Do Infinitely Many Prime Pairs Exist?

[nnnnnnnn]

There is (as far as I know) no proof-for or against- that there are infinately many prime pairs such as 3, 5 or 29, 31...

Anyway, is it intuitive to assume that there should be infinitely many pairs just b/c of the fact that there are infinitely many numbers? or does this have nothing to do with it?

 

[HallsofIvy]

Well, "intuitive" is not a very good "mathematics" term!

Is it "intuitive to assume" that there are infinitely many even primes b/c of the fact that there are infinitely many numbers.

 

[nnnnnnnn]

Its funny that you say that because talking about math is the only time I would say intuitive...

Anyways, I can't think of a good example but I can think of an example:
in a class for real numbers I had to prove that 1>0. I knew this to be true because it was intuitive but it was tricky to prove...

 

[matt grime]

You can't prove that 1>0, unless you assume certain things...

 

[mathwonk]

My intuition is that there are indeed infinitely many prime pairs, but it is based on nothing I can describe clearly.

I.e. to me it would just be odd for there to exist a largest prime pair. There is a tendency of mathematical facts to be "natural" and not so odd.

To me at least it would seem less surprizing or odd for there to be an infinite number of prime pairs.

I.e. for there to be a largest one, I would thionk there needs to be a "reason" for that. Whereas if there are infinitely many, then there is no special one, and that is more expected to me.

But all mathematicians have different intuition, so no one need agree with me.

 

[shmoe]

You're right that it's currently unkown whether or not there are infinitely many prime pairs.

There is the twin prime conjecture which claims that the number of prime pairs less than x is asymptotic to Cx(\log{x})^{-2}, where the C is explicit (about 1.32..). There are heuristic arguments to support this, but of course no one can prove it yet.

There's a partial victory by J.R. Chen which implies there are either infinitely many prime pairs, or there are infinitely many primes p where p+2 is the product of two primes (possibly both are true).

 

[nnnnnnnn]

You can't prove that 1>0, unless you assume certain things...

thats not the point but we were working with real numbers so didn't need to assume anything- just to follow the established rules...

 

[mathwonk]

I like shmoe's asymptotic formula. it gives substance to a prediction there are infinitely many.

I.e. if one has actual data up to a few billion billion billion... or so, that there is a pattern to the density of prime pairs, then it seems believable that the density will not suddenyl go to zero after some point.

 

[mathwonk]

try this: in the spirit of proving that 1>0, prove that any function f defined on the real numbers and satisfying f(x+y) = f(x)f(y), (think exponential function), is either identically zero, or always positive.

 

[shmoe]

Billions and billions of data points can look convincing, but can really come back to bite you in number theory. Like Merten's conjecture (that |\sum_{n\leq x}\mu(n)|\leq x^{1/2} where \mu is the mobius function), or the conjecture that the prime counting function is strictly bounded above by the logarithmic integral. Both were proven false, but the first counterexamples are huge (afaik, none are known explicitly in either case, just some scary upper bounds). These are a bit different then the twin primes though, I don't think there was really much to support these false conjectures besides computations. The twin prime conjecture has other convincing evidence.

 

[Hurkyl]

I love this theorem: the Frivolous theorem of Arithmetic.

 

[Icebreaker]

I love this theorem: the Frivolous theorem of Arithmetic.

What the...

 

[hypermorphism]

What the...

Let b be the natural number such that b + n for any natural n is considered to be a very, very, very large number. Let B be the set of all naturals less than or equal to b. Then the cardinality of B is finite, while the cardinality of the complement of B within the set of all naturals is the same as the cardinality of the set of all naturals. QED.

 

[Icebreaker]

http://mathworld.wolfram.com/StrongLawofSmallNumbers.html

"The first strong law of small numbers (Gardner 1980, Guy 1988ab, Guy 1990) states 'There aren't enough small numbers to meet the many demands made of them.' "

"The second strong law of small numbers (Guy 1990) states that 'When two numbers look equal, it ain't necessarily so.' "

What the...

 

[marteinson]

Actually I think we can prove there are infinitely many prime pairs. But I'm very rusty on formal proofs, so maybe one of you experts could formalize on what I'm saying will work.

There are three types of primes: (a) 2 and 3, (b) those which operated on by mod(6) = 5, and (c) those which under mod (6) = 1.

In other words, every multiple of six, 6n, has a pair of potential primes at 6n +/- 1, as noticed by eratosthenes.

However, no one seems to have used modular arithmetic as I suggest in my paper
http://www.chass.utoronto.ca/french/as-sa/ASSA-14/article7en.html
to separate the primes above 3 into two series, equalling 1 and 5 in mod6, or, you could think of them as equalling 7 and 5 in mod6. There is no interdependency between the primeness of the terms of the two series, 6n+1 and 6n-1, and both series display the only candidates for primeness, and contain all primes, and all their members -are- primes unless factorizable by an inferior member of the same series.

