Structure and randomness in the prime numbers

7 February, 2008 in math.NT, talk, travel | Tags: number theory, powerpoint, randomness, structure | by Terence Tao

This Thursday I was at the University of Sydney, Australia, giving a public lecture on a favourite topic of mine, “Structure and randomness in the prime numbers“. My slides here are a merge between my slides for a Royal Society meeting and the slides I gave for the UCLA Science Colloquium; now that I figured out to use Powerpoint a little bit better, I was able to make the latter a bit more colourful (and the former less abridged).

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9 January, 2011 at 3:37 pm

Pete Quinn

If anyone is interested, I’ve tried to explain the ideas more concisely in the form of a draft paper. Note it’s written by an engineer, not a mathematician, and therefore is written intuitively in the way the ideas developed, but not necessarily in a particularly elegant fashion.

http://petequinnramblings.wordpress.com/2011/01/09/draft-paper-on-patterns-in-prime-numbers/

I’d appreciate if anyone could tell me if the ideas have been published before, and please note it’s a rough draft, so references in particular need to be tightened up, and the formatting is a little icky.

10 January, 2011 at 8:01 am

Hugh

Sorry, the attempt twin prime proof is here:

http://barkerhugh.blogspot.com/2011/01/twin-primes-and-polignac-conjecture.html

24 February, 2011 at 10:57 pm

mobiusfunction

Two recurrences:
T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (0 + (sum from i = 1 to k-1 of T(n-i,k-1))) mod 2.

T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (1 + (sum from i = 1 to k-1 of T(n-i,k-0))) mod 2.

Mahonian numbers modulo 2

Mahonian numbers mod 2

21 March, 2011 at 8:02 am

petequinn

Hopefully this is not seen as using up unnecessary bandwidth.

Acknowledging again this is just a bit of a hobby project, for fun. We’ve been playing with this a little further, and can offer a couple of observations perhaps a little more concisely.

Within any primorial, there is a symmetric pattern of “non-primes” that repeats for all repetitions of that primorial to infinity. These numeric positions can never be prime, as they all represent multiples of primes (namely the factors of that primorial). There is also an asymmetric, non-repetitive component of non-primes within each primorial. This asymmetric component represents all additional “non-primes” that need to be deleted to leave behind only primes.

The repetitive pattern, whose symmetry turns out to be fairly easy (I think) to prove, involves only the primes that are factors of the primorial. The asymmetric, non-repetitive, component involves (prime) multiples of all primes higher than the prime factors but less than the square root of the primorial.

Each of the higher primes not involved in the symmetric pattern within a primorial has a repetitive, symmetric pattern that eventually emerges as the numbers get high enough. But this does not emerge until you reach the first primorial involving that prime as a factor.

Of interest (if perhaps only to me), is there are “long” sequences of non-primes leading up to and following every primorial, and every multiple of every primorial. If the nth primorial is Pn#, then (Pn# – 1) and (Pn# + 1) could be prime (common knowledge), but every other number between (xPn# – P(n+1)+1) and (xPn# + P(n+1)-1) must not be prime, where x is some arbitrary integer (also, I think, fairly simple to prove). So if for some reason you need an arbitrarily long string of sequential non-primes, simply go to the primorial corresponding to the closest higher prime (or any multiple thereof).

15 April, 2011 at 3:09 am

Mats Granvik

The prime zeta function (Wikipedia) is defined as:
$P(s)=\sum\limits_{p\,\in\mathrm{\,primes}} \frac{1}{p^s}$

It is known that:
$\log(\zeta(s))=\sum\limits_{n=1}^\infty \frac{\Lambda(n)}{\log(n)}\frac{1}{n^{s}}$

We may consider:
$a_n=\prod\limits_{d\mid n} \frac{\Lambda(d)}{\log(d)}$
where zero is not included in the multiplication and $a_1=1$ .

