Terence Tao claims:

For instance, we have an exact formula for the $n^\text{th}$ square number – it is $n^2$ – but we do not have a (useful) exact formula for the $n^\text{th}$ prime number $p_n$!
“God may not play dice with the universe, but something strange is going on with the prime numbers.” (Paul Erdős, 1913–1996)

However there exist an exact formula for the nth prime

A double sum for the nth prime pn is $$p_n=1+\sum_{k=1}^{2(\lfloor n\ln n\rfloor+1)}\Biggl[1-\Biggl\lfloor\frac{\sum_{j=2}^k 1+\lfloor s(j)\rfloor}n\Biggr\rfloor\Biggr],\tag{13}$$ where $$s(j)\equiv-\frac{\sum_{s=1}^j \bigl(\bigl\lfloor\frac js\bigr\rfloor-\bigl\lfloor\frac{j-1}s\bigr\rfloor\bigr)-2}j\tag{14}$$

(See Prime formulas at MathWorld.) There exist even an exact formula for prime counting function $\pi (x)$. So why are mathematicians trying to prove the Riemann hypothesis to find a better estimate for $\pi(x)$ and $p_n$ when they have those exact formulas?

Will Orrick
  • 18,856
  • 1
  • 44
  • 78
  • 863
  • 1
  • 7
  • 3
  • 40
    Because those formulas are meaningless. They are basically the mathematical equivalent of a truism. – Lucian Sep 21 '14 at 15:58
  • 42
    Let $P$ be the set of prime numbers. Then $\pi(x) = \sum_{x \in P} 1$. Useless formulas aren't always as obviously useless as what I just wrote. Drawing the line between such "formulas" and useful/interesting formulas is not easy. (This is discussed in Wilf's paper "What is an answer?") – Hugh Denoncourt Sep 21 '14 at 16:21
  • 2
    Short answer is that those formulas are not nice to be studied. They don't show anything relevant. They are slow to compute etc. – Ali Caglayan Sep 21 '14 at 16:48
  • So you think those formulas have a better form? So why instead of solving the Riemann hypothesis we search for a new form for those same formulas (13)+(14)? – user177691 Sep 21 '14 at 17:03
  • 15
    If someone found a formula which computed the primes efficiently it would be a significant breakthrough. The formulae people have created so far tend to be coded versions of tests for primality which are harder to apply than the original tests (the code adds a layer of complexity), or depend on knowing the primes (e.g. one can build constants whose decimal digits reveal the primes, but this is circular) or coded versions of algorithms to generate primes (again the code adds nothing but complexity). – Mark Bennet Sep 21 '14 at 21:13
  • 31
    @HughDenoncourt Your useless formula has other issues besides being useless... $\pi(x)=\sum_{p\in P,\ p\le x} 1$ would work. –  Sep 22 '14 at 05:11
  • 4
    These formulas are akin to "There exist some algorithm to compute the $n$th prime in finite time" – Hagen von Eitzen Sep 22 '14 at 13:20
  • possible duplicate of [Is there a known mathematical equation to find the nth prime?](http://math.stackexchange.com/questions/1257/is-there-a-known-mathematical-equation-to-find-the-nth-prime) – David K Sep 22 '14 at 20:27
  • 25
    You ignored the word "useful" in Tao's comment: here, "useful" means something that allows us to compute the $n$th prime significantly faster than what follows straightforwardly from its definintion. – ShreevatsaR Sep 22 '14 at 20:49
  • Time complexity **matters**. Even on a very fast computers finding the nth prime may take from a couple of days to a couple of months – Dunno Sep 23 '14 at 12:46
  • 2
    @Thursday Thanks, good catch! – Hugh Denoncourt Sep 23 '14 at 14:38
  • 2
    Here is one more formula: $p_n = $ "The nth prime number". As other comments have noted, there can be a differemce between a formula (and a computable one) and a definition (truism) presented as a (alternative computable) formula. Another $p_n$ = "nth number not divisible by $p_{n-1}$, $p_{n-2}$, $p_{n-3}$ .." – Nikos M. Sep 23 '14 at 22:28
  • https://math.stackexchange.com/questions/4299011/a-formula-that-counts-exactly-the-twin-prime-pairs-occuring-in-an-interval-a-b – Abstract Space Crack Nov 07 '21 at 21:10

9 Answers9


If there is a mathematical body of knowledge that allows you to manipulate formulas (13) and (14) into a better form, then great! The Riemann zeta function has already proved its worth in this field. It was instrumental in the proofs by Hadamard and de la Vallée-Poussin of the Prime Number Theorem. What allowed that to happen was that the zeta function is a meromorphic function, and so complex variable theory could be applied to its study. One never knows with certainty whether further study of the zeta function will pay off, but people are betting that it will, based on past successes.

I don't know whether anyone is promoting (13) and (14) as useful tools in the study of prime numbers, but if so, the burden is on those people to exhibit mathematical tools that allow one to manipulate (13) and (14) effectively.