See the new "modulus 6 clock spiral" which I propose to replace Ulam's spiral, in the article, and you'll see what I mean.

Peter

 

[shmoe]

Considering primes mod 6, and indeed primes in more general arithmetic progressions, is an old concept.

That all prime pairs (except 3 and 5) are of the form 6n-1, 6n+1 is nothing new either, nor does it show there are infinitely many prime pairs. It just tells you (vaguely) where to look for them.

"...and all their members -are- primes unless factorizable by an inferior member of the same series."

This is false, 25=1 mod 6 but 25=5*5, and 5 is not 1 mod 6.

The other way is true, if n=5 mod 6 and n is composite then it has a prime divisor congruent to 5 mod 6 (though it may have prime divisors congruent to 1 mod 6 as well)

 

[Hurkyl]

Looking at primes of various modulo classes is done, and not just modulo 6.

You've made a mistake, BTW -- A number of the form 6n+1 can have all of its nontrivial factors of the form 6m-1. (e.g. 25) Also, A number of the form 6n-1 can have factors of the form 6m+1.

 

[shmoe]

http://www.chass.utoronto.ca/french/as-sa/ASSA-14/article7en.html

I've only skimmed some of it, an excerpt:

"Conversely, it can easily be demonstrated that each of the three even series on the spiral can be generated by some combination of two primes, either both in the five o'clock series, both in the seven o'clock series, or one in each, without exception, using simple modular arithemtic. I leave the formally correct proof to real mathematicians, however."

There are certain things that you can wave away with "can easily be demonstrated". Goldbach's conjecture is not one of them.


From your "Simple Algorithm":

" c) test each candidate by dividing it by each prime ≤√m, and by each previously rejected candidate ≤√m"

This is just the sieve of Eratosthenes, after 'pre-sieving' by 2 and 3, except you've added this unnecessary bit that I've highlighted in bold. If m is composite that it has a prime divisor less than or equal to it's square root, so it's sufficient (and faster) to only consider primes less that sqrt(m) here.

 

[marteinson]

I appreciate your insights, and it's helpful that real mathematicians can correct me when I'm wrong. But I don't see why the math community makes such a big deal out of Ulam's spiral's "strikingly non-random appearance!" when its non-randomness can be explained in terms of the 6n+/-1 observation by Eratosthenes, as I have done in the graphical illustration of the mod6 clock spiral.

Clearly, the literature is missing the forest for the tree, in failing to recognize that primes are indivisible precisely because they are adjacent to highly divisible numbers I have nicknamed 'prim' numbers, i.e. such things as multiples of 2 and 3, or 2 and 3 and 4, and so on.

And the two series, if you explore the modular arithmetic of all six series, still do demonstrate the Goldbach conjecture, just loook at them.

Once again, thanks for all your good points.

 

[marteinson]

As to hurkyl's pointing out my first 'error', I think he's incorrect. I never said a number of the form 6n+1 can't have factors on the series 6n-1. On the contrary, I said 6n+1, when factorizable and therefore not prime, either has both non-trivial factors in the form 6n-1 or both in the form 6n+1, or one from each series. On the second error, the "unnecessary bit", he is absolutely right and I stand corrected. It's easy to lose track of common sense when thinking in the abstract, and vice versa. I have taken that part out of the algorithm, which is, as he rightly says, just an Eratosthene sieve with 2 and 3 already taken out.

Many thanks.

 

[marteinson]

And thanks to smoe for his comment, but it's a misunderstanding due to the fact that "series" has the same spelling in the singular and plural. When I write "from the same series" I mean the "same two series".

But thanks for the feedback, it's humbling but useful.

 

[marteinson]

Here is a link to the modulus-6 spiral I drew, for people who want to have a quick look at it.
http://www.chass.utoronto.ca/french/as-sa/ASSA-14/modulus6-spiral90.jpg

Maybe you can see directly from it the properties in mod 6 that I'm trying to point out.

 

[shmoe]

I appreciate your insights, and it's helpful that real mathematicians can correct me when I'm wrong. But I don't see why the math community makes such a big deal out of Ulam's spiral's "strikingly non-random appearance!" when its non-randomness can be explained in terms of the 6n+/-1 observation by Eratosthenes, as I have done in the graphical illustration of the mod6 clock spiral.

They don't really make what I'd consider a big deal out of it. How exactly do you think it's explained by the 6n+/-1?

Clearly, the literature is missing the forest for the tree, in failing to recognize that primes are indivisible precisely because they are adjacent to highly divisible numbers I have nicknamed 'prim' numbers, i.e. such things as multiples of 2 and 3, or 2 and 3 and 4, and so on.