Doctormatt rewrote $a_n$ as:
$a_n=\prod\limits_{p^k \mid \mid n} \frac{1}{k}$

Can it be proven that:
$P(s)=\log\sum\limits_{n=1}^\infty \frac{a_n}{n^s}$

23 July, 2011 at 8:07 pm

spaghetti monster

Ian Robinson: You said…..
“The image of the physical makeup of the Whirlpool Galaxy looks identical to my drawing, I have deliberately withheld an important piece of the jigsaw until a later date.
The truth is I require someone with a bit of mathematical clout to give these findings the validity and respect they deserve, no one else is aware of my findings.”
—————————————-

You might find the link below more helpful than Terry Tao’s blog.
http://math.ucr.edu/home/baez/crackpot.html

20 September, 2011 at 3:53 pm

petequinn

In case anyone is interested, I extended the previous bit of numerical experimentation to twin primes, and found (perhaps unsurprisingly) some similar patterns. Some details here:

http://wp.me/p1h3oO-O

Was just playing around a bit to try to understand the bounds of the twin prime conjecture problem. No breakthroughs to share (surprise!) but I find the patterns and trends to be interesting, and help me, as a visual thinker, to put the challenge in perspective.

31 October, 2011 at 8:05 am

petequinn

CRACKPOT ALERT :-)

More on primes and twin primes from a complete amateur. Looking for help answering a question…

What I’ve tried to describe in previous posts are some patterns that develop within primorials. For the n-th primorial, Pn#, the distribution of primes can obviously be fully developed by repetition of all primes less that sqrt(Pn#). The removal (or sieving) of all non-primes can be divided into two broad components: (1) a symmetric pattern of non-primes removed through repetition of all primes from 2 to Pn (symmetric since these all divide evenly into Pn#), and (2) a non-symmetric distribution of non-primes generated by all other primes from P(n+1) to the highest prime < sqrt(Pn#).

Numerical experimentation shows that the symmetric component of non-primes projects outward, presumably to infinity, for every subsequent repetition of Pn#, so that primes can never fall on those positions, and primes appear, at "random" (clearly not random, but apparently random) intervals at the other positions that were not "non-primes" within Pn# screened by the primes to Pn.

Interestingly, if only to me, is that the number of "possible primes" (i.e. not "non-primes" screened by the primes to Pn), which I will call Nn, within any given primorial, Pn# ( obviously defined as the product of all primes to Pn), is related to the primorial as follows: Nn = (product for all i=1 to n) [Pn – 1]

(sorry I don't have LaTEX)

In my last post I linked to some notes describing results of numerical experimentation that show that similar patterns develop for twin primes. Within any given primorial, all possible twin primes can be sieved via symmetric removal due to primes to Pn, and then non-symmetric removal by higher primes to sqrt(Pn#). The symmetric pattern of "non-twins" removed by, and lingering "possible twins" left alone by, the sieving of primes to Pn, then repeats to infinity for all repetitions of Pn#, so that higher twins will always only occur at these "possible twin" locations.

Even more interesting (probably still only to me, haha) is that the number of "possible twins" within any given primorial is ALSO related to the primorial, similar to the way the "possible primes" are, but with the following form, where Tn is the number of "possible twins" within any given repetition of a primorial:

Tn = (product for all i=2 to n) [Pn – 2]

Alternatively, one can describe the number of "possible twins" removed when going from Pn# to P(n+1)# as:

T(n+1) = P(n+1)*Tn – 2 * (product for all i=2 to n) [P(n-1) – 2]
= P(n+1)*Tn – 2 * Tn
= (Pn – 2)*Tn

Using a couple of random examples to illustrate this observation:

consider P4 = 7, where P4# = 210.

The number of possible positions for primes within repetitions of 210 projecting out to infinity is 48, a result you can check for yourself fairly easily (work out Pi(mod 210) for as many primes as you like and you will find exactly 48 unique values). This is (7-1)*(5-1)*(3-1)*(2-1)

Similarly, the number of possible twin prime pairs within repetitions of 210 is T4=15, which is (7-2)*(5-2)*(3-2). You can readily see that T3 would be 3, or (5-2)*(3-2).