Because of all the floor functions, (13) and (14) don't have very nice analytical properties. As other posters have pointed out, they encode an elementary algorithm for identifying primes. In particular, the expression $$ \left\lfloor\frac{j}{s}\right\rfloor-\left\lfloor\frac{j-1}{s}\right\rfloor $$ that appears in the numerator of (14) equals $1$ if $s$ evenly divides $j$ and equals $0$ if it does not. Summing this quantity for $s$ from $1$ to $j$ gives you the number of divisors of $j.$ Prime numbers have exactly two divisors; composite numbers all have at least three. The numerator in (14) is therefore $0$ when $j$ is prime, and positive when it is composite. Dividing by $j$ and multiplying by $-1$ results in the quantity $s(j),$ which is $0$ when $j$ is prime and lies in the interval $(-1,0)$ when $j$ is composite. The floor of $s(j$) therefore equals $0$ when $j$ is prime and $-1$ when $j$ is composite. Adding $1$ to $\lfloor s(j)\rfloor$ gives the characteristic function of the primes.

Notice that using the formula (14) requires one to do more divisions than even a naive brute-force algorithm to compute the number of divisors would require. That price might be worth paying if the formula had nice mathematical properties that allowed one to extract additional information from it, but I don't see that this has been demonstrated. To me, it looks like a cumbersome arithmetic encoding of an inefficient algorithm. I would be happy to be proved wrong about this!

Formula (13) likewise looks like a cumbersome arithmetic encoding of a brute-force method for computing the $n^\text{th}$ prime, and seems likely to require more calculation than state-of-the-art algorithms. Again, it would be nice if someone could show how to extract useful information from the formula, but it doesn't look very promising.

Added: I think it worth pointing out that some non-trivial analytic number theory enters into formula (13) in the choice of upper bound on the summation over $k.$ You need to ensure that the $p_n$ lies in the range of summation, but that $p_{2n}$ does not. The reason is that the summand equals $1$ when $k<p_n,$ equals $0$ for $p_n\le k<p_{2n},$ and becomes negative for $k\ge p_{2n}.$ So you need the summation over $k$ to include all $p_n-1$ terms where the summand equals $1$, but none of the terms where it is negative. This will ensure that the sum evaluates to $p_n-1.$

It appears to me that you need something like (3.12) and (3.13) in this paper of Rosser and Schoenfeld to justify the upper bound. Formula (3.12) is Rosser's Theorem which Dietrich Burde's answer to this post suggests requires deep knowledge of zeta function zeros. Interestingly, the author of your formulas (13) and (14) (Sebastián Martín Ruiz) does not rigorously justify the upper bound in his article, giving only a heuristic argument and and stating that it is “very probable” that the necessary inequalities are satisfied.

Perhaps one could play with the constant in front of $\lfloor n\log n\rfloor$ and try to use Chebyshev's bounds (Theorem 3.1 here) instead of Rosser's Theorem. (The derivation of Chebyshev's bounds doesn't require knowledge of zeta function zeros.) But this seems possibly rather delicate, and may not work.

Second addition: Some discussion of the formula, and Ruiz's view of it, can be found here.

Will Orrick
  • 18,856
  • 1
  • 44
  • 78

Mathematicians are not really interested in particular values of the counting function. (They already know tons of them, obtained by running efficient sieves.)

What they are after is insight on its asymptotic behavior (how precisely does it vary as $x$ grows). Better, they would like to understand the intriguing irregularity in the distribution and put some order into it.

The given formula, as well as many other similar ones, is of no use for computing the function values (it is far too inefficient) and does not bring any insight (it is just a rewrite of a trivial algorithm).

The relation to the zeroes of the Zeta function is much deeper.

  • 2
    So why is Zeta function much deeper? – user177691 Sep 21 '14 at 17:53
  • So why instead of solving the Riemann hypothesis we search for a new form for those same formulas (13)+(14)? – user177691 Sep 21 '14 at 18:02
  • 19
    These formulas are just a restatement of "try all divisors". The Zeta function bridges to the theory of analytic functions and brings nontrivial results. Anyway, any attack strategy is allowed. –  Sep 21 '14 at 20:12

Formulas for primes is a lovely 1983 paper by Underwood Dudley in Mathematics Magazine. It explains how to cook up impressive-looking but trivial/useless formulas for the $n$th prime or the prime-counting function, and gives a smattering of references to unfortunate cases in the literature of such formulas being published.

  • 271,033
  • 27
  • 280
  • 538
Dan Petersen
  • 6,524
  • 2
  • 23
  • 29

Usually when someone says "we have a formula for $f(x)$", they mean that we have a way to compute $f$ faster than it's direct definition (or by the definition, if that is fast enough).