This is false. There will be numbers adjacent to multiples of 2, 3, 4, 5, etc. (any number of factors you like) that are not prime. In otherwords, being adjacent to a 'highly divisible' number (for whatever definition of 'highly divisible' you like) does not make your number prime. Pick any 'highly divisible' number k you like (actually any nonzero number at all that you like), the sequence kn+1 (and also kn-1) will contain infinitely many composite numbers.

And the two series, if you explore the modular arithmetic of all six series, still do demonstrate the Goldbach conjecture, just loook at them.

Again, "just look at them" doesn't even begin to resemble a proof. Give some details on why you think this is true.

And thanks to smoe for his comment, but it's a misunderstanding due to the fact that "series" has the same spelling in the singular and plural. When I write "from the same series" I mean the "same two series".

This would go for Hurkyl's objection too. When you said the "same series" it really looks like the 'same' is there to distinguish between the series 6n+1 and 6n-1, saying which one the factors of 6n+1 (or 6n-1) would have to come from.

Though if you've factored a number of the form 6n+1 into 2 factors, you can't have one factor congruent to 1 and the other congruent to 5, they are both 5 or both 1.

 

[neurocomp2003]

The question above is twin primes ...not paired primes...i thought paries primes was
primes of : (p,p+x)

 

[matt grime]

Here's one for the amateur Goldbach sleth.

Offer a definition of "highly divisible" and I almost certain I can guarantee a string of n "highly divisible" consecutive numbers (ie all the ones lying close to it, again in any sense you care to actually define, are not prime).

 

[marteinson]

Well, only he that knows the way can see that there's a path, and when you don't want to accept something you are inclined not to see it. I NOWHERE said that being adjacent to highly divisible integers automatically makes numbers prime, the point is that prime numbers are prime because they are adjacent to highly divisible integers, and those in that position which are not prime nevertheless have very few factors, also for the very reason that they are adjacent to highly divisible numbers. I suggest you read the entire text thoughtfully and speculatively before disagreeing. The posts here are brief paraphrases, intended to signal what is in the article, nothing more, and you have now, several times, misinterpreted them by skimming and shooting.

The point I'm making, which you don't want to see, is that highly divisible numbers may be thought of as depriving their immediate neighbors of factors, by what I described as a "displacement principle." The neighbors of multiples of six, for instance, are frequently prime, and in fact are the only places you can find any primes above 3, and even when they are not prime, they frequently have only one pair of non-trivial factors, even at orders of magnitude where most integers have several, even dozens, of non-trivial factors. In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

So, in essence, you're reading the posts quickly and turning what I am saying in the article around, almost backwards, in order to "hastily" refute it.

Thanks for contributing, however. I'm not surprised to find resistance to the idea. I'll think about the answers to your other questions as well.

 

[Data]

In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

This is just not true. I can find you a number with the form 6n \pm 1 with arbitrarily many prime factors (as a matter of fact, every composite number [and, of course, every integer in general] not divisible by 2 or 3 has this form).

The contrapositive also helps: The reason that every prime greater than 3 has the form above is precisely that every positive integer without that form is divisible by either 2 or 3. This says nothing about how many prime factors such integers have (besides having more than one as long as they're greater than 3 !).

 

[Hurkyl]

The neighbors of multiples of six, for instance, are frequently prime

False. As the numbers grow larger, the odds that a neighbor of a multiple of 6 is prime decreases to zero.

Remember the Frivolous Theorem of Arithmetic: almost all natural numbers are very, very, very large. You've only looked at small numbers, (and will ever only look at small numbers!) and have no reason to think large numbers behave like small numbers.

 

[matt grime]

Let's do it in more depth shall we?

As best we can tell:

1) n is a prim number if it is a multiple of 6 and has more factors than n-3,n-2,n-1,n+1,n+2,n+3.

2) Numbers adjacent to prim's are "likely" to be prime.

Though there is no proof of this, and you've only looked at small examples. This almost surely isn't going to hold in general. Obvisouly in the part you've graphed the result is merely a result of the smallness of the numbers you're looking at.

You offer no actual proof that studying "prim" numbers will lead to primes, only some very verbose argument that effectively states that all primes are +-/1 mod 6 except 2 and 3, and that modular arithmetic is under used. This isn't true, and can be used to provide a compellingly short proof of the statement all primes greater than 3 are congruent to +/-1 mod 6.

Simply put you are working with numbers that are far too small to have any interesting behaviour.

For instance, on that chart of pink and yellow highlighted primes and prims, it is no surprise that the "prims" have more factors than their immediate neighbours owing to the size of the numbers involved: it is difficult for numbers less than 32 to have lots of factors if they aren't mutliples of 6.


The "displacement principle" section

"Stated otherwise, all integers adjacent to multiples of 6 have zero factors and are therefore prime, unless by ‘coincidence’ they themselves have a prime pair by which they are divisible. All numbers in this particular n ±1 position either have zero (or very few) factors other than one and themselves."

"very few"? What does that mean? And "coincidence"?