Note that T4 = 15 = (P4 – 2)*T3 = 5*3

I can't work out the math to prove these results, but they seem far to simple to be accidental, and to not continue to infinity. Can anyone suggest a way to demonstrate these results algebraicly, or otherwise?

Thanks for humouring me.

Pete

31 October, 2011 at 8:08 am

petequinn

dangit, no editing function… there are a few silly typos that should be obvious by context, like some n’s in products should be i’s

1 November, 2011 at 12:10 pm

petequinn

Goodness I left numerous irritating little typos in that post, sorry. To correct a few:

“Possible primes” within the nth primorial, Pn#, is:

Nn = (product for all i=1 to n) [Pi – 1]

“Possible twins” within the nth primorial is:

Tn = (product for all i=2 to n) [Pi – 2]

T(n+1) = [P(n+1) – 2]*Tn

For a few moments, let’s suspend disbelief and pretend these definitions for Nn and Tn are proven.

I now propose that the density of primes, within all natural numbers, is less than Nn/Pn#, and converges on Nn/Pn# as n approaches infinity. I will claim this without proof for the moment and maybe elaborate later. More disbelief to suspend, sorry. :-)

I also propose that the density of twin pairs, within all natural numbers, is less than Tn/Pn#, and converges on Tn/Pn# as n approaches infinity.

The ratio of the density of twin pairs to the density of primes, within all natural numbers, which is the same as the density of twin pairs within all primes, can be taken as (Tn/Pn#)/(Nn/Pn#) = Tn/Nn, n approaching infinity

Let’s now compare the density of twin pairs within the set of primes to the density of primes within the set of natural numbers, as:

[Tn/Nn]/[Nn/Pn#], as n approaches infinity.

I believe it is fairly straightforward to show that this ratio converges on about 1.320…, the twin prime constant.

This result tells us, I think, that the set of twin primes corresponding to the set of primes is 1.32 times larger than the set of primes corresponding to the set of natural numbers (with apologies, if this isn’t written correctly from a mathematical perspective). Since the set of primes is infinite, hence the set of twin primes is also infinite.

Does that make sense?

I’ve claimed a couple of things in here without proof – the first being the existence and definitions of Nn and Tn, which I’ve demonstrated (to myself anyway) through numerical experimentation but not proven mathematically. I would think that, given the simplicity of the relationships they should be easy to develop algebraically, but admit that they are beyond my ability. The second unproven claim is that the ratios converge to the values indicated as n approaches infinity. I think those claims are much easier to rationalize, and the fact that they lead to the twin prime constant would seem to suggest they must be correct.

Thoughts?

Thanks for reading. :-)

Pete

1 November, 2011 at 9:43 pm

petequinn

One more crackpot post, thanks for the indulgence Prof. Tao.

I’d appreciate if someone would comment, even if only to make fun of me. This is the internet after all. :-)

I know this is a famous problem and it’s bad form to hastily claim a proof (or “near-proof,” as in this case I think) of a famous problem, but since I am not a mathematician, I have no reputation to lose by exposing myself to ridicule for this, and I can easily slink back to the day job when someone pokes a big hole through this. :-)

Let me elaborate on two of the items I’ve claimed without proof in the previous post, and then also show why the ratio of twin pair/primes density to primes/natural numbers density is 1.32, or at least converges and is not less than 1. That will be in the next post to keep this one relatively short.

To find all prime numbers, we need to strip away all composite numbers. We can imagine doing this by progressively removing, or sieving, all prime multiples, working our way from the lowest prime through to infinity.

Let’s start with by removing all multiples of 2, the even numbers, from the set of natural numbers. This strips out half of all natural numbers as prime candidates, leaving only the odd numbers as possible primes. Then remove all multiples of 3, 5, 7 and so on through to Pn, the nth prime, with n approaching infinity.

