What way would you find numbers with a set number of factors?
If prime numbers are numbers that only factor into 1 and themselves, then what would 3 factors including one and itself be?
[spoiler] It turns out that the middle factor must follow a rule: 1*n^2=n*n=x The factors of n^2 must be 1, n, n^2. Since n has to be prime, that means that the answer is the set of all the squares of primes. [/spoiler] Try this for 4, 5, 6, ... factors for x including 1 and x. 
Let Pi be distinct primes for all i:
4 factors: [spoiler](P1)^3 or (P1)*(P2)[/spoiler] 5 factors: [spoiler](P1)^4[/spoiler] 6 factors: [spoiler](P1)^5 or (P1)^2*(P2)[/spoiler] 7 factors: [spoiler](P1)^6[/spoiler] 8 factors: [spoiler](P1)^7 or (P1)^3*(P2) or (P1)*(P2)*(P3)[/spoiler] so on, so forth... 
At 1 and 0 you get the identies 1 and 0. 0 can't be divided at all and 1 has only one possible factor including itself and itself. Hehe we'll just say "period" to make more sense.
At 2 total factors there's only primes. (works both ways: A>B and B>A) The solutions for 2 factors are a kind of 'key' to the higherfactorcount sets in the 3D tree since obviously primes are the simplest factors possible. If there's an infinite number of primes, is there an infinite number of the 3factor results? 4factor? All? [spoiler] See 2nd post to see why there must be. Yes, there are infinite primes but no telling how long you'll have to wait to find the next one! I left this in the spoiler so those who want to can do the work themselves to find and understand the proof.[/spoiler] 
This is very closely related to something I was thinking about this afternoon.
As there are an infinite number of primes, and as all primes are either 1(mod 6) or 5(mod 6), are there an infinite number of primes 1(mod 6)? 
Yes, this is a special case of the "prime number theorem for arithmetic progressions." Simply put, it says that if you have an arithmetic progression a+b*x with gcd(a,b)=1 and x∈[B]N[/B], you get infinitely many primes. What is more, each such progression for different values of a (but the same b) gets an "equal share" of the primes. See Crandall and Pomerance, Prime Numbers, Theorem 1.1.5.
Alex 
Thank you.

Amazon has one copy of the 2nd edition if you've got $70 (new hard cover math books aren't cheap!) to expand your library. Since I'm near several libraries, colleges and universities, I think I'll be cheap and just go spend some time at a desk in one of them. This should be as interesting as the books by Howard Anton covering some of the more interesting parts of vectors and matrices  assuming you're like me and read that kind of stuff for 'fun'.
ISBN: 0387252827 (there's an older first edition #0387947779 for a little less) 
Some things have been added/changed in the second edition, most notably it now includes the AKS algorithm. But the first edition is still perfectly worthwhile to have and you may be able to get a second hand copy inexpensively now. Maybe check university .market newsgroups or online auctions?
Alex 
[quote=nibble4bits]assuming you're like me and read that kind of stuff for 'fun'.[/quote]
I do indeed. Last night I read a chapter of "The Art of Calculus" wrapped up in bed with a big mug of cocoa. I spent this morning working through the exercises. This afternoon I did some work on my maths course, and am now relaxing with a chapter of William LeVeque's "Elementary Theory of Numbers". I don't know that it's fun, exactly, but I do find it very rewarding. Crandall & Pomerance, and of course Knuth were on my wish list for Christmas (again), but sadly Santa saw fit to leave me a mouse and some aftershave. Maybe if I manage to sell another picture before Easter then Amazon will be getting a call. 
[QUOTE=Numbers]Crandall & Pomerance, and of course Knuth were on my wish list for Christmas (again), but sadly Santa saw fit to leave me a mouse and some aftershave.[/QUOTE]You could always feed the mouse to the cat and drink the aftershave...share the joy, I say.

...
Poor cat trying to eat that plastic peripheral. 
Actually, the aftershave came from my girlfriends mother. Rather than buy me the brand I have used for the past twenty years she got something quite nasty (perhaps she was trying to tell me something). So I would probably be better off feeding the after shave to the cat so that it wouldn't notice that the mouse has a flashing light in its bottom!

