-1 \rangle) \setminus \langle p_{i} \rangle T_{i},\\ p_{i+1}&=& U (T_{i+1}). \end{array}$

By the induction, it is easy to prove that $T i$ is the reduced residue system $\bmod\,m_{i}$ , the reduced residue system is closed under the multiplication of residue classes, so that the least number except 1 in the reduced residue system $T i$ is the prime $p i$ .

One plugs i nto this formula, this formula will produce the i-th prime without testing. This primitive recursive function actually exhibits infinitely many primes, it provides a constructive proof of Eucilid's result. Now the primes appear in a highly regular pattern, the occurrence of individual primes among the natural numbers follows well-defined laws also.

References

Andersen, Jens Kruse (2008), Primes in Arithmetic Progression Records, http://users.cybercity.dk/~dsl522332/math/aprecords.htm, retrieved on 2009-01-27 .
Bowyer, Adrian (n.d.), Formulae for Primes, http://people.bath.ac.uk/ensab/Primes/, retrieved on 2008-07-09 .
Jones, James P.; Sato, Daihachiro; Wada, Hideo; Wiens, Douglas (1976), "Diophantine representation of the set of prime numbers", American Mathematical Monthly 83: 449–464, doi:10.2307/2318339 .
Jones, James P. (1982), "Universal diophantine equation", Journal of Symbolic Logic 47: 549–571 .
Rivera, Carlos (n.d.), Problem 38. Sebastián Martín Ruiz- Prime formulas, http://www.primepuzzles.net/problems/prob_038.htm, retrieved on 2008-07-09 .
Rowland, Eric S. (2008), "A Natural Prime-Generating Recurrence", Journal of Integer Sequences 11: 08.2.8, ISSN 1530-7638, http://www.cs.uwaterloo.ca/journals/JIS/VOL11/Rowland/rowland21.html .

External links

Eric W. Weisstein, Prime Formulas at MathWorld.
Eric W. Weisstein, Prime-Generating Polynomial at MathWorld.
Eric W. Weisstein, Mill's Constant at MathWorld.
A Venugopalan. Formula for primes, twinprimes, number of primes and number of twinprimes. Proceedings of the Indian Academy of Sciences – Mathematical Sciences, Vol. 92, No 1, September 1983, pp. 49-52. Page 49, 50, 51, 52, errata.

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Formula for primes

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