I'm a 31 year old engineer, and I've recently came up with a way to exactly predict the probability of the number of prime numbers between two different integers.

For example using my way, the number of prime numbers between $0$ and $100$ is between $0$ and $50$. And it turns out that it is correct, since there are $25$ primes between $0$ and $100$.

But is this of any real importance that would lead me to publish a paper? Also my way is purely elementary and so I suspect that mathematicians would even bother to give it a look.

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    $\begingroup$ Do you know of the prime number theorem? $\endgroup$
    – Ryan Reich
    Commented Dec 30, 2013 at 18:36
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    $\begingroup$ There are already a lot of estimates about counting the number of primes below a given number, such as the prime number theorem. Without saying more about your estimate, it's impossible to assess. $\endgroup$
    – user61527
    Commented Dec 30, 2013 at 18:36
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    $\begingroup$ No, my way is different from the prime number theorem pi(x)~x/ln(x), since this last one gives only an approximation whereas my method gives the right answer and precise. $\endgroup$ Commented Dec 30, 2013 at 18:37

5 Answers 5


The method in question is noting that no primes other than 2 and 5 are divisible by 2 or 5, and thus any ten numbers $10n,10n+1,\ldots10n+9$ (really, any ten consecutive numbers) contain at most 4 primes, assuming the first term is larger than 5.

This can be improved by removing multiples of 3, so that no 30 consecutive numbers can have more than 8 primes (as long as the smallest is more than 5). It's an improvement since the basic method would give only a bound of 12 for such an interval. Of course this can be increased by throwing in new primes like 7.

Indeed, with more subtlety you can prove things about intervals smaller than the least common multiple of the primes used, and this has been done for intervals of lengths up to perhaps 3000. For larger intervals analytic methods have been used which restrict the number of primes which can appear.

This is not a new result, but the family of results is important! In particular, they were an ingredient in improving Zhang's theorem, a major step toward the twin prime conjecture. So what you have is just a piece of the puzzle, but it's a very large, intricate puzzle. Be glad you have seen part of it!

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    $\begingroup$ Thank you for your answer!! $\endgroup$ Commented Dec 30, 2013 at 19:09
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    $\begingroup$ @Charles: I haven't seen many answers on MSE that I would call "gracious" but this has got to be one of them. The mathematics community is blessed to have someone as eloquent, patient, and gracious as you. $\endgroup$ Commented Dec 31, 2013 at 1:16
  • $\begingroup$ how does charles know what the "method in question" is? comments? that may be so but to me this makes some kind of hidden assumption (it should try to cite something for this conclusion). also note in general these types of analyses are called "sieving algorithms/methods" and they go all the way back to Eratosthenes.... there is a large developed literature on sieving theory (and yes, under active research/development as with the recent Zhang breakthrough) ... also, roughly, anything that could be described in stackexchange comments is unlikely to have any new scientific relevance... $\endgroup$
    – vzn
    Commented Dec 31, 2013 at 16:39
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    $\begingroup$ @vzn: Yeah, it's this comment in particular math.stackexchange.com/questions/622498/… $\endgroup$ Commented Jan 1, 2014 at 7:27

From the description in the comments — which I encourage putting into the post itself — your method is essentially trivial. You provide a bound of $\frac4{10}n$ for the number of primes in the range $[x, x+n]$, independent of $x$; that is, your bound is linear in the length of the interval. But in fact, much better bounds are known - it can be shown that $\lim\limits_{n\to\infty}\frac1n(\pi(x+n)-\pi(x)) = 0$, whereas your method only shows that this limit is $\leq\frac4{10}$. You've crafted a marginal improvement on the 'simple' bound of $\frac12=\frac5{10}$, obtained by just observing that primes greater than two are odd, by also 'sieving' out factors of $5$; note that you could have gotten an even better bound by not working in decimal and instead sieving out factors of $3$, giving a bound of $\frac12-\frac16=\frac13$ — or by sieving out $3$ and $5$ (for instance, considering intervals of length 30 instead of length 10), giving a bound of $\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac15\right) = \frac4{15}$.

On the other hand, in some sense you're 'on the right track' to some of the fundamentals of number theory: the generalized version of this product — $\prod_{p=2}^k\left(1-\frac1p\right)$ — is essentially the probability that a number is divisible by no prime less than $k$; and in fact, it can be shown that the product itself diverges to $0$ as we take $k\to\infty$. If you look at it closely, this provides another proof that there are infinitely many primes; if there weren't, then clearly the product would have to 'converge' to some finite value, because it would terminate! Estimates on how this product diverges are one way of crafting the Prime Number Theorem alluded to in the other answers and comments, which is arguably the most fundamental result in analytical number theory. This particular 'discovery' may not be of any merit, but if you're interested in the subject I encourage you to continue down this path; you've got a lot of learning ahead of you, but there's plenty of interesting material to be had!

