Not eleven smooth (four consecutive integers >11)) - Page 2

SmartMersenne · 2019-06-02, 10:10

Quote:

Originally Posted by 2M215856352p1

The statement is true, however the assumption of the result that the largest pair of consecutive 11-smooth numbers is 9800 and 9801 was made, hence we have yet to complete the proof.

I have written a Python 3 script which verified the conjecture up to 10³⁰. The script could take a few seconds to run, hence you would need to be patient. Please notify me if the program has a bug.

The Python script does not generate a proof because there still remains the possibility of a solution beyond 10³⁰. To really prove the conjecture, the only method I have now is to solve the 31 Pell's equations involved, which is going to be very tedious.

To run the script, please change the file extension from .txt to .py. I don't know why .py file extension is not supported.

If 10¹⁰ was an overkill, isn't 10³⁰?

SmartMersenne · 2019-06-02, 10:11

Quote:

Originally Posted by retina

It's no fun if I just give you the answer. But suffice to say that nothing more than basic high school mathematics is needed.

When something sounds that easy, there is usually a mistake somewhere.

This is no competition. You can take all the credit. So, I invite you to present your proof.

retina · 2019-06-02, 10:20

Quote:

Originally Posted by SmartMersenne

When something sounds that easy, there is usually a mistake somewhere..

I certainly admit there might be a mistake. I'll post my proof later if no one seems to figure it out and then everyone can point out the holes in it.

Quote:

Originally Posted by SmartMersenne

This is no competition. You can take all the credit. So, I invite you to present your proof.

I'm long past the days of caring about credit. I'm much more interested to see what others here come up with.

2M215856352p1 · 2019-06-02, 11:06

Quote:

Originally Posted by SmartMersenne

If 10¹⁰ was an overkill, isn't 10³⁰?

Sorry for not being clear. 10¹⁰ was overkill under the assumption that we can use the OEIS result, but in actual fact, we cannot use that result to computationally prove the result.

axn · 2019-06-02, 11:32

Searching all 11-smooth numbers under 10^40, the largest run of 3 consecutive 11 smooth (above 11) is 98-100, and largest run of 2 consecutive 11 smooth is (still) 9800, 9801.

There are appr. 1 billion 11-smooth numbers under 10^100.

Dr Sardonicus · 2019-06-02, 11:44

The proof I came up with only uses well-known results (especially in this Forum) of elementary number theory.

I don't know about "basic high school mathematics," but perhaps there is a simpler proof than mine. But the proof I found is simple enough that I thought it justified posting the assertion as a puzzle.

If you find a proof, for now you can post a declaration to the effect "I have a proof" to give others time to think about it.

To me, numerical verification is particularly useful for assertions I'm not sure about. A quick check that turns up a small numerical counterexample can prevent a lot of wasted effort trying to prove something that isn't true.

OTOH, once you have verified something to the point where you're convinced it's actually true, it's time to start trying to prove it
:-D

Regarding the further assertions on 3 and 2 consecutive 11-smooth numbers, I'm not sure how to prove them. The proof I have for 4 consecutive numbers won't work for 2 or 3 consecutive numbers.

SmartMersenne · 2019-06-02, 11:44

Quote:

Originally Posted by 2M215856352p1

Sorry for not being clear. 10¹⁰ was overkill under the assumption that we can use the OEIS result, but in actual fact, we cannot use that result to computationally prove the result.

Don't worry, I am just teasing you

SmartMersenne · 2019-06-02, 11:55

Here is a summary of what I am thinking of:

The proof I am thinking involves assigning primes to 4 consecutive numbers. We are only allowed to use {2,3,5,7,11}. Our options are:

1) There must be at exactly two numbers with factor 2.
2) There is either one or two numbers with factor 3.
3) There can be at most one number with factor 5, 7 or 11.

Now we need to consider all possibilities of assigning those primes to numbers and then consider the possibilities for powers. However, since we have a limited set to use from it would boil down to assigning 5, 7 and 11 to separate numbers, because otherwise, we will have very limited number of possibilities for the other numbers.

2M215856352p1 · 2019-06-02, 15:57

Quote:

Originally Posted by Dr Sardonicus

Regarding the further assertions on 3 and 2 consecutive 11-smooth numbers, I'm not sure how to prove them. The proof I have for 4 consecutive numbers won't work for 2 or 3 consecutive numbers.

You could simply use the procedure described in https://projecteuclid.org/download/pdf_1/euclid.ijm/1256067456 and https://en.wikipedia.org/wiki/Størmer%27s_theorem. To be frank, I cannot understand all the math behind the proof of correctness of this procedure.

c10ck3r · 2019-06-03, 02:48

Quote:

Originally Posted by retina

There is a much easier method to prove it.

Would this involve expressing numbers mod 2310?

retina · 2019-06-03, 02:51

Quote:

Originally Posted by c10ck3r

Would this involve expressing numbers mod 2310?

I think that you are on the right path.

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2019-06-02, 11:32	#16
axn Jun 2003 2²·3·421 Posts	Searching all 11-smooth numbers under 10^40, the largest run of 3 consecutive 11 smooth (above 11) is 98-100, and largest run of 2 consecutive 11 smooth is (still) 9800, 9801. There are appr. 1 billion 11-smooth numbers under 10^100.

2019-06-02, 11:44	#17
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	The proof I came up with only uses well-known results (especially in this Forum) of elementary number theory. I don't know about "basic high school mathematics," but perhaps there is a simpler proof than mine. But the proof I found is simple enough that I thought it justified posting the assertion as a puzzle. If you find a proof, for now you can post a declaration to the effect "I have a proof" to give others time to think about it. To me, numerical verification is particularly useful for assertions I'm not sure about. A quick check that turns up a small numerical counterexample can prevent a lot of wasted effort trying to prove something that isn't true. OTOH, once you have verified something to the point where you're convinced it's actually true, it's time to start trying to prove it :-D Regarding the further assertions on 3 and 2 consecutive 11-smooth numbers, I'm not sure how to prove them. The proof I have for 4 consecutive numbers won't work for 2 or 3 consecutive numbers.

2019-06-02, 11:55	#19
SmartMersenne Sep 2017 2×53 Posts	Here is a summary of what I am thinking of: The proof I am thinking involves assigning primes to 4 consecutive numbers. We are only allowed to use {2,3,5,7,11}. Our options are: 1) There must be at exactly two numbers with factor 2. 2) There is either one or two numbers with factor 3. 3) There can be at most one number with factor 5, 7 or 11. Now we need to consider all possibilities of assigning those primes to numbers and then consider the possibilities for powers. However, since we have a limited set to use from it would boil down to assigning 5, 7 and 11 to separate numbers, because otherwise, we will have very limited number of possibilities for the other numbers.