Is the Product of Primes Significant in Number Theory?

[Loren Booda]

Consider all primes

2, 3, 5, 7, 11, 13...

and their products such that

2x3=6, 2x3x5=30, 2x3x5x7=210, 2x3x5x7x11=2310, 2x3x5x7x11x13=30030...

Is this latter series used in number theory?


Likewise, can one determine

lim (2+3+5+7+11+13...p_n-1)/(2+3+5+7+11+13...p_n)
n-->_[oo]

in analogy to phi of Fibonacci numbers?

 

[Hurkyl]

This function is called the primorial function.

 

[Loren Booda]

Thanks for the helpful hyperlink, Hurkyl.

Do you or anyone else have a hint about the second series I mentioned, the prime Fibonacci analog, and its limit:

lim (2+3+5+7+11+13...p_n-1)/(2+3+5+7+11+13...p_n)=?[/size]
n-->_[oo]

 

[Hurkyl]

I'm not sure why you call it a Fibonacci analog...

Anyways, your fraction can be rewritten as:

$$\frac{2 + 3 + \dots + p_{n-1}}{2 + 3 + \dots + p_n}<br /> = 1 - \frac{p_n}{2 + 3 + \dots + p_n}<br /> = 1 - \frac{1}{1 + \frac{2 + 3 + \dots + p_{n-1}}{p_n} }$$

So solving your limit reduces to finding

$$\lim_{n\rightarrow\infty}\frac{2 + 3 + \dots + p_{n-1}}{p_n}$$

No proof of any value for this leaps to mind, however.

 

[Hurkyl]

I don't know how I managed not to read the big thread on TeX.

(PS you got some groupings wrong)

 

[chroot]

Originally posted by Hurkyl
I don't know how I managed not to read the big thread on TeX.

(PS you got some groupings wrong)

Whoops! Feel free to edit my posts to reflect the correct TeX, or delete them altogether if you wish.

Also, please note that I don't intend to coerce people into using TeX if they are already comfy and happy with basic HTML.

- Warren

 

[Hurkyl]

I need to learn LaTeX eventually anyways, might as do it here where I can get some practical benefit out of it. The only LaTeX I've written thus far was for writing a tutorial to use some code I had written, so I haven't gotten to play with any of the math stuff!

 

[Hurkyl]

All right, here goes.

Define

$$S(n) := \frac{2 + 3 + ... + q_n}{n}$$

where $$q_n$$ is the largest prime less than $$n$$. I aim to prove:

$$\lim_{n\rightarrow\infty} S(n) = \infty$$

From which we can deduce

$$\lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{p_n}<br /> = \infty$$

and thus

$$\lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{2 + 3 + \dots + p_n}} = 1$$


It's clear that

$$\lim_{n\rightarrow\infty} S(n) = \lim_{\substack{ n\rightarrow\infty \\ n~{\it even}} } S(n)$$

So I will restrict my attention to the case where $$n$$ is even.


The general approach is to estimate the numerator of $$S(2n)$$ by just looking at the primes in the range $$[n, 2n)$$, and underestimating $$S(2n)$$ by $$n$$ times the number of primes in this range. To do this, I will use Chebyshev's bound on the prime counting function:

$$\frac{7}{8} < \frac{ \pi(n) }{ \frac{n}{ln~n} } < \frac{9}{8}$$

So here goes:

$$\begin{equation*}<br /> \begin{split}<br /> S(2n) &= \frac{2 + 3 + \dots + q_{2n}}{2n} > \frac{1}{2n} (\pi(2n) - \pi(n)) n \\<br /> &> \frac{1}{2} \left (\frac{7}{8} \frac{2n}{ln~2n} - \frac{9}{8} \frac{n}{ln~n} \right)<br /> \end{split}<br /> \end{equation*}$$

A bit of elementary calculus proves that

$$\lim_{n\rightarrow\infty} \frac{1}{2} \left( \frac{7}{8} \frac{2n}{ln~2n} - \frac{9}{8} \frac{n}{ln~n} \right) = \infty$$

And we're done.

 

[Loren Booda]

Hurkyl,

What experience do you have in math? You seem the most competant of a talented bunch here at PF. I hope you have seen the http://www.quantumdream.net at my website, my greatest accomplishment in mathematics.

LB

 

[Hurkyl]

It's the LaTeX. It makes one look smarter.

I'm relatively fresh out of school, actually. I got my BS's in math and computer science two years ago, and started work this January. I've been hired as a mathematician, but my work thus far has been leaning more towards the programming.

However, math has been my hobby since I was little, so I have experienced a lot more than these credentials would suggest.

 

[QuantumNet]

Sorry.

But is this right then?

2^p - 1 = p

2 - 1
4 - 1
8 - 1
32 - 1
128 - 1
2048 - 1


Am I wrong?

 

[Sariaht]

p*p

A multiple between two primes is always right in the middle of two primes.

1, 2*1, 3, 2*2, 5, 2*3, 7, /, 3*3, 2*5, 11, /, 13 etc.

3*3 is right in the middle of 7 and 11
2*5 is right in the middle of 7 and 13
etc.

(11*3 is right in the middle of 37 and 29)


Erik-Olof Wallman

 

[HallsofIvy]

But is this right then?

2^p - 1 = p

2 - 1
4 - 1
8 - 1
32 - 1
128 - 1
2048 - 1


Am I wrong?

