Will this number be in that series?

[musicgold]

Hi,
This is not homework. I am reading a book: "The art of infinite: The Pleasure of Mathematics" and pages 66-67 discuss a series and the author seems to be making some assumptions that are not clear to me. So I want to make sure that I understand this.

1. Homework Statement
A 3-rhythm starting with 2 looks like this: 2, 5, 8, 11, 14, 17, ... p, ...
It can be specified as 3n-1 where n goes from 1 to infinity.

Now the author makes the following claim that number m is also on this series.
$m = 3 . (2 . 5 . 8...p ) -1$

Homework Equations

$3n -1$

The Attempt at a Solution

I can test the author's claim for the first three numbers in the series using Excel and 239 is indeed on the series.

I am more interested in finding a proof that shows that the product of first x numbers in the series, multiplied by 3 and after subtracting 1 from it, gets us another number on the series.

Thanks

 

[DrClaude]

When you write

$m = 3 . (2 . 5 . 8...p ) -1$

you mean $m = 3 (2 \times 5 \times 8 \times \cdots \times p ) -1$.

In that case, you only need to prove that $(2 \times 5 \times 8 \times \cdots \times p )$ is a valid choice for $n$. If you can calculate $m$ for all $n \in \mathbb{N}$, the proof is quite simple.

 

[musicgold]

When you write

you mean $m = 3 (2 \times 5 \times 8 \times \cdots \times p ) -1$.

Yes

In that case, you only need to prove that $(2 \times 5 \times 8 \times \cdots \times p )$ is a valid choice for $n$. If you can calculate $m$ for all $n \in \mathbb{N}$, the proof is quite simple.

I see what you are saying. Any natural number can work as an input to the function 3n-1.

Upon further review, it appears that the output of the function can't included two types of numbers: any multiples of 3 and mutiples of 3 +1.

Is that a fair way of thinking about it?

Also, the author makes another claim which I am not clear how he could make.

He asks what factors each term in the original series could have and then says the choices can only have the form 3n-1, 3n, and 3n+1.
Then he goes on to explain that that 3n and 3n+1 can't work.

My questions is how could the author come up with those choices? How can one look at a series and its formula and say that each term in the series can only have factors in this form.

 

[Mark44]

Also, the author makes another claim which I am not clear how he could make.

He asks what factors each term in the original series could have and then says the choices can only have the form 3n-1, 3n, and 3n+1.
Then he goes on to explain that that 3n and 3n+1 can't work.

My questions is how could the author come up with those choices? How can one look at a series and its formula and say that each term in the series can only have factors in this form.

The author is using modular arithmetic, specifically modulo 3, which converts an integer into one of three equivalence classes. Which equivalence class you get is determined solely by what the remainder is when the number is divided by 3.

Given any integer n, when you divide it by 3, the remainder will be 0, 1, or 2.
If $n \equiv 0 \mod 3$ then n = 3k, for some integer k. Examples include 0, 3, 6, 9, and so on.
If $n \equiv 1 \mod 3$ then n = 3k + 1, for some integer k. Examples include 1, 4, 7, 10, and so on.
If $n \equiv 2 \mod 3$ then n = 3k + 2, for some integer k. Examples include 2, 5, 8, 11; that is, the numbers in your sequence. Note that 3k + 2 gives you the same numbers as 3k - 1, except offset by one number in the sequence.

 

[musicgold]

The author is using modular arithmetic, specifically modulo 3, which converts an integer into one of three equivalence classes. ...

Thanks Mark.
I am generally familiar with the Modulo arithmetic. What I am surprised with is the author's claim that at least one prime factor of any member of the series will have the 3n-1 form. I verified that. Either the member itself or one of its factors has the 3n-1 form.

Then I went ahead and checked if this can be applied universally, i.e. for a series formed by a rule, every member of the series at least have one prime factor of that form.

I tested a series created by the rule 4n+1, and it turns out that the hypothesis doesn't work.
5, 9, 13, 17, 21, 25, 29, 33...

The prime factors of 21 are 7 and 3 and neither conform to the form 4n+1. The same thing is true with 33. What am I missing?

 

[RPinPA]

Am I missing something? This seems trivial.

The sequence is the set of numbers $3n - 1$ where n is a positive integer. Every number of that form is in the list.
$m = 3N - 1$ where $N = 2 \times 5 \times 8 \times \cdots \times p$.
$N$ is a positive integer.
Therefore $m$ is in the list, at position $N$.

