Dividing n consecutive integers by (1,2,3,...,n)

[Max Alekseyev]

On Thu, May 15, 2008 at 7:44 AM, Don Reble <djr at nk.ca> wrote:
>> Don, could you please shed some light on how you got all these values?
>
>    I test each integer in turn, to see whether it's the value.
>    Just careful coding in a machine-friendly language.

Did you follow the same approach as Robert?

>> Can your approach help to compute more values?
>
>    Yeah, I could have run the program longer.

Actually, I meant to ask if your approach can be adapted to counting
the number of feasible permutations. Can it?

>> It turns out that there are no so many feasible permutations, their
>> number for n=1,...,16 is:
>>
>> 1, 2, 6, 8, 48, 24, 216, 120, 240, 128, 2544, 336, 11520, 3168, 1536, 480
>
>    Furthermore, sometimes two permutations map onto the same starting
>    point, modulo lcm(1..n). The number of starting points is
> 1, 2, 5, 8, 32, 11, 72, 64, 66, 17, 213, 37, 417, 120, 67, 60, 694, 217

That is a bit unclear to me.
In particular, for n=3, your calculations indicate that there are only
5 different starting points while there are 6 feasible permutations.
But it turns out that solutions to the corresponding systems of congruences:

m = 0 (mod r(1))
m = -1 (mod r(2))
m = -2 (mod r(3))

(where r goes over all 6 permutations of {1,2,3}) are exactly all
different residues modulo 6.
In other words, each of the feasible permutations here results in an
unique solution modulo lcm(1,2,3)=6. But how then the number of
starting points can be equal 5?
Maybe, I've misunderstood what is the "starting point"? Please clarify.

>    That's a good way to find all solutions; but if you want only the
>    least solution, it may be a waste of time.

I cannot agree with this statement right away, in particular, since
feasible permutations can be constructed in a fast way using
branch-and-bound approach. A straightforward implementation seems to
be able to produce the result in reasonable time for n up to 20 at
least. But, maybe, there exists even a smarter way to construct such
permutations so that it will beat Robert's and your approach(es) to
computing Leroy's sequence.

Regards,
Max



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Eugene McDonnell