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"The set $\{f(n)\}, n=1,2,\ldots$ includes all primes except a finite number of exceptions."

This statement is true for $$f(n)=\sqrt{1+24n},$$ for which the exceptions are 2 and 3. It also generates non-integers and non-prime integers, however. (spoiler alert) The proof uses the fact that the two neighbours of all primes (except 2 and 3) contain the factors 2, 3 and 4 between them (2*3*4=24).

Since no more factors than 2^3 and 3 can be found generally in the two neighbours of primes, the expression $f(n)=\sqrt{1+24n}$ is the most restrictive expression of that particular functional form that still generates all primes. That is, it generates all primes and the smallest number of non-primes.

Other expressions that fit the statement above are $f(n)=\sqrt{1+4n}$, $f(n)=2n-1$ and of course $f(n)=n$, but they produce more non-primes.

My question: Is it known which closed-form expression $f(n)$ generates all primes and the smallest number (in some sense) of non-primes? Is it perhaps $f(n)=\sqrt{1+24n}$?

andreasdr
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  • The question isn't well-defined, since as you rightly point out $\sqrt{1+24n}$ doesn't generate all primes, so you'd have to say something about how many exceptions are allowed or how to combine the numbers of false positives and false negatives, e.g. look for the lowest sum of the two. Also, $\sqrt{1+24n}$ generates infinitely many non-primes, so it can't generate the smallest number of non-primes in the sense of cardinality -- are you thinking in terms of asymptotic densities? – joriki Oct 07 '11 at 09:16
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    The number of primes in ${1,2,\ldots,N}$ is $\pi(N)$ (where $\pi$ is the prime counting function). Since $f(n)=O(\sqrt{n})$, the number of primes in ${f(n):1 \leq n \leq N}$ is at most $\pi(f(N))<\pi(N)$ for sufficiently large $N$. So, $f(n)=n$ produces fewer non-primes than $f(n)=\sqrt{1+24n}$. – Douglas S. Stones Oct 07 '11 at 09:21
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    Also note that there are expressions which generate exactly the primes. These would be the optimal solutions unless you exclude things like sums or the floor function from your notion of "closed-form expression". – joriki Oct 07 '11 at 09:25
  • @Douglas: That depends on whether you compute the density with respect to the arguments or the function values. This underscores my point that the question isn't well-defined. – joriki Oct 07 '11 at 09:26
  • http://math.stackexchange.com/q/21748/15660 – Pedja Oct 07 '11 at 10:07
  • You are all pointing out stuff that I was nervous about when posting the question :). I'm not a mathematician, but mustn't there be a sense in which $\sqrt{1+24n}$ generates "fewer" primes than $f(n)=n$? Intuitively, I would say that it does. I think there is a interesting and well-posed question here somewhere. Any suggestions on how to edit mine to get to it? :) – andreasdr Oct 07 '11 at 10:54
  • Also, by "closed-form expression", I mean that floor functions and infinite sums are excluded (doesn't that follow from the definition, or is it such a loose concept?) – andreasdr Oct 07 '11 at 10:55
  • pedja: Thanks, I found that question also, but the discussion there is about whether $f(n) = \sqrt{1+24n}$ generates only primes, which the simple counterexample $n=26$ (generates the non-prime integer 25) shows that it does not. – andreasdr Oct 07 '11 at 11:00
  • Aha, now I see that $f(n)=n$ actually generates a lower number of non-primes than $f(n) = \sqrt{1+24n}$ pretty much any way you see it. I was completely "out bicycling" as we say in Sweden. I'll rephrase the question. – andreasdr Oct 07 '11 at 11:20
  • How do I close the question? Due to the problems with this one, I'd like to post a new one, but won't have time to do so until tomorrow. – andreasdr Oct 07 '11 at 11:26
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    For the same reasons that all primes $>3$ occur as values of this formula, all primes greater than $5$ occur as values of $\sqrt[4]{120n+1}$ and all primes greater than $7$ as values of $\sqrt[12]{10920n+1}$. The nonprime integer values of these formulas are a bit sparser than for $\sqrt{24n+1}$. – David E Speyer Dec 06 '11 at 15:41

3 Answers3

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A number is relatively prime to $24$ if and only if it is a square root of unity mod $24$. So $f(n)$ generates all integers that are not divisible by $2$ or $3$, including all primes greater than $3$. It is possible to do better: $g(x) = \sqrt{\frac{5 + 33 \cdot (-1)^n + 15 n}{2}}$ generates all primes greater than $5$ without including any integers divisible by $2$, $3$, or $5$.

Dan Brumleve
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It was shown in the answer at: Does the formula $\sqrt{ 1 + 24n }$ always yield prime?

that the expression doesn't generate all primes. There isn't any known algebraic expression that generates an infinite number of consecutive primes. So there can't be a "simplest" such expression. There are of course expressions that generate long strings of primes then fail.

Community
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MaxW
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  • I think you are misinterpreting the question. It is not the same as the one you linked, but some of the answers there apply here as well. – Dan Brumleve Jan 27 '12 at 07:45
  • Dan you're right. But the question "Is it known which closed-form expression f(n) generates all primes and the smallest number (in some sense) of non-primes?" doesn't make sense now that I read it carefully. F(n) doesn't generate "all primes." The word "and" is a conjunction not a qualifier. So is andreasdr looking for some measure on how much an equation deviates from the ideal of generating all the primes in sequence, then ranking equations against that measure? – MaxW Jan 27 '12 at 21:35
  • I think we're looking for functions such that all primes occur as values for some input while minimizing the asymptotic density of composite integers that also occur. – Dan Brumleve Jan 28 '12 at 01:48
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I wrote these two small source codes in Maple:

$1)$

"for n from 1 to 100 by 1 do

if isprime(2n-3) then

print(n);

end if;

end do;"

$2)$

"for n from 1 to 100 by 1 do

if type(sqrt(24n+1),prime) then

print(n);

end if;

end do;"

The first code is giving $44$ values for n when result of expression is prime number,and second code is giving only $13$ values for n.

So we may conclude that closed form $f(n)=2n-3$ is "richer" than closed form $f(n)=\sqrt{24n+1}$

EDIT :

In one of my previous posts I have shown that :

$M_p \equiv 1 \pmod {6\cdot p}$ , where $M_p$ is Mersenne number .

So :

$2^{p}-1=6np+1 \Rightarrow 2^p=6np+2 \Rightarrow 2^{p-1}=3np+1$

If we solve this equality for $p$ we will get the next formula :

$$p=-\left(\frac{3n\cdot W\left(-\frac{2^{-1-\frac{1}{3n}}\cdot \ln 2}{3n}\right)+\ln 2}{n\cdot \ln 8}\right)$$

where $W$ is Lambert W function

This closed-form expression generates all primes greater than $3$ and the small number (in sense of asymptotic density ) of the non-prime integers .

Correction :

Actually this closed expression produces integer values for lower branch of Lambert W function which is denoted as $W_{-1}(x)$ so formula is :

$$p=-\left(\frac{3n\cdot W_{-1}\left(-\frac{2^{-1-\frac{1}{3n}}\cdot \ln 2}{3n}\right)+\ln 2}{n\cdot \ln 8}\right)$$

Pedja
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