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

[Max Alekseyev]

On Wed, May 14, 2008 at 7:00 PM, Don Reble <djr at nk.ca> wrote:
>> %I A000001
>> %S A000001 2,3,4,6,6,20,24
>> %N A000001 a(n) = the least integer > n such that
>> r(1)|a(n), r(2)|(a(n)+1), r(3)|(a(n)+2),..., and
>> r(n)|(a(n)+n-1), where (r(1),r(2),r(3),...,r(n))
>> is some permutation of (1,2,3,...,n).
>
> %S A000001 2,3,4,6,6,20,24,48,48,110,110,110,243,403,402,2504,2352,12219,25200,
> %T A000001 60458,14256,95760,120120,582090,582096,186120,3299404,11060250,
> %U A000001 28648620,376576202,9469950,832431604,832431603,962161203,1403352722
> %Y A000001 Cf. A071373.

Don, could you please shed some light on how you got all these values?

btw, I thought about limiting my original approach of iterating over
all permutations to only feasible ones.
I call a permutation (r(1),...,r(n)) feasible only if for every pair
of indices i and j, gcd(r(i),r(j)) divides (i-j). Only for such
permutations, the system of congruences

m = 0 (mod r(1))
m = -1 (mod r(2))
...
m = -(n-1) (mod r(n))

is solvable.
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

Can your approach help to compute more values?

Thanks,
Max


X-NetKnow-OutGoing-4-69-9-1-MailScanner-Watermark: 1211295578.32778 at SWEU2powpnPmo/xljUEsHw