Consider the second primorial, P2# = 6, after first removing the even numbers, so we have 1, 3 and 5 as prime candidates, and 2, 4 and 6 as confirmed “non-primes” between 1 and 6. Note that this pattern, which is symmetric within 6, repeats ad infinitum, or in other words, after sieving the even numbers, all remaining numbers (mod 6) are either 1, 3 or 5.

Now remove all multiples of 3 from the remaining set of “possible primes.” This removes one possible prime candidate between 1 and 6, leaving 1 and 5 as possible primes. Note again this leaves a symmetric pattern, which, since we have now only removed multiples of 2 and 3, repeats ad infinitum, removing 2/3 of all natural numbers as non-primes and leaving 1/3 for further consideration as possible primes. After having sieved the 2s and 3s, the remaining “possible prime” candidates (mod 6) are all either 1 or 5.

We can continue in the same fashion to remove all remaining multiples of 5, 7, 11, 13, …. to Pn. Each time we do this, we will be left with a set of non-primes and “possible primes” that is symmetric within the respective primorial, and which repeats ad infinitum. The density of “possible primes” within the set of natural numbers, after sieving all primes to some arbitrary Pn, is equal to the number of “possible primes” within the respective primorial divided by the value of the primorial.

The complete set of prime numbers is revealed when we have sieved all possible composite numbers by repeating to infinity each prime, and striking out all these repetitions as non-prime. Therefore, if we continue this process as described, ad infinitum, to remove all multiples of all primes to Pn, where n approaches infinity, then we will be left with only primes. Hence, if we call Nn the number of “possible primes” within Pn#, then Nn/Pn# converges on the density of primes within the set of natural numbers as n approaches infinity, since as n approaches infinity, Nn becomes the set of primes and Pn# becomes the set of natural numbers.

Now let’s examine how many natural numbers are removed as candidates for any given prime, Pn. Choose an arbitrary prime, Pn, and consider the number of multiples of Pn within Pn#. This is clearly P(n-1)#. If no multiples of primes smaller than Pn had yet been removed, then repetitions of Pn to infinity would eliminate Pn prime candidates out of every Pn# natural numbers. However, assuming Pn > 2, and following our sieving process sequentially with monotonically increasing n, by the time we are set to remove remaining candidates that are multiples of Pn, there are a number of multiples that have already been removed, namely all those that are multiples of lower primes < Pn. If you conduct a simple accounting exercise, starting with the smallest prime numbers and working up through a few, you should be able to convince yourself that of the P(n-1)# possible multiples of Pn within Pn#, [P(n-1)# – N(n-1)] have already been eliminated by the multiples of Pn with lower primes, where N(n-1) means the number of possible primes left within the lower primorial, P(n-1)#, after sieving of all primes up to P(n-1), and Nn is equal to, as claimed before:

Nn = (product for all i=1 to n) [Pi – 1] —– eqn [1]

To demonstrate… removal of all even numbers eliminates half of all numbers between 1 and 2 (= P1#), and by extension removes half of all numbers within every multiple of P1# to infinity, or half of all natural numbers.

Removal of all remaining multiples of P2 = 3 between 1 and 6 (= P2#) eliminates the number 3 as a possible prime within subsequent repetitions of P2#. Note that 6 was already removed, since it is a multiple of 2. So there were 2 (= P1#) multiples of 3 within P2#, but 1 (i.e. {P1# – N1}, or {P1# – eqn [1] for n-1, where n = 2} had already been removed by multiples of 3 by 2, leaving only 1 possible prime candidate to be removed at this sieving pass, to leave 2 remaining “possible prime” candidates (1 and 5) within P2#, or 2/6 = 1/3 “possible primes” within the set of all natural numbers after repeating this pattern an infinite number of times. Hence all primes (mod 6) greater than P2 must equal 1 or 5, a well known result.

Now if we remove the remaining multiples of P3 = 5 between 1 and 30, we find that of the 6 (i.e. P2#) possible multiples of 5, 4 had already been removed, leaving only 2 (i.e. {P2# – eqn [1] for n-1 where n = 3}), namely 5 and 25, to be screened by repetitions of 5. Thus of the 10 possible primes within 30 existing after sieving by 2 and 3, two more candidates (5 and 25) have been removed, and all primes (mod 30) greater than P3 must equal 1, 7, 11, 13, 17, 19, 23 or 29 (i.e. exactly 8 possibilities).

