Harmonics and distribution of prime numbers

[hprime]

Hello all

 

I'm new to this forum, I don't know if this topic or related ones have been discussed somewhere else, but a quick search did not bring me any relevant result.

I'm investigating on geometric progressions with initial value 1 and ratio p/(p-1), where p is a prime number. For two different values of p, it's trivial that the two progressions will have no common term (apart of the initial 1). For example powers of 2, 3/2, 5/4 ... are all distinct.

 

My question is : given two such progressions, can they contain terms arbitrarily close to each other? Put in more formally:

 

Let u_n(p)=(p/(p-1))ⁿ, where p is a prime integer and n a positive integer (n >1)

 

If p₁ and p₂ are distinct primes, is 0 the lower bound of |u_n(p₁) - u_k(p₂)| ?

 

I would be happy to have the answer even for the first prime numbers p₁ = 2 and p₂ = 3, that is : can one find a power of 3/2 arbitrarily close to a power of 2?

My conjecture is that the lower bound is strictly greater than 0, and for the case of 2 and 3, the lower bound might be 0.25 = (3/2)² - 2

 

Any related works that could point to a solution - or related open issues, there are many on prime numbers

 

EDIT : "Harmonics" in the title might be not obvious. The original question comes from how we cope in (musical) harmony with the fact that 2⁷ differs from (3/2)¹², but not too much, so that according instruments is possible at all, dealing with the wolf interval in various ways or temperaments. (7/6)⁹ and 2² have less difference than the above pair, but harmonics based on the division of a string in seven parts are just ignored in occidental music. It led me to the general question of how close such harmonics can get ...

Edited May 16, 2014 by hprime

[Unity+]

Hello all

 

I'm new to this forum, I don't know if this topic or related ones have been discussed somewhere else, but a quick search did not bring me any relevant result.

I'm investigating on geometric progressions with initial value 1 and ratio p/(p-1), where p is a prime number. For two different values of p, it's trivial that the two progressions will have no common term (apart of the initial 1). For example powers of 2, 3/2, 5/4 ... are all distinct.

 

My question is : given two such progressions, can they contain terms arbitrarily close to each other? Put in more formally:

 

Let u_n(p)=(p/(p-1))ⁿ, where p is a prime integer and n a positive integer (n >1)

 

If p₁ and p₂ are distinct primes, is 0 the lower bound of |u_n(p₁) - u_k(p₂)| ?

 

I would be happy to have the answer even for the first prime numbers p₁ = 2 and p₂ = 3, that is : can one find a power of 3/2 arbitrarily close to a power of 2?

My conjecture is that the lower bound is strictly greater than 0, and for the case of 2 and 3, the lower bound might be 0.25 = (3/2)² - 2

 

Any related works that could point to a solution - or related open issues, there are many on prime numbers

Ah, you have an unsolved problem that actually may be added to the list.

 

In order to solved this problem, we would need to find if there are infinitely many twin primes because using calculus we can determine that:

 

[math]u_{n}\left ( p_{1} \right )=\lim_{p_{1} \to \infty }\frac{p_{1}}{p_{1}-1}=1[/math]

 

[math]u_{n}\left ( p_{2} \right )=\lim_{p_{2} \to \infty }\frac{p_{2}}{p_{2}-1}=1[/math]

 

In order for this to work, both have to be only 1 difference apart. In order to determine this to be true, we would have to prove that there infinitely many twin primes with a difference of 1.

 

EDIT: Could you clarify why you did "0.25 = (3/2)² - 2"?

 

Also, if you proved this conjecture before the twin prime conjecture, you would have proven the twin prime conjecture because this conjecture requires the twin prime conjecture to be solved.

Edited May 16, 2014 by Unity+

[hprime]

Thanks for the quick reaction, Unity+

 

But I think you misunderstood the question, or it was not clearly set. When I write :

 

If p₁ and p₂ are distinct primes, is 0 the lower bound of |u_n(p₁) - u_k(p₂)| ?