Would you like to test your hypothesis?

the number n=25,194=2^3*3^3*13*17*19

(nb I've not checked the divisors of n+1, n-1, but I can, given enough time, find n a multiple of 6 such that n, n-1 and n+1 have at least as many prime divisors you care to give me via the chinese remainder theorem, or by trial and error.)


is such that n-1 is divisible by 7 and n+1 by 5. Do you think that is going to be a prim number? It has a lot of factors but what about its neighbours? Do your small examples give you any feel for the large ones?

You also state that "multiples of 6 are the most highly divisible numbers" What does that mean? why not multiples of 10 or of 15? What degree of multiplicity are you measuring that by?

 

[shmoe]

I NOWHERE said that being adjacent to highly divisible integers automatically makes numbers prime,...

Funny, because you say this exact thing in the next part of this sentence:

... the point is that prime numbers are prime because they are adjacent to highly divisible integers, and those in that position which are not prime nevertheless have very few factors, also for the very reason that they are adjacent to highly divisible numbers.

Earlier you said: "...primes are indivisible precisely because they are adjacent to highly divisible numbers...", I can't see any way to interpret this besides a belief that being next to a 'highly divisible' number will make you prime. (Of course 'highly divisible number' is a vague poorly defined term here, but that's a different issue.). This isn't the first time the words you've said don't match what you claim to mean mathematically.

The point I'm making, which you don't want to see, is that highly divisible numbers may be thought of as depriving their immediate neighbors of factors, by what I described as a "displacement principle."

Something solid can be made of this. If r>1 is a divisor of m, then r is not a divisor of m-1 or m+1. This is not new or in anyway complicated. However, for any \epsilon>0, the number of divisors* of a number m is O_{\epsilon}(m^\epsilon), while the number of primes less than m is about m/log(m), so as m grows this idea of numbers hogging all the prime factors becomes insignifigant because there are so many more other primes available.

*edit-I had meant to add that the number of distinct prime divisors of m is at most log(m)/log(2). So m 'uses up' an even smaller proportion of the primes.

The neighbors of multiples of six, for instance, are frequently prime, and in fact are the only places you can find any primes above 3, and even when they are not prime, they frequently have only one pair of non-trivial factors, even at orders of magnitude where most integers have several, even dozens, of non-trivial factors.

There's a reason that all the neighbours of multiples of 6 have at most two factors in your little spreadsheet. It's quite simple, 5*7*11>300. What you're observing will go away as you look out furthur and the impact of not being able to be divisible by 2 or 3 diminishes.

In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

Data and matt have already mentioned being able to find numbers in 6n+/-1 that have arbitrarily many factors. Let me also mention that you can find an integer k where the numbers 6k+1, 6(k+1)+1, 6(k+2)+1,...6(k+1000000)+1 and 6k-1, 6(k+1)-1, 6(k+2)-1,...6(k+1000000)-1 are all composite. The choice of 1,000,000 here was arbitrary-you can find such a string as long as you like. I'd hardly call something which can occur billions upon billions of times in a row a "coicidence".

So, in essence, you're reading the posts quickly and turning what I am saying in the article around, almost backwards, in order to "hastily" refute it.

By the time of my last post in this thread I had read your article, so I find this insinuation unwarranted, false, and a little insulting. Don't take the fact that I'm not rushing around the streets shouting "Goldbach's has fallen!" as evidence that I don't understand or haven't read it, take it as evidence that I've read it and found it lacking anything interesting (that isn't a trivial observation) or new and correct. You'll often hear mathematicians call the sequence of primes an untameable beast. While from their perspective this is true, the primes are still understood in ways that you haven't begun to imagine. Mathematicians just aren't satisified yet, but this doesn't mean some pretty powerful results aren't known.


I've said that primes in arithmetic progressions have been studied before. You should look into Dirichlet's theorem on the matter to see exactly what it says. So far you've only been discussing 6n+/-1 but you can look at qn+r. For each choice of q these sequences divide up the integers into q 'bins' (depending on your choice of r). Dirichlet will give you an asymptotic relation for the number of primes in each bin and say how they are distributed amongst them. If you consider what happens in the case q=12 for example, you might find it suprising that primes are in fact no more likely to be found in 12n+/-1 than they are in 12n+/-5. I say this might be suprising under the belief that you might feel 12 is 'more highly divisible' than 6.

I don't want to discourage you from investigating prime numbers (or any mathematics at all), but you have to realize that there is a wealth of background information that you haven't seen yet. Without enough background you should consider the possibility that what you're doing has already been done before or is just rubbish. Actually this is something to keep in mind regardless of your background, but it gets easier to tell as you progress. Also, an inability to actually put in precise mathematical terms the things you wish to discuss is only going to cause confusion and frustration.

 

[ExecNight]

I see you are talking about 6x+1 and 6x-1;

I got some info about it,

First of all;

y=1,t=1
y and t increases y+1=2,t+1=2 and so on..