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    $\begingroup$ Thank you for your answer! $\endgroup$ Commented Dec 30, 2013 at 19:11
  • $\begingroup$ We posted our answers at the same time, you just a minute or two behind me. My answer is a bit simpler (by design) but your answer does a better job explaining the connections to higher math for those who can follow it. $\endgroup$
    – Charles
    Commented Dec 30, 2013 at 19:11
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    $\begingroup$ @Charles Agreed - the two answers say a lot of the same things but I think they complement each other nicely. $\endgroup$ Commented Dec 30, 2013 at 19:14
  • $\begingroup$ @StevenStadnicki Should the first limit be as $x \to \infty$? $\endgroup$
    – user88319
    Commented Dec 31, 2013 at 7:38
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    $\begingroup$ @hardmath: Products which go to 0 are said to diverge there, as they correspond (via logarithms) to sums going to $-\infty$. $\endgroup$
    – Charles
    Commented Jan 1, 2014 at 18:53

If your method is indeed new, then mathematicians will certainly be interested.

However, there is a lot of existing work in this area — estimates and bounds for the density of primes within intervals. Have you read up on this work at all? As with any potential new discovery, it is your responsibility before anyone else’s to put some time and effort in working out if it is new — certainly there’s a point where it’s worth asking other people with more knowledge of the existing literature, but to convince them it’s worth their time and effort, you should be able to show you’ve put a fair bit of effort in yourself as well.

Without seeing your method, we can’t judge whether it’s new or not. I suspect it’s unlikely to be, I’m afraid, just since this is an area that’s been worked on by so many people for so long. However, don’t be too disheartened if it isn’t new — many mathematicians’ first discoveries turn out to be rediscoveries, and it still means you figured out something good, even if other people had also figured it out previously. Having worked it out independently, you’ll be in a better position to read up on related existing work, and then perhaps have a starting point to find new things!

  • $\begingroup$ Thanks for your really inspiring answer. To see my method, check the comments and please tell me if it is a rediscovery. Regards. $\endgroup$ Commented Dec 30, 2013 at 18:49

Just to spark your curiosity, I'll prove you an elementary lower bound on $\pi(m)$ on the name of the community, as this does not answer the question. Essential theorem used here: every number $n$ is of the form $ab^2$ with squarefree $a$. Check for yourself using FTA that $a=p_1p_2\cdots p_k$, so no squares of primes appear in $a$. You can see this intuitively when considering $\sqrt{n}$, it is always of the form $b\sqrt{a}$.

That said, choose $x$ from the set $S=\{1,2,3,4,\ldots, m\}$ such that $1\leqslant x=ab^2\leqslant m$, with $a$ squarefree. Clearly, we have $m$ possiblilities to choose $x$. That's fine; now, we'll restrict the number of possible $a$'s and $b$'s we can choose. We will show that there are at most $2^{\pi(m)}$ such $a$'s and at most $\sqrt{m}$ such $b$'s.

We know we have $\pi(m)$ primes between $1$ and $m$. Therefore, $a$ is of the form $p_1^{\varepsilon_1}p_2^{\varepsilon_2}p_3^{\varepsilon_3}p_4^{\varepsilon_4}\cdots p_{\pi(m)}^{\varepsilon_{\pi(m)}}$, where $\varepsilon_k$ is either $0$ or $1$ depending or whether $p_k$ appears in the prime factorization of $a$. We see, then, that there are $2^{\pi(m)}$ such possible $a$'s to choose. And how many $b$'s? The key restriction is that $b\leqslant\sqrt{m}$, since $a\geqslant 1$.

So we can choose at most $2^{\pi(m)}$ possible $a$'s and $\sqrt{m}$ possible $b$'s to form $x=ab^2\leqslant m$. However, not all the numbers formed are smaller than $m$, some might exceed it. Nevermind that now; we may refine that later.

That way, we establish $m\leqslant 2^{\pi(m)}\sqrt{m}$. Now, manipulate to get $$\sqrt{m}\leqslant 2^{\pi(m)}\implies m\leqslant 2^{2\pi(m)}\implies\log_2(m)=\frac{\log(m)}{\log(2)}\leqslant 2\pi(m)\\\implies\frac{\log(m)}{2\log(2)}\leqslant \pi(m).$$

This might not be groundbreaking, but now you know a proof involving $\pi(x)$ and logarithms, so hey, them don't seem unrelated after all. There is a whole class of elementary proofs bounding $\pi(x)$ using combinatorics and related techniques, you can see them in books like Hardy and Wright's or Ireland and Rosen's.


Yes, probabilistic primality testing and related gadgets are quite common. They assist encryption, for example, by providing candidate primes.

However, it's hard to say whether your particular way of doing it is worthy of being published, especially without seeing it. This is quite an extensively studied area.

  • $\begingroup$ Thanks for your answer, please check my last comment. Regards. $\endgroup$ Commented Dec 30, 2013 at 18:43

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