I mean : the lower bound for a given couple (p₁,p₂), not over all values of (p₁,p₂)

 

For the latter question, seems to me the answer is trivial and indeed 0, independently of the twin primes conjecture, we need only to know that there is an infinity of primes, which seems to be true

 

Quick proof :

For any e > 0, there exists some p such asu₁(p) < 1 + e

Then for all p' > p, u₁(p') < 1 + e and then u₁(p) - u₁(p') < e

 

Regarding your last question

Could you clarify why you did "0.25 = (3/2)² - 2"?

 

This is the lowest |u_n(p₁) - u_k(p₂)| value I could find "experimentally" for p₁ = 2 and p₂= 3/2, that is for n =1 and k= 2.

My conjecture is that the two progressions don't come closer to each other for great values of n and k. But infinity is great, mainly towards the end

Edited May 16, 2014 by hprime

[Unity+]

Thanks for the quick reaction, Unity+

 

But I think you misunderstood the question, or it was not clearly set. When I write :

 

If p₁ and p₂ are distinct primes, is 0 the lower bound of |u_n(p₁) - u_k(p₂)| ?

I mean : the lower bound for a given couple (p₁,p₂), not over all values of (p₁,p₂)

 

For the latter question, seems to me the answer is trivial and indeed 0, independently of the twin primes conjecture, we need only to know that there is an infinity of primes, which seems to be true

 

Quick proof :

For any e > 0, there exists some p such asu₁(p) < 1 + e

Then for all p' > p, u₁(p') < 1 + e and then u₁(p) - u₁(p') < e

 

Regarding your last question

Could you clarify why you did "0.25 = (3/2)² - 2"?

 

This is the lowest |u_n(p₁) - u_k(p₂)| value I could find "experimentally" for p₁ = 2 and p₂= 3/2, that is for n =1 and k= 2.

My conjecture is that the two progressions don't come closer to each other for great values of n and k. But infinity is great, mainly towards the end

Yes, I misunderstood. My bad.

 

Well, at least that is clarified. Thank you.

[imatfaal]

Hello all

 

I'm new to this forum, I don't know if this topic or related ones have been discussed somewhere else, but a quick search did not bring me any relevant result.

I'm investigating on geometric progressions with initial value 1 and ratio p/(p-1), where p is a prime number. For two different values of p, it's trivial that the two progressions will have no common term (apart of the initial 1). For example powers of 2, 3/2, 5/4 ... are all distinct.

 

My question is : given two such progressions, can they contain terms arbitrarily close to each other? Put in more formally:

 

Let u_n(p)=(p/(p-1))ⁿ, where p is a prime integer and n a positive integer (n >1)

 

If p₁ and p₂ are distinct primes, is 0 the lower bound of |u_n(p₁) - u_k(p₂)| ?

 

I would be happy to have the answer even for the first prime numbers p₁ = 2 and p₂ = 3, that is : can one find a power of 3/2 arbitrarily close to a power of 2?

My conjecture is that the lower bound is strictly greater than 0, and for the case of 2 and 3, the lower bound might be 0.25 = (3/2)² - 2

 

Any related works that could point to a solution - or related open issues, there are many on prime numbers

 

 

You are correct |u_n(3) - u_k(2)| must be greater than zero. Cannot be equal to zero for any integers - proof by contradiction follows

Edited May 16, 2014 by imatfaal
thats 2^m+n

[hprime]

@imatfaal

 

Your proof is correct and its quite trivial result I took for granted, sorry

But the point is : if every value of |u_n(3) - u_k(2)| is greater than 0 it does not imply the lowest bound of the set of such values is greater than 0.

It is this lowest bound I'm interested in.

[Enthalpy]

Take the logarithm of both sequences: now you have successive multiples of two irrational numbers, and the question amounts to:

"are there two integers that multiply the two irrationals, so that the multiples are close enough".

 

And my answer is: the bound equals zero, because these two integers can always be found, however small the difference shall be.

 

The existence of these two integers is exactly equivalent to finding a ratio of integers as close as desired to a ratio of both irrationals, which is nothing else than a real number - and we now that there are rational numbers arbitrarily close to any real number.