So have you gotten to the point where you know how and why e and Pi (3.14..., not Pi[x]) are relevant to primes?
The constant e is the amount of interest+principle you'de get for compounded constantly vs flat/single cycled 100% interest. Another way to put that is that it's what happens when feedback in a system is so randomized in it's order and timing and occours so often, that it appears to equivilent to constantly compounded interest. P_n+1 is most certainly based on P_n, P_n1... It's [U]forbidden[/U] from having those as factors. This is why it at least seems that Pi(x) is almost the same as n/(a + log n) especially for larger numbers. Yes, I know that n/(1 + log n) is not the best estimate. :) Pi the constant is related to the trigonometry used in to find those 1/2 Riemann zeros and run some multiplication methods. However, that's probally a pretty long stretch. I was interested the talk about "how the primes are unusual in that they appear to have added noise" which seems to imply that they'de otherwise be easy to generate for a function like Prime(x)=the x'th prime that works in linear time relative to x. Anyone have any good links for more details? I'm thinking that the primes are not the only sequence to have this behaviour. What if I defined a set that instead of x not dividing by the set of 2 to x1, required some other kind of condition for a number to be next in the series? Something like x/(2 to x1)=2 for some value of the denominator would be a start  at the very least, it'll give me an excuse to burn CPU cycles. heh 
I'm afraid I found this whole post very confusing.
[quote=nibble4bits]P_n+1 is most certainly based on P_n, P_n1... It's forbidden from having those as factors.[/quote] Bearing in mind that there is no break between your "explanation" of random feedback in a system being related to e, the base of natural logarithms and this, which appears to be saying that of three consecutive integers the third cannot have either of the other two as factors (which is not true) and which you seem to think is connected to pi(x) the prime counting function, is there any chance you could squeeze in some mention of the probability of rolling dice as well? I have no idea what you are trying to say here. [quote=nibble4bits]I was interested the talk about "how the primes are unusual in that they appear to have added noise"[/quote] You include a phrase in inverted commas as though it were a quote, but I have no idea who or what you are quoting or how this is connected to anything that has gone before. And what do you mean by "random noise"? Sorry, I'm just confused. Then you suggest [quote=nibble4bits]some other kind of condition for a number to be next in the series? Something like x/(2 to x1)=2[/quote] Well I'm afraid this function [B]never[/b] = 2. The maximum value of this function occurs when x = 1.4426947, y = 1.0614757 so once again I have no idea what you are talking about. Were you smoking something when you typed this? :o) 
1) It's hard to use this font to represent prime number 1(2=), 2(=3), 3(=5) and so on. Of course the very definition of prime implies that lower primes may not be factors of a higher prime. Very basic stuff but P_(N+1) is not P_N + 1. I'm sorry about that!
2) There's an exact quote: "God may not play dice with the universe, but something strange is going on with the prime numbers."  Paul Erdos His other famous quote is about mathematicians and coffee. :) 3) Yeah I know there's no solution but I find it hard to believe the set of primes is ever remotely unique. If A*B=Prime then either A=1 and B=the prime number, or the other way around. I'm wondering what kinds of solutions you get when C*D+E=Prime. C and D being two factors that are a constant E away from a prime. It seems that any such conditions are always based on the prime numbers. Are there counterexamples that don't rely on primes? I think any set based on division is going to require some mention of primes. 4) Nah, I'm just trying to figure this out and it helps if I can see how far away I am from understanding specific points. x/(2)...x/(x1) < I think that's much clearer than x/(2 to x1). 
[quote=nibble4bits]There's an exact quote: "God may not play dice with the universe, but something strange is going on with the prime numbers."  Paul Erdos
His other famous quote is about mathematicians and coffee. :)[/quote] A mathematician is a machine for turning coffee into theorems. I had a banner printed with this on it to stick on the wall above my computer. My girlfriend got a guy she knows to engrave it into a nice piece of slate and mount it onto a polished mahogany base. She gave it to me for christmas and it now sits on my desk and the tatty paper banner has come down. [quote=nibble4bits]I'm wondering what kinds of solutions you get when C*D+E=Prime. C and D being two factors that are a constant E away from a prime. It seems that any such conditions are always based on the prime numbers.[/quote] Since you have defined your variables C, D and E in terms of the prime numbers, it is hardly surprising that your solutions are found in terms of the prime numbers. It's a bit like wondering why you get change in dollars when you gave the girl in the shop dollars in the first place. If you want to view your solutions in terms of something other than prime numbers then you need to define your variables appropriately. That is where such simple identities as p = 6q+r come from. They allow us to define our variables in terms of something other than p, and yet afterwards convert q and r into something that relates back to the original problem. [quote=nibble4bits]x/(2)...x/(x1) < I think that's much clearer than x/(2 to x1).[/quote] That is certainly clearer. I thought you meant [tex]\Large \frac{x}{2^{x1}}[/tex] Enjoy. 
Well that function is interesting too for graphing... but not what I meant. Yeah I didn't mean x/(2^(x1))=(2x)/(2^x)
BTW How do I make that algebraic font? 
[quote=nibble4bits]BTW How do I make that algebraic font?[/quote]
If you see anything on the forum and ask yourself "how did he do that?" click on the "Quote" button on the post and you can read how it was done. Maths symbols are posted using LaTex, and you can find a comprehensive set of maths symbols, functions, operators etc. [url=http://www.mersennewiki.org/index.php/Help:Math_Formulas]here[/url]. Just remember that the page you are reading was written for the wiki, so you need to enclose your maths in the tex tags [ tex ][I]maths in here[/I][ /tex ] rather than in the <math> </math> tags given on that page. Googling for LaTex will also give hundreds (at least) of hits, the one I use the most is [url=http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/]St. Andrews University[/url] but there are many others. HTH. 
[QUOTE=Numbers]Enjoy.[/QUOTE]
[tex]Oh^{my}_{goodness}![/tex] 
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