?? Yes, "2^p-1= p" is wrong. In fact, the example you give show that :
4-1= 2²-1= 2 NOT 2.
8-1= 2³-1= 7 NOT 3, etc.

What exactly did you intend to say?

 

[suyver]

I think he means to suggest that for every p, 2^p-1 is a prime number? These are the so-called Mersenne-numbers and not all of them are prime. Although the largest prime numbers found to date are typically Mersenne-numbers, not every p generates a prime number. Simplest example: 2¹¹-1 is composite (23*89).

 

[Sariaht]

Originally posted by suyver

That's what I, eh, he ment.

Best wishes Erik-Olof Wallman!

 

[Loren Booda]

I had never seen that before

 

[suyver]

Originally posted by Loren Booda
I had never seen that before

The Mersenne-numbers are very interesting because of the so-called Lucas-Lehmer Test, which is a (relatively) easy method of deciding if any Mersenne-number is prime or composite.

 

[Sariaht]

p₁ = 1
p₂ = 2
p₃ = 3
p₄ = 5
p₅ = 7
p₆ = 11

All numbers within the serie:
p₁*p₂*p₃*p₄*...*p_n +/- 1 are primenumbers.

In this serie, p_n can be raised to all primenumbers between 1 and p; the serie will still be giving primes.

All primes, that in the serie are raised to zero, multiplied with each other becomes the degree of conjugative.

I can prove that this equation is new:

p₁*p₂*p₃*p₄*...*p_n = ( can't we call it p?, when n! = 1*2*3*...*n ? ).

It's to good to be old!

 

[suyver]

Two observations:

1) The smallest prime number is 2, not 1.

2) p1*p2*p3*p4*p5*p6*p7 = 2*3*5*7*11*13*17 = 510510
510510 - 1 = 510509 = 8369 * 61

Sorry, but if things were so simple...

 

[Sariaht]

I'm sorry

Originally posted by suyver
Sorry, but if things were so simple... [/B]

I'm sorry. Thanks anyway. but... wait.. 510 510 is a double query.

Maybe it don't work on them? THANX!

by the way, 510 is 11111110 binary.

Maybe both sides don't have to be primes either? Or?

 

[Sariaht]

The smallest primenumber is one.

Originally posted by suyver
The smallest prime number is 2, not 1.

That's not true, though.

1 is the smallest primenumber.

 

[Hurkyl]

1 is not a prime number.

 

[Sariaht]

Originally posted by Hurkyl
1 is not a prime number.

förlåt mig. I Sverige säger vi:

Primtal är alla tal som bara är delbara med 1 och sig självt.

sorry. In Sweden we say:

Primes are all numbers that you only can divide with 1 and itself.

1/1 = 1

1/1 = 1

 

[suyver]

Originally posted by Sariaht
In Sweden we say:

Primes are all number that you only can divide with 1 and itself.

This may be the case, but that is not the normally accepted definition of a prime number. Normally the smallest prime number is said to be 2...

 

[Sariaht]

Originally posted by suyver
This may be the case, but that is not the normally accepted definition of a prime number. Normally the smallest prime number is said to be 2...

What about the equation?

What is your definition of a prime?

 

[suyver]

Originally posted by Sariaht
What about the equation?

That makes no difference of course: 1*x = x for all x. So including your p1=1 doesn't change the fact that your equation is incorrect...

Originally posted by Sariaht
What is your definition of a prime?

The commonly accepted definition for a prime number is any number having no factor except itself and one. From this rule it follows that 1 is not a prime number, but 2 is.

 

[Sariaht]

2*3 = 6 | +/- 1 | 5, 7

2*3*5 = 30 | +/- 1 | 29, 31

2*2*3 = 12 | +/- 1 | 11, 13

2*3*3 = 18 | +/- 1 | 17, 19

2*3*5*5 = 150 | +/- 1 | 149, 151

2*3*5*7*7 = 1470 | +/- 1 | 1469, 1471

allright.

Only one primesquare is aloud.

3*5*7*7 = 735 | +/- 2 | 733, 737
3*5*5*7 = 525 | +/- 2 | 523, 527
3*3*5*7 = 315 | +/- 2 | 313, 317
5*7*11 = 385 | +/- 2*3 | 379, 391
7*7 = 49 | +/- 30 | 19, 79

etc?

 

[suyver]

Originally posted by Sariaht
2*3 = 6 5 7

2*3*5 = 30 29 31

2*2*3 = 12 11 13

2*3*3 = 18 17 19

2*3*5*5 = 150 149 151

2*3*5*7*7 = 1470 1469 1471

allright.

Only one primesquare is aloud.

You should really learn to write your ideas (or whatever they are) more clearly. I have NO IDEA what you are trying to say...

 

[Sariaht]

Originally posted by suyver
You should really learn to write your ideas (or whatever they are) more clearly. I have NO IDEA what you are trying to say...

Were does my equation error?

 

[suyver]

The prime numbers have a RANDOM distribution over the natural numbers. You will not succeed in finding such an (easy or not) algorithm to always generate a new prime number from a set of already known ones. However, the numbers tend to become large, making it difficult to see that they are not prime.

If you find a simple algorithm and PROVE that this algorithm works for at least the first 100 prime numbers, then I will look at it again. But like this it's becomming a waste of my time: you're just guessing new algorithms without any idea about why they should work in the first place...