IF x = 5y-1 then
6x+1 is not a prime

IF x = 5y+1 then
6x-1 is not a prime

The above can be divided by 5!

The problem with the above formula is, new prime divisors comes to place to make the formula obselete. For example at 49 which is 7*7, 7 is added to the serie as a new divisor..And then you can notice that 11,13,17 starts joining the new divisors and this goes on as "x" increases...

Of course there might be a rule(if you ask me there is not) when these numbers become divisors of 6x + 1 or 6x - 1..If there is one good luck in finding it =)

 

[Data]

of course there's a "rule" - the smallest (composite) integer of the form 6n \pm 1 = m such that p_i | m where p_i \neq 2, 3 is prime is just 5p_i. For example, 7 is not "added to the series" at 49, this occurs at 35 = 7(5) \equiv -1 (mod 6).

 

[ExecNight]

IF x = 5y+1 then
6x-1 is not a prime

The above can be divided by 5!

Sorry but 35 can't pass my first logical test as,

When y=1
x = 6
5y+1 = 5+1 = 6
6x-1 is not prime
6*6-1 = 35 is not a prime

While the number didn't pass my test, No, 7 becomes the divisor for the first time when 6*8 + 1 = 49

And all of the prime numbers becomes the MAIN divisor for the first time when 6x+1 is the square of that prime number..

That is my observation at least...

 

[Data]

I'm sorry... I don't understand your posts very well at all. I understand your tests for divisibility by 5 of numbers of the form 6n \pm 1 - those are fine. I was responding precisely to this:

Of course there might be a rule(if you ask me there is not) when these numbers become divisors of 6x + 1 or 6x - 1..If there is one good luck in finding it =)

 

[ExecNight]

For example, 7 is not "added to the series" at 49, this occurs at 35 = 7(5) -1 (mod 6).

Then can you open this one a bit.

 

[shmoe]

And all of the prime numbers becomes the MAIN divisor for the first time when 6x+1 is the square of that prime number..

What exactly is a "main divisor"?

I also have no idea what you are talking about in the rest of your posts. 5 is a divisor of 6(1)-1, so why isn't 5 "added to the list" at this point? Could you perhaps elaborate on what you are trying to say?

 

[ExecNight]

Shmoe you should read my other prime post

To see why 7*5 is not added

MAIN divisor means its lowest divisor for example


77 : 11 is divisor, 7 is divisor but 7 is the MAIN divisor because it is the lowest possible number that can divide 77...

or

287 : 41 is divisor but as you can see above 7 is the MAIN divisor again..

 

[shmoe]

MAIN divisor means its lowest divisor for example

You mean lowest divisor besides 1? You know this isn't a standard term. If you want to use your own terminology and expect people to understand you it's a good idea to provide your definitions.

So it seems that 7 is the main divisor of 7, why wouldn't you add 7 to your list at 7 then? (I had meant 7, not 5 in my last post, whoops) Or does "main divisor" mean "smallest divisor of a number other than itself or one"?

Shmoe you should read my other prime post

Do you mean this one https://www.physicsforums.com/showthread.php?t=38437 ?

 

[ExecNight]

Shmoe i understand why you don't understand..And that is because you are not thinking in the ways of an algorithm.

As you can see your observations has no computational use..And you still show me states which my algorithms first few lines has already eliminated. if you still don't know what i am talking about here it is again then;

y=1
y increases y+1=2 and so on..

IF x = 5y-1 then
6x+1 is not a prime

IF x = 5y+1 then
6x-1 is not a prime

As you can see above, The number 35 was eliminated by the algorithm already..So it primality is out of question already,

Number "1" divides all, so it is absurd to take it into consideration..

And why i don't take 7 at 7 is because it is a prime number, i am talking about non-primes that breaks the 6x+1, 6x-1 rule. Before making comments you should first look what I am talking about...

 

[shmoe]

Shmoe i understand why you don't understand..And that is because you are not thinking in the ways of an algorithm.

As you can see your observations has no computational use..And you still show me states which my algorithms first few lines has already eliminated. if you still don't know what i am talking about here it is again then;

How about you actually write out your algorithm clearly then if you expect people to understand it.

It looks like you are hoping to cross of composite numbers n by checking for prime divisors that are less than sqrt(n) (this is a generous interpretation). If this is the case, why on Earth don't you just say this is the sieve of Erathosthenes (after "pre-sieving" 2 and 3)rather than make up your own terms and presenting things in such an incomplete and obscure way as you have here.

 

[ExecNight]

10 y=1, x=1
20 If 5y-1<x Go to 60
30 If 5y+1<x Go to 70
40 If 5y-1=x Go to 90
50 If 5y+1=x Go to 100
60 Print "6x+1" is a possible prime
70 Print "6x-1" is a possible prime
80 x=x+1 Go to 20
90 Print "6x+1" is not a prime it is divisible by 5. Go to 50
100 Print "6x-1" is not a prime it is divisible by 5
110 y=y+1 Go to 80


I am waiting for your comments on the algorithm Shmoe, Thanks...