[Enthalpy]

Fun: I used to play an instrument tuned as 3/2.

 

Good that you told the application. I had nearly thought you worked on the discrete logarithm

[hprime]

@Enthalpy

 

Thanks fo the hint, but I don't buy your proof, sorry. This was my first try at it

The fact that the lower bound of |log(x) - log(y)| is 0 implies that the ratio x/y can be arbitrarily close to 1, that does not mean their difference can be arbitrary small.

 

Trivial counter-example is log(n+1) - log(n) ...

 

Actually, I'm pretty convinced that the lower bound in that case is NOT 0 in general and I'm searching a proof of that at least for 2 and 3/2. My hunch is that it has something to do with the Catalan's conjecture (which is not a conjecture any more).

http://en.wikipedia.org/wiki/Catalan%27s_conjecture

 

My conjecture is that having two powers of 2 and 3/2 arbitrarily close would break Catalan's theorem at some point. But arithmetic is tough

Edited May 20, 2014 by hprime

[imatfaal]

hprime

 

Out of curiosity - how far did you test by hand?

[hprime]

@imatfaal

 

I've no computing power at hand, just checked with MS Excel up to 2^1024 which is close to (3/2)^1750 or 10^308 where it breaks.

If you define u_n as the increasing sequence of powers of 2 and 3/2 (1, 3/2, 2, 9/4, 27/8, 4 ...

The difference between two consecutive terms of this sequence is growing with a law such that log(u_n+1 - u_n)/log(u_n) seems to converge towards 1.

 

Which means (u_n+1 - u_n) is growing with the same order of magnitude than u_n itself.

 

Of course my conjecture is based on a very small number of terms (less than 3000) but I take it very improbable that this law has sudden exceptions for larger values of n. And there again i'm pretty convinced it has something to do with Catalan's theorem or related conjectures, but I have not found a proof of equivalence so far.

Edited May 21, 2014 by hprime

[Enthalpy]

@Enthalpy

Thanks fo the hint, but I don't buy your proof, sorry [...]

 

Yes, I realized that meanwhile. Thanks for explaining, I don't have to do it by myself.

[Enthalpy]

Ô Hprime, did you find a way to let Excel compute with arbitrary or extended precision? Up to now, I observed only 16 digits accuracy from Excel, that isthe normal accuracy of 64 bits floats. A 0.25 difference would then limit the comparison to numbers like 2.5e15, that is, the 87 first members of the sequence.

 

You may compute some more terms if using an approximation for (3/2)ⁿ-2^k, through [(3/2)ⁿ/2^k-1]*2^k, where you assimilate e^x-1 to x. Though, this game too will stop early.

 

----------

 

Also, you wrote "I have no computing power", but if your Excel runs on Windows, you must have JScript and VBScript available. Type a .js file with text editor, launch it. Similar to C, with typing less annoying than C++ (and 64 bits precision from what I've seen, unles a multiprecision library exists)

http://wsh2.uw.hu/index.html

Excel can also be programmed in VBA (visual Basic for applications).

 

Besides Maple and similar algebraic programmes, you could write your own programme in Lisp, which has built-in arbitrary precision and rational numbers. Some Lisp interpreters are free or demos, but I don't remember which one I used among

CLisp

CormanLisp

LispWorks (the personal edition has a heap size limit, bad for this use)

there are many more.

You could check Lush as well, this one is free (for Linux or for Cygwin on Windows) but I didn't try it

http://2.lisp-lush-development.lisptalks.info/

http://sourceforge.net/projects/lush/files/

Edited May 26, 2014 by Enthalpy

[hprime]

Enthalpy thanks for all the wise advices !

 

Regarding Excel as a quick-and-dirty exploration tool, the differences computed are well over values where accuracy could be problematic.

 

When I write "I have no computing power" I should have written more honestly "I have no programming skills", beyond my rusty BASIC from the 80's, and the XSLT I use daily in my job for data conversion, but this is way off anything useful here.

 

Meanwhile, I've also set the question on the OEIS forum with some interesting results.

http://list.seqfan.eu/pipermail/seqfan/2014-May/013065.html