# New prime series?

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?

Hurkyl
Staff Emeritus
Gold Member
This function is called the primorial function.

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

Hurkyl
Staff Emeritus
Gold Member
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.

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Hurkyl
Staff Emeritus
Gold Member
I don't know how I managed not to read the big thread on TeX. (PS you got some groupings wrong)

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chroot
Staff Emeritus
Gold Member
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
Staff Emeritus
Gold Member
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
Staff Emeritus
Gold Member
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.

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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

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Hurkyl
Staff Emeritus
Gold Member
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?

2p - 1 = p

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

Am I wrong?

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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

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HallsofIvy
Homework Helper
But is this right then?

2<sup>p</sup> - 1 = p

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

Am I wrong?

?? 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?

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).

Originally posted by suyver

That's what I, eh, he ment.

Best wishes Erik-Olof Wallman!

I had never seen that before

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.

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!

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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...

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?

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

That's not true, though.

Hurkyl
Staff Emeritus
Gold Member
1 is not a prime number.

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

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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...

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...