 

[shmoe]

It has several problems, but from the looks of it you are just trying to remove multiples of 5 from the sequences 6n+/-1? You've included nothing about adding numbers to the list of divisors, which is where most of the confusion was.

If all you want to do is knock of multiples of 5 from the sequences 6n+/-1 there is a much simpler way to do it. Just use a divisibility test. If 6n+/-1 is divisible by 5, chuck it, otherwise move on.

Alternatively, if you've already generated the sequences 6n+/-1, and they're in order, say like 5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55..., you can remove 25, go 3 down the list, remove 35, go 7 down the list, remove 55, go 3 down the list, etc. Keep alternating between removing the 3rd and the 7th on the list and this will strike off every multiple of 5 (this pattern follows from where multiples of 5 that aren't divisible by 2 or 3 live mod 30).

Or you could instead consider all numbers of the form 30n+/-1, 30n+/-7, 30n+/-11, and 30n+/-13. All primes greater than 5 will be of this form (of course not every number of this form is prime) and none of the numbers of this list will be divisible by 5 (or 3 or 2). (this isn't really much different from the last paragraph though)

 

[ExecNight]

well shmoe if i had found a rule when the prime divisors join, i would have alrady created a perfect formula that always generates a prime number.

 

[shmoe]

well shmoe if i had found a rule when the prime divisors join, i would have alrady created a perfect formula that always generates a prime number.

I guess I'm totally confused as to what you are trying to accomplish. As it stands your algorithm appears to be an attempt to find the numbers of the form 6n+/-1 that are not divisible by 5 (in other words numbers not divisible by 2, 3, or 5). Is this not a correct assessment?

Or are you considering this some kind of incomplete algorithm for finding all primes (up to a certain finite point I'd assume)? If so have you investigated and understood the simple sieve of Erathosthenes? Do you understand how it determines the next number to sieve by once it hits the end of the list?

 

[Rahmuss]

I love this theorem: the Frivolous theorem of Arithmetic.

How true that theorem is.

I was thinking about some stuff and had some questions and thoughts.

1 - The distance between prime numbers seem to be mostly prime numbers themselves (although not always). Does this pattern hold that most are that way? Or maybe the higher the prime number goes, the more likely you'll have a non-prime number between them. Does anyone have a big list of primes with facts about things like this?

2 - Testing numbers by dividing them by primes. I'm sure there are better ways to check for prime numbers than just checking every single number. I figure numbers could move from one group to another to verify further and further whether they're prime or not. Any thoughts?

3 - Why are there prime pairs? What other interesting characteristics are there about primes? What do you mean by divisible by one and itself? Really what does that mean (deeper).

Is infinity a prime number? I know that sounds silly since infinity isn't really a number, and it's not really divisible by itself the same way that "regular" numbers are; but maybe prime numbers can be termed as some abstract quantity just as infinity is. I'm guessing this is pretty dumb; but just throwing thoughts out.

4 - The largest prime is (infinity/2) -1? Again... infinity... oh well. I'm not sure you could prove that there is a limit to prime numbers since you have that big old infinity there; but would there be a point where prime numbers would be in greater abundance as they get bigger? Like a wave function Cos(x). So at first you have a lot of primes... then the number drops rapidly... but could it go back up? Why or why not?

Please ignore my questions if they are too dumb.

 

[shmoe]

1. This is false, the distance between any two odd numbers is even, and the only even prime is two. Essentially the distance between two primes is primes when you have a pair of twin primes. (note this covers the distance between an odd prime and 2 being prime as well).

2. For a start, you only need to check for divisors less than a numbers square root. There's the Sieve of Erathosthenes, one of the simplest yet still effective ways to produce a list of primes. There are much better primality tests than thie naive divisibility test though, Elliptic Curve methods, AKS algorithm, and many many more, including specialized tests for numbers of a certain form (such as Mersenne primes).

3. Shucks, that's a tough one. In some sense the primes a pretty random, so it's not suprising if they sometimes get close to one another. No one knows yet if there are infinitely many of them, so it's not really possible to give a reason why they are there.

Being prime means you cannot be factored in anything but the most trivial way. You can think of primes as the building blocks of all natural numbers. Remember the Fundamental Theorem of Arithmetic?

Infinity is not a natural number, therefore it doesn't stand a chance at being called prime. First thing you have to do to be considered into the club of primeness is become a bonafide natural number, which infinity is really bad at doing.

4. There is no largest prime, there are infinitely many (you should be able to find the usual proof attributed to Euclid without much difficulty). The do become scarcer as you go out. If you were to grab some random number near a large number x, the probability that it's prime is about 1/log(x) (that's the natural logarithm). The prime number theorem says that is the number of primes less than or equal to x is about x/log(x).