Repeating the same logic now for multiples of P4 = 7 up to 210, of the 30 (= P3#) multiples of 7 to be removed, 22 ({P3# – product for i = 1 to 3 of [Pi – 1]}) were already screened by multiples of 7 with 2, 3 or 5, leaving 8 other multiples of 7 to be eliminated at this stage. Thus of 8 x 7 = 56 possible primes within 210 remaining after sieving by 2, 3 and 5, the further sieving by 7 removes 8 other possible prime candidates, leaving 48 “possible primes” within repetitions of 210, and therefore 48 unique values for all primes higher than 7 (mod 210), which I won’t list here.

Now let’s examine why Nn takes the form I’ve claimed, using P4 as a working example.

If we want to count the number of multiples of 7 with 2, 3 and 5, we should expect:

3 x 5 multiples of 2 x 7,
2 x 5 multiples of 3 x 7,
2 x 3 multiples of 5 x 7,
2 multiples of 3 x 5 x 7,
3 multiples of 2 x 5 x 7,
5 multiples of 2 x 3 x 7, and
1 multiple of 2 x 3 x 5 x 7

In order to identify unique multiples of 7 already removed by 2, 3 and 5, we obtain [(3 x 5) + (2 x 3) + (2 x 3)] – (2 + 3 + 5 + 1) = 22, which we can re-organize and re-write as {5 x 3 x 2 – [(5 – 1) x (3 – 1) x (2 – 1)]} = 22. Or, the number of unique multiples of 7 removed within multiples of 210 is 8 = (5 – 1) x (3 – 1). And hence the number of possible primes remaining within multiples of 210 is 8 (from P3#) x 7 – 8, or 48, or 8 x 6, or 8 x (P4 – 1), or (P4 – 1) x (P3 – 1) x (P2 – 1) x (P1 -1).

Hence, Nn = (product for all i=1 to n) [Pi – 1]

Phew. :-)

I think that is fairly straightforward, and I expect the logic to generate the product formula for Tn is very similar, but this came to me on today’s 9 mile run, and I think I need another good run or two to come around to figuring that one out. :-)

To wrap this post, I think I’ve defended the prior claimed definition of Nn, the number of “possible prime” candidates within the nth primorial (which by extension serves to define the density of “possible prime” candidates within the set of natural numbers after sieving by primes up to Pn). And, I think I’ve defended the claim that this density of “possible primes” within the nth primorial converges to the density of primes within the set of natural numbers as n approaches infinity.

OK, on to producing the twin prime constant, or something like it, in one more post…

1 November, 2011 at 10:32 pm

petequinn

So let’s start with the assumption that the claimed definitions of Nn and Tn are proven. I think I’ve proven the first, and can only say I believe the second from numerical experimentation, and will try to defend it, given more time and more miles. :-)

Let’s pick up from two posts back with the idea that the ratio of the density of twin pairs within the set of primes to the density of primes within the set of natural numbers can be expressed as:

D(t:p)/D(p:nat) = [Tn/Nn]/[Nn/Pn#] = Tn(Pn#)/(Nn)^2, with n approaching infinity

Where D(t:p) means the density of twin pairs within the set of primes, and D(p:nat) = density of primes within the set of natural numbers.

This can be re-written as:

D(t:p)/D(p:nat) = product for i = 1 to infinity {Pi x P(i – 2)/[P(i-1)]^2} ——- eqn [2]

Where we include P0 = 1 to allow us to simplify the examination of the products with no change to the result.