 

[WWGD]

Thanks Mark.
I am generally familiar with the Modulo arithmetic. What I am surprised with is the claim that at least one prime factor of any member of the series will have the 3n-1 form. <SNIP>. What am I missing?

Isn't that supposed to be about factors of your _original series_ , of the form 3n+1?

 

[RPinPA]

He asks what factors each term in the original series could have and then says the choices can only have the form 3n-1, 3n, and 3n+1.

That has nothing to do with the sequence of numbers. That's just a statement about numbers. Multiples of 3 are spaced at intervals of 3, for instance 6, X, X, 9. Every number is either a multiple of 3 (it's of the form 3n), or it's 1 more than a multiple of 3 (it's of the form 3n + 1) or it's two more than a multiple of 3, which means it's also 1 less than the NEXT multiple of 3 (so it has the form 3n' - 1).

But I'm confused about your discussing factors of numbers in the series. Nothing's being factored here. We're just talking about multiples of 3, and numbers which are next to them. I think maybe that's the source of some of the confusion, you're trying to make statements about prime factors, but nothing you've quoted from the book so far says anything about prime factors.

 

[RPinPA]

He asks what factors each term in the original series could have and then says the choices can only have the form 3n-1, 3n, and 3n+1.

Ah, OK. This is a statement about factors. I stand corrected. OK, what I said before is true. That statement is true because ALL INTEGERS have to be either a multiple of 3, one less than a multiple of 3, or one more than a multiple of 3.

Then he goes on to explain that that 3n and 3n+1 can't work.

No number in that sequence is a multiple of 3, because if you divide $3n - 1$ by 3 you get $n - (1/3)$, which is not an integer for any $n$. Therefore no number in that sequence is a multiple of $3m$ for any $m$.

But they could be multiples of $3n + 1$. For instance 4 is of the form $3n + 1$. And 8, which is in the list, is certainly a multiple of 4.

So surely he didn't say no number in that list can have a factor of the form $3n + 1$. You're trying to prove a statement which isn't true.

 

[Mark44]

Now the author makes the following claim that number m is also on this series. $m = 3 \dot (2 \cdot 5 \cdot 8 \dots \cdot p ) -1$

I am more interested in finding a proof that shows that the product of first x numbers in the series, multiplied by 3 and after subtracting 1 from it, gets us another number on the series.

Actually, the numbers 2, 5, 8, etc. are a what is called a sequence.

Am I missing something? This seems trivial.

I agree with @RPinPA that this is really a trivial problem. $2 \cdot 5 \cdot 8 \dots \cdot p$ is just an integer, so multiplying any integer by 3 and then subtracting 1 will produce another number in the sequence.

I tested a series created by the rule 4n+1, and it turns out that the hypothesis doesn't work.

Well, this rule generates a completely different sequence. In mod 4 arithmetic, there are four possible remainders, 0, 1, 2, and 3, so it's not likely that what works for the sequence in post #1 will also work for this new sequence.

 

[musicgold]

Well, this rule generates a completely different sequence. In mod 4 arithmetic, there are four possible remainders, 0, 1, 2, and 3, so it's not likely that what works for the sequence in post #1 will also work for this new sequence.

Maybe I should provide a bit more background. What the author is doing with that series is to prove that even such a trivial series can have an infinity of prime numbers. He proves by contradiction that $m$ is either prime or destroys the claim that $p$ was the last prime in the series.

So I thought I should be able to prove the same using a different series. Looks like I am doing something wrong.

 

[WWGD]

Maybe I should provide a bit more background. What the author is doing with that series is to prove that even such a trivial series can have an infinity of prime numbers. He proves by contradiction that $m$ is either prime or destroys the claim that $p$ was the last prime in the series.

So I thought I should be able to prove the same using a different series. Looks like I am doing something wrong.

I think you are referirng to a somewhat-advanced result that any arithmetic series with the value of r relatively prime to the first term a1. will contain infinitely-many primes. Here your a1 is 2 and your r is 3. I think the result is by Dirichlet.

 

[musicgold]

I think you are referirng to a somewhat-advanced result that any arithmetic series with the value of r relatively prime to the first term a1. will contain infinitely-many primes. Here your a1 is 2 and your r is 3. I think the result is by Dirichlet.

Yes! Thanks.

Interestingly, this Wiki page says that the 4n+1 series has infinite prime nubmers.

 

[WWGD]

Yes! Thanks.

Interestingly, this Wiki page says that the 4n+1 series has infinite prime nubmers.

Yes, I think that is a pretty advanced result. Try it and we will help you if you need so.