However this isn't to say that there aren't quite a few of the little guys running around. You know \sum_{n=1}^{\infty}1/n diverges but \sum_{n=1}^{\infty}1/n^2 converges, so in some sense there are far fewer perfect squares than there are natural numbers. Now \sum_{p\ prime}1/p=1/2+1/3+1/5+1/7+1/11+\ldots diverges (the sum is over all primes), so there are still a considerable number in some sense.

 

[Rahmuss]

1. This is false, the distance between any two odd numbers is even, and the only even prime is two.

Oops! You're right... What I meant was the amount of numbers between them. i.e. between 7 and 11 there are three numbers (8, 9 and 10). Sorry about that.

2. For a start, you only need to check for divisors less than a numbers square root. There's the Sieve of Erathosthenes, one of the simplest yet still effective ways to produce a list of primes. There are much better primality tests than thie naive divisibility test though, Elliptic Curve methods, AKS algorithm, and many many more, including specialized tests for numbers of a certain form (such as Mersenne primes).

Wow! I had no idea. Could you use all of these tests? For example, use one test and all numbers that pass use the second test on them and so forth and see if there is a pattern or a quicker way to find primes? I'll have to look at some of those. Any discussions on PF for those?

3. Shucks, that's a tough one. In some sense the primes a pretty random, so it's not suprising if they sometimes get close to one another. No one knows yet if there are infinitely many of them, so it's not really possible to give a reason why they are there.

So this is more coincidental and can't really be reasoned mathematically?

Being prime means you cannot be factored in anything but the most trivial way. You can think of primes as the building blocks of all natural numbers. Remember the Fundamental Theorem of Arithmetic?

Primes are the building blocks in that they are more fundamental since they are only divisible by one and that number? What's the next step up? Having numbers divisible by 1, 2 and that number? What would those be called? Next divisible by 1, 2, 3 and that number? I guess in that sense there should be numbers that are more rare than prime numbers that would be more fundamental than prime numbers then. What would those be? What about numbers divided by fractions? I'm just rambling here, sorry.

Infinity is not a natural number, therefore it doesn't stand a chance at being called prime. First thing you have to do to be considered into the club of primeness is become a bonafide natural number, which infinity is really bad at doing.

LOL... yeah... sounds like infinity won't fit the picture very well. So in that sense you can't have an infinite amount of prime numbers since infinite is not a concept that would match a prime number sequence?

4. There is no largest prime...

I was just joking about that.

...there are infinitely many (you should be able to find the usual proof attributed to Euclid without much difficulty). The do become scarcer as you go out. If you were to grab some random number near a large number x, the probability that it's prime is about 1/log(x) (that's the natural logarithm). The prime number theorem says that is the number of primes less than or equal to x is about x/log(x).

Ok, that makes sense, even to someone as limited as I am in my knowledge of math and science.

However this isn't to say that there aren't quite a few of the little guys running around. You know \sum_{n=1}^{\infty}1/n diverges but \sum_{n=1}^{\infty}1/n^2 converges, so in some sense there are far fewer perfect squares than there are natural numbers. Now \sum_{p\ prime}1/p=1/2+1/3+1/5+1/7+1/11+\ldots diverges (the sum is over all primes), so there are still a considerable number in some sense.

Sum over all primes diverges... that sounds odd to me... would that mean that there are NOT an infinite amount of primes?

 

[shmoe]

1. Ahh, I see. This could be a function of looking at only small numbers where the distances between two primes is even smaller, and an unnaturally large percentage of "small" odd numbers are prime. Up to the first 100 or so primes the largest gap is 17 numbers long. If all these gaps were somehow randomly chosen odd numbers from 1 to 17, you'd expect about 6/9 of them to be prime, which turns out to be close to the truth.

2. If a test declares a number prime, why would you want to apply another primality test to it? There are some "probabilistic" primality tests where you do repeated applications, but I don't think that's what you meant. Google "primality testing", you'll probably find heaps of info.

3. It's probably not coincidental. There are convincing heuristic arguments (read-arguments with gaps in them) that would indicate there are infinitely many prime pairs. It is also known that there are infinitely many pairs p, p+2 where one is prime and the other has at most 2 prime factors, that's coming tantalizingly close to a solution (yet so far out reach still).

Next best thing to primes would be a low number of prime factors. In some ways they can be hard to separate (see paragraph above). Prime powers are also pretty decent too.

not sure what you mean about "infinite is not a concept that would match a prime number sequence", since you seem to agree there is no largest prime and therefore infinitely many of them.

4. Quite the opposite about the sum of recipricals of the primes diverging, showing this sum diverges is one way to prove there are infinitely many of them. If there were a finite amount, then this sum would have a finite value. However by adding up enough terms, you can make this sum as large as you like. It turns out if you add up 1/p for all primes less than say x, you'll get about log(log(x)), which although going to infinity very slowly, it still goes.