This can be written as:

D(t:p)/D(p:nat) = limit as n approaches infinity {Pn x P(n – 2)/[P(n-1)]^2} x … x {13 x 11/12^2} x … x {3 x 1/2^2} x {2 x 1/1^2}

Numerical experimentation shows this series converges quite nicely to 1.320… as claimed. I won’t attempt to work out the arithmetic, but by inspection, as n increases, the nth term steadily converges on 1, being always < 1, and with n = 1, 2, 3 and 4 we get 2, 1.5, 1.406…, and 1.367…

Without knowing the actual value of this infinite product, we can obtain a lower bound by examining:

Product for i = 1 to infinity {i x (i – 2)/(i – 1)^2}

which converges on 1.

This latter product includes all the terms in eqn [2], but also includes additional terms less than 1 for values of i that are not prime. Hence it decreases more quickly than eqn [2], and is always lower than eqn [2], after the first two terms, when the same number of terms has been included in both products. Therefore in the limit, eqn [2] converges on a positive number that is less than 1.367… (by inspection) and not less than 1.

So, if we believe the definition of Tn as claimed, I think we have shown that the set of twin prime pairs is infinite, since its density within the infinite set of primes is at least equal to the density of primes within the infinite set of natural numbers.

Of course the definition of Tn is not yet proven. A few more miles to be run first… :-)

Questions, comments, rotten tomatoes welcome.

Hopefully no fatal typos in this post, fingers crossed…

2 November, 2011 at 11:34 am

petequinn

And of course there were some critical typos. Corrections:

D(t:p)/D(p:nat) = product for i = 1 to infinity {Pi x (Pi – 2)/[(Pi-1)]^2} ——- eqn [2]

Where we include (P1 – 1) = 1 (instead of 0)…

——

I think that’s the worst of it.

3 November, 2011 at 7:34 am

petequinn

Goodness even the corrections need corrections:

Where we include (P1 – 2) = 1…

The idea there is simply to keep 1 as a placeholder (instead of (2 – 2 = 0) for when we group the individual product terms. It’s not a necessary element, just a convenience to make the elaboration of the infinite product easier to manage and look at.

I see there are still some other minor typos in the preceding posts: a (2 x 3) instead of (2 x 5), and at least one Pn instead of the correct P(n-1)#. Sorry for those.

I think I’ve just about got the correct explanation for Tn, will try to elaborate after the end of the work day. Meanwhile, I should probably explain one thing I’ve glossed over that may have some readers stuck or rejecting my logic.

In the preceding posts I’ve made no specific mention of “actual primes,” but have rather focussed on a discussion of confirmed “non-primes” (as ruled out in any given sieving by a given Pn) and “possible primes” that remain after each sieving step. I’ve suggested that these values can be extrapolated to infinity to obtain the ultimate densities of primes and twin primes.

But what about the pesky ACTUAL primes that develop as we go, namely 2, 3, 5, …. Pn? These actual primes occupy positions I have called “non-primes.” Is this a fatal flaw in my logic? I’m fairly confident this has no bearing, and here is why…

If we take Nn and Tn to be the number of possible primes and possible twins within the nth primorial, where “possible” implies “still standing as candidates after sieving by primes to Pn,” then these values represent the numbers of possibilities within all primorials after the first one (accepting for the moment that Tn remains to be developed/defended).

Within the first primorial, we know that we have n ACTUAL primes, which are sitting on n “non-prime” positions. If we want to be precise, then, we should perhaps state that the number of “possible primes, “within the first Pn# only, is actually Nn + n, and thereafter remains Nn. We may also have some number, which must be less than n, of actual twin primes.

It is trivial to show to show that (Nn + n)/Pn#, the density of primes within the first primorial, converges to Nn/Pn# as n grows. It is also trivial to show that this addition of n “possible primes” within the first primorial has no bearing on the density of “possible primes” in the set of natural numbers after the nth sieving step, when we recall that the first primorial containing the extra n “possible primes” is only one of an infinite number, with the rest all containing exactly Nn “possible primes.”

Hence the density of “possible primes” within the set of natural numbers after n sieving steps remains Nn/Pn#, and the density of twin pairs Tn/Pn#.

