Not eleven smooth (four consecutive integers >11))
 

The entry for the number 11 in the [u]Penguin Dictionary of Curious and Interesting Numbers[/u] 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.

It is not a proof but it would be interesting to see if this property holds up to, say 10^6.

Regards,
Matt

[QUOTE=MattcAnderson;518316]It is not a proof but it would be interesting to see if this property holds up to, say 10^6.

Regards,
Matt[/QUOTE]

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.

This looks like a solid conjecture.

[QUOTE=SmartMersenne;518318]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.[/QUOTE]

According to [URL="http://oeis.org/A117581"]http://oeis.org/A117581[/URL], 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 [URL="https://en.m.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem"]https://en.m.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem[/URL].

[QUOTE=2M215856352p1;518326]According to [URL="http://oeis.org/A117581"]http://oeis.org/A117581[/URL], 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 [URL="https://en.m.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem"]https://en.m.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem[/URL].[/QUOTE]

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.

[QUOTE=SmartMersenne;518327]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.[/QUOTE]

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 10[SUP]30[/SUP]. 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 10[SUP]30[/SUP]. 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.

[QUOTE=2M215856352p1;518334]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.[/QUOTE]There is a [i]much[/i] easier method to prove it.

[QUOTE=retina;518335]There is a [i]much[/i] easier method to prove it.[/QUOTE]

How do we do that? I would like to hear from you.

[QUOTE=2M215856352p1;518336]How do we do that? I would like to hear from you.[/QUOTE]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.

[QUOTE=retina;518337]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.[/QUOTE]

A good enough hint for me. Thanks. Should I spoil the solution myself if I find it?