New prime series?

1. Nov 15, 2003

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

in analogy to phi of Fibonacci numbers?

2. Nov 16, 2003

Hurkyl

Staff Emeritus
This function is called the primorial function.

3. Nov 16, 2003

Loren Booda

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

4. Nov 16, 2003

Hurkyl

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

Anyways, your fraction can be rewritten as:

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

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.

Last edited: Nov 16, 2003
5. Nov 16, 2003

Hurkyl

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

(PS you got some groupings wrong)

Last edited: Nov 16, 2003
6. Nov 16, 2003

chroot

Staff Emeritus
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

7. Nov 16, 2003

Hurkyl

Staff Emeritus
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!

8. Nov 16, 2003

Hurkyl

Staff Emeritus
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

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

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:

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

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.

Last edited: Nov 16, 2003
9. Nov 16, 2003

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

Last edited by a moderator: Apr 20, 2017
10. Nov 16, 2003

Hurkyl

Staff Emeritus
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.

11. Nov 25, 2003

QuantumNet

Sorry.

But is this right then?

2p - 1 = p

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

Am I wrong?

Last edited by a moderator: Dec 2, 2003
12. Dec 8, 2003

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

Last edited: Dec 8, 2003
13. Dec 8, 2003

HallsofIvy

Staff Emeritus
?? Yes, "2<sup>p</sup>-1= p" is wrong. In fact, the example you give show that :
4-1= 2<sup>2</sup>-1= 2 NOT 2.
8-1= 2<sup>3</sup>-1= 7 NOT 3, etc.

What exactly did you intend to say?

14. Dec 8, 2003

suyver

I think he means to suggest that for every p, 2p-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).

15. Dec 8, 2003

Sariaht

16. Dec 8, 2003

Loren Booda

I had never seen that before

17. Dec 9, 2003

suyver

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.

18. Dec 9, 2003

Sariaht

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

All numbers within the serie:

In this serie, pn 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 eachother becomes the degree of conjugative.

I can prove that this equation is new:

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

It's to good to be old!

Last edited: Dec 9, 2003
19. Dec 9, 2003

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...

20. Dec 9, 2003

Sariaht

I'm sorry

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?

Last edited: Dec 9, 2003