New prime series?

  • #1
3,077
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Hurkyl
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This function is called the primorial function.
 
  • #3
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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...pn-1)/(2+3+5+7+11+13...pn)=?
n-->[oo]
 
  • #4
Hurkyl
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I'm not sure why you call it a Fibonacci analog...

Anyways, your fraction can be rewritten as:

[tex]
\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} }
[/tex]

So solving your limit reduces to finding

[tex]
\lim_{n\rightarrow\infty}\frac{2 + 3 + \dots + p_{n-1}}{p_n}
[/tex]

No proof of any value for this leaps to mind, however.
 
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  • #5
Hurkyl
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I don't know how I managed not to read the big thread on TeX. :frown:

(PS you got some groupings wrong)
 
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  • #6
chroot
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Originally posted by Hurkyl
I don't know how I managed not to read the big thread on TeX. :frown:

(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
 
  • #7
Hurkyl
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I need to learn LaTeX eventually anyways, might as do it here where I can get some practical benefit out of it. :smile: 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
Hurkyl
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All right, here goes.

Define

[tex]
S(n) := \frac{2 + 3 + ... + q_n}{n}
[/tex]

where [tex]q_n[/tex] is the largest prime less than [tex]n[/tex]. I aim to prove:

[tex]
\lim_{n\rightarrow\infty} S(n) = \infty
[/tex]

From which we can deduce

[tex]
\lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{p_n}
= \infty
[/tex]

and thus

[tex]
\lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{2 + 3 + \dots + p_n}} = 1
[/tex]


It's clear that

[tex]
\lim_{n\rightarrow\infty} S(n) = \lim_{\substack{ n\rightarrow\infty \\ n~{\it even}} } S(n)
[/tex]

So I will restrict my attention to the case where [tex]n[/tex] is even.


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

[tex]
\frac{7}{8} < \frac{ \pi(n) }{ \frac{n}{ln~n} } < \frac{9}{8}
[/tex]

So here goes:

[tex]
\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*}
[/tex]

A bit of elementary calculus proves that

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

And we're done.
 
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  • #9
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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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  • #10
Hurkyl
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It's the LaTeX. It makes one look smarter. :smile:

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
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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  • #12
356
0
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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  • #13
HallsofIvy
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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?
 
  • #14
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0
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
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0
Originally posted by suyver

That's what I, eh, he ment.

Best wishes Erik-Olof Wallman!
 
  • #16
3,077
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I had never seen that before
 
  • #17
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0
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.
 
  • #18
356
0
p1 = 1
p2 = 2
p3 = 3
p4 = 5
p5 = 7
p6 = 11

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

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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  • #19
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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...
 
  • #20
356
0
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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  • #21
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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.
 
  • #22
Hurkyl
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1 is not a prime number.
 
  • #23
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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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  • #24
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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...
 
  • #25
356
0
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?
 

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