For example, the function $f(x) = x^2$ can be computed by if $x$ is given in binary (or any radix) by the way you probably learned in elementary school. On the other hand, if you were asked to factor a large number, you would quickly learn that there is no known "formula", in other words, nothing is much faster than simply checking all numbers (the fastest known approaches are about as fast as checking the factors on a number with 1/3 the number of digits IIRC).

  • 22,195
  • 5
  • 35
  • 67
  • We are not interested in the speed of computation, I'm just interested in if I want to find the n th prime without looking into prime tables then with that formula I can compute even if it takes me time. – user177691 Sep 21 '14 at 16:44
  • 22
    @user177691 If speed of computation is not a concern, you can easily check whether a given integer is prime by checking if any smaller integer divides it. – Andrea Sep 21 '14 at 17:19
  • 3
    The general number field sieve is actually much better than you describe: its running time is bounded by $C e^{2 \left ( \ln n \right )^{1/3}}$. (This is not sharp, but it is close enough for this comment.) What you describe would be something like $e^{1/3 \ln n}$, and it is clear that $1/3 \ln n$ is much larger than $2 (\ln n)^{1/3}$ for large $n$. It is still not fast, of course, which is the point of your argument. – Ian Sep 21 '14 at 19:54
  • @Ian Ah thanks for the correction. – DanielV Sep 21 '14 at 20:14
  • 22
    @user177691: since you don't care about time, of course you can compute the nth prime number without looking into prime tables. Just start testing each integer in turn whether it is prime or not (by trial division), and if so count 1, until your count reaches n. And if your response is "well, that doesn't count as directly computing the nth prime": what I just described is pretty much what the formula does! The main difference is the formula encodes the whole process as a (nested) summation, whereas I described it with conditions and incrementing counters. Same process in different language. – Steve Jessop Sep 21 '14 at 23:41
  • 2
    @user177691 How do you think they make the tables? – DanielV Sep 22 '14 at 00:58
  • No one has mentioned it here, but a number $n$ is prime iff $\exists a\le\sqrt{n}, a\in\mathbb N_{\ge 2}$ such that $a\mid n$. Knowing that you only need to check the divisors upto the square root makes it a lot more efficient. – user26486 Sep 22 '14 at 07:02
  • @mathh Your \exists should be \nexists. – Ari Brodsky Sep 23 '14 at 01:20
  • @AriBrodsky yes, sorry. – user26486 Sep 23 '14 at 14:15
  • +1, look at my own comment on top as well, which takes a similar route – Nikos M. Sep 23 '14 at 22:32

Those formulae don't help us prove anything of use. For example, can you use those formulae to derive Dirichlet's theorem (that every arithmetic sequence $ax+b$ where $\gcd(a,b)=1$ has an infinite number of primes)*? No. Those formulae don't make that easier at all.

*It's a very hard theorem to prove.

There are also plenty of conjectures (such as the Bunyakovsky conjecture) that are currently unsolved, and those formulae don't make it any easier for us to solve them.

Akiva Weinberger
  • 19,084
  • 1
  • 31
  • 84

The emphasis in Tao's statement should be on useful: the formula you quoted is neither computationally fast nor gives any theoretical insight, it is simply a fancy rearrangement of a well-known fact about primes (Wilson's theorem, I believe in this case). The Riemann Hypothesis on the other hand allows a much deeper understanding of the primes, and the hope is that a proof would yield new tools and approaches beyond the mere knowledge of its truth.

Markus Shepherd
  • 713
  • 4
  • 13

If this formula could say something about primes, for example how to factor sums of primes, it would be interesting.

  • 13,268
  • 4
  • 23
  • 72

Hardy and Wright in "An Introduction to the Theory of Numbers", give a short and simple formula in Theorem 419 for the n-th prime however they explain that to use the formula it is necessary to know the first n primes, which defeats the purpose of the formula.

They say that if the value of the inputs to their formula could be expressed independently of the primes then its status would be different, however, "There seems to be no likelihood of this, but it cannot be ruled out as entirely impossible".

  • 255
  • 1
  • 7

Well in the context of computing the $n^{th}$ prime number, I can almost guarantee you from looking at that formula that if I now write the code to enumerate both it and another patch of code based entirely on checking the divisibility of the output of a random number generator, the RNG with beat the code written for that formula by a long shot.

Likewise for all of the studies I've been conducting on variations and alterations to Wilson's theorem, you see the computational expense of factorization is not going to disappear upon knowing explicit formula for the $n^{th}$ prime number, it is minimized by applying a number of considerations from a number of differing fields, but the one that is most familiar for me is Modular Arithmetic, particularly things like Euler's Theorem. In fact one thing I realised tonight, they will never ever run out of factorization exploits for which to ensure the security of encryption from those that all stem from that one theorem alone, and no doubt there are more out there that are too advanced for me at this point, so it really isn't as big of a deal as everyone makes it out to be in my opinion.

Adam Ledger
  • 1,180
  • 6
  • 16