Now if only we could prove the definition of Tn, we might have the problem licked. :-)

(or I could be completely wrong, and given the fact that I’ve never done anything elegant in math on my own, this latter outcome seems somehow more likely, yet this continues to be fun and intellectually stimulating to me, so thanks again for the continued indulgence)

3 November, 2011 at 8:30 am

Terence Tao

I think any further discussion of these matters should take place on your own web site, where you have the ability to edit your own posts.

3 November, 2011 at 8:32 am

petequinn

OK fair enough, thanks for letting it go on this far.

All the best!

6 November, 2011 at 12:28 am

petequinn

One final post to let anyone who may have been following this know the story continues to develop here:

http://petequinnramblings.wordpress.com/2011/11/05/prime-triples/

Further numerical experimentation shows predictable numbers of candidate k-tuples, other than primes and twin primes, within the primorials.

Thanks again Prof. Tao, I won’t post here on this topic again, I promise.

Cheers,

Pete

11 February, 2012 at 5:27 am

jose javier garcia

Click to access 1111.0105v4.pdf

the Riemann Xi function can be written as $ \xi(s)=Cdet(H+s(s-1)+1/4) $

Here ‘H’ is a Hamiltonian operator whose potential is $ V^{-1}(x) = A \sqrt D n(x) $ where N(x) is the eigenvalue staircase :)

13 February, 2012 at 12:40 pm

jose javier garcia

my operator is correct but none put attention because i am no famous :( , you can check the energies and see that they agree with the square of the riemann zeros :D

16 July, 2014 at 3:14 am

0_Lesh_a

$\Longrightarrow \ -\ {\rm means}{\rm \ }{\rm that\ the}{\rm \ }{\rm number\ is\ represented}{\rm \ }{\rm as\ a\ product\ of}{\rm \ }{\rm two\ numbers}$
$\longrightarrow \ -\ means\ that\ the\ number\ is\ not\ represented\ as\ a\ product\ of\ two\ numbers$
$\displaystyle {\mathbb O}\ \Longrightarrow mm\longrightarrow nn\longrightarrow nm$
$\displaystyle {\mathbb D}\Longrightarrow nn\longrightarrow nm\longrightarrow mm$
$\displaystyle {\mathbb G}\Longrightarrow nm\longrightarrow mm\longrightarrow nn$
$\displaystyle {\mathbb P}\longrightarrow nn\longrightarrow mm\longrightarrow nm$
$\displaystyle mm-product\ of\ an\ odd\ number\ and\ an\ odd$
$\displaystyle nn-\ 2^{a_i};;\ a_i\ (1,2,3,4,5\dots )$
$\displaystyle nm-{\rm the\ product}{\rm \ }{\rm of\ an\ even\ number}{\rm \ }{\rm and\ odd}\$$ latex \displaystyle {\mathbb P}\in \left(2,3,5,7,11\right)-prime\ numbers$
To exhaust all combinations
$\displaystyle {\mathbb S}\Longrightarrow nn\Longrightarrow mm\longrightarrow nm$
$\displaystyle {\mathbb T}\Longrightarrow nn\Longrightarrow nm\longrightarrow mm$
$\displaystyle {\mathbb K}\Longrightarrow mm\Longrightarrow nm\longrightarrow nn$
$\displaystyle {\mathbb Z}\Longrightarrow nn\Longrightarrow mm\Longrightarrow nm$

The question is a mirror image of the entire sequence of numbers or through an extension of the product operator. Or the first and the second time. And most importantly, how it can help to understand the nature of prime numbers, I do not know

16 July, 2014 at 7:22 pm

Anonymous

@0_Lesh_a: Will you please stop your comments? They are obviously gibberish.

17 July, 2014 at 4:58 am

0_Lesh_a

It’s just your opinion

11 June, 2019 at 6:21 am

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Structure and randomness in the prime numbers

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The Polymath Blog

76 comments

Leave a reply to Terence Tao encuentra un fallo en la demostración de Li de la Hipótesis de Riemann [ENG] Cancel reply

For commenters