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Old 2019-06-01, 22:43   #1
Dr Sardonicus
 
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Default Not eleven smooth (four consecutive integers >11))

The entry for the number 11 in the Penguin Dictionary of Curious and Interesting Numbers says, "Given any four consecutive integers greater than 11, there is at least one of them that is divisible by a prime greater than 11."

Your mission, should you decide to accept it, is to prove this.
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Old 2019-06-02, 00:56   #2
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It is not a proof but it would be interesting to see if this property holds up to, say 10^6.

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Matt
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Old 2019-06-02, 01:41   #3
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It is not a proof but it would be interesting to see if this property holds up to, say 10^6.

Regards,
Matt
So, the proposed property says "There are no 4-consecutive 11-smooth numbers larger than 11".

I checked all the 4-consecutive integers in the range [12,10^10] and couldn't find a violation of this property.
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Old 2019-06-02, 01:41   #4
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This looks like a solid conjecture.
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Old 2019-06-02, 03:36   #5
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Originally Posted by SmartMersenne View Post
So, the proposed property says "There are no 4-consecutive 11-smooth numbers larger than 11".

I checked all the 4-consecutive integers in the range [12,10^10] and couldn't find a violation of this property.
According to http://oeis.org/A117581, the largest pair of consecutive 11-smooth numbers is 9800 and 9801. Since you made the check up to 10^10 (which is overkill), this provides a strong case that the conjecture is true.

To prove the conjecture, we can solve 31 Pell’s equations using the procedure described in https://en.m.wikipedia.org/wiki/St%C...er%27s_theorem.

Last fiddled with by 2M215856352p1 on 2019-06-02 at 04:13
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Old 2019-06-02, 05:16   #6
SmartMersenne
 
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According to http://oeis.org/A117581, the largest pair of consecutive 11-smooth numbers is 9800 and 9801. Since you made the check up to 10^10 (which is overkill), this provides a strong case that the conjecture is true.

To prove the conjecture, we can solve 31 Pell’s equations using the procedure described in https://en.m.wikipedia.org/wiki/St%C...er%27s_theorem.
If the largest pair of consecutive 11-smooth numbers is 9800 and 9801, doesn't that prove that there can't be four consecutive ones after 9801? So we are done.
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Old 2019-06-02, 09:24   #7
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If the largest pair of consecutive 11-smooth numbers is 9800 and 9801, doesn't that prove that there can't be four consecutive ones after 9801? So we are done.
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.
Attached Files
File Type: txt 11smooth.txt (843 Bytes, 102 views)
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Old 2019-06-02, 09:32   #8
retina
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Originally Posted by 2M215856352p1 View Post
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.
There is a much easier method to prove it.
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Old 2019-06-02, 09:35   #9
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There is a much easier method to prove it.
How do we do that? I would like to hear from you.
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Old 2019-06-02, 09:37   #10
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Originally Posted by 2M215856352p1 View Post
How do we do that? I would like to hear from you.
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.
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Old 2019-06-02, 09:46   #11
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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.
A good enough hint for me. Thanks. Should I spoil the solution myself if I find it?

Last fiddled with by 2M215856352p1 on 2019-06-02 at 09:57
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