 

[Rahmuss]

1. Ahh, I see. This could be a function of looking at only small numbers where the distances between two primes is even smaller, and an unnaturally large percentage of "small" odd numbers are prime. Up to the first 100 or so primes the largest gap is 17 numbers long. If all these gaps were somehow randomly chosen odd numbers from 1 to 17, you'd expect about 6/9 of them to be prime, which turns out to be close to the truth.

I ran my program from about 1,000,973 to 1,002,361 and came up with these results:

7, 17, 3, 13, 5, 3, 13, 27, 11, 5, 1, 3, 13, 15, 29, 5, 13, 3, 13, 5, 21, 17, 29, 11, 11, 11, 7, 9, 1, 3, 19, 5, 15, 11, 5, 1, 11, 9, 19, 15, 11, 7, 23, 9, 25, 3, 17, 1, 11, 5, 17, 5, 27, 7, 9, 19, 9, 13, 3, 25, 9, 19, 39, 13, 3, 5, 1, 11, 9, 7, 71, 21, 7, 5, 5, 23, 3, 1, 5, 27, 31, 11, 11, 3, 5, 7, 9, 7, 11, 21, 5, 1, 21, 17, 35, 13, 5, 9, 1, 3, 25, 9, 41, 1, 3, 1, 9, 1, 15

The ratio here is presicely 27/109 (or about 1/4). Less than 2/3; but fairly close. Interesting that the ones that most come up which are not prime numbers are divisible by prime numbers... some big endless loop I'm sure, since you take those (ie. 9, 15, 21, 27) and we find that they are divisible by three and some other prime number... and it continues... I don't dare push my program too far. It already takes almost 0.10 seconds to calculate terms over 5,000,000. I did it in C++ if anyone wants the code (it's pretty small).

EDIT: Another quick note. The amount of numbers listed here always seems to be divisible by a prime number. Do they take that into account when they attempt to find a prime?

ADDITIONAL EDIT: They are divisible by any prime that I can calculate, except for the number 2 (of course), so I would submit that 2 is not in the same category... not that it couldn't still be a prime; but it's not the same type.

2. If a test declares a number prime, why would you want to apply another primality test to it? There are some "probabilistic" primality tests where you do repeated applications, but I don't think that's what you meant. Google "primality testing", you'll probably find heaps of info.

I did mean "probabilistic" there. You're right, it wouldn't make much sense using another prime number test on a prime number.

3. It's probably not coincidental. There are convincing heuristic arguments (read-arguments with gaps in them) that would indicate there are infinitely many prime pairs. It is also known that there are infinitely many pairs p, p+2 where one is prime and the other has at most 2 prime factors, that's coming tantalizingly close to a solution (yet so far out reach still).

Wow! Sounds like they're on the right track. I'm not great with number crunching. I just do the thinking.

Next best thing to primes would be a low number of prime factors. In some ways they can be hard to separate (see paragraph above). Prime powers are also pretty decent too.

Interesting. I wouldn't have thought of prime powers.

not sure what you mean about "infinite is not a concept that would match a prime number sequence", since you seem to agree there is no largest prime and therefore infinitely many of them.

I'm not sure either... I guess what I was getting at was that since infinity isn't really a defined quantity; but a concept, or limit, or... help me out here... anyway... since it's not a "natural number" (I think that's what you've called it), then saying that numbers (which happen to be "natural numbers") have an infinite limit (limit that is not "natural") seems to conflict. Though I'm sure it doesn't but it sounds like it.

4. Quite the opposite about the sum of recipricals of the primes diverging, showing this sum diverges is one way to prove there are infinitely many of them. If there were a finite amount, then this sum would have a finite value. However by adding up enough terms, you can make this sum as large as you like. It turns out if you add up 1/p for all primes less than say x, you'll get about log(log(x)), which although going to infinity very slowly, it still goes.

Interesting as well... I must have missed something in my math classes. I, for some reason, am not aware of the relationship between primes and log(x); but many people seem to have mentioned it. What is that relationship exactly?

 

[shmoe]

1. If a number isn't prime then it's divisible by a prime. Notice how small the gaps are (in your range 39 or less). The prime number theorem will tell us that the average gap sioze for primes less than x will be about log(x) (as there are x/log(x) primes). So if your looking at a bunch of "randomish" odd numbers that aren't much larger than log(x), you'd expect "many" primes. Furthurmore, the ones that aren't prime will have a prime factor no greater than their square root, so it's no surprise you see 3 often there. The only composite number less than 39 without 3 as a factor is 25 (which did occur in your range)

3. There's no problem with saying there are infinitely many primes. You accept that there are infinitely many integers? It's just a statement about cardinality of sets.

4. See the prime number theorem. The log there is really the cause of so many other logs. I'll have to think about a simple way of explaining why it makes an appearance there.