Difference boxes/polygons
zak seidov
zakseidov at yahoo.com
Thu Jun 9 15:56:49 CEST 2005
** Ref: A. Behn et. al.,
** The convergence of difference boxes,
** Amer. Math. Month. v. 112 (May 2005), pp 426-439.
Dear SEQFANs,
This should be known, but not for me.
Put any four numbers, a, b, c, d in the vertices of
the square;
in the middle of each side write unsigned difference
between numbers in the ends of the side, and get
another four numbers a, b, c, d in the vertices of the
smaller square.
Repeat the procedure until a=b=c=d=0.
It's evident that for any initial a, b, c, d, the
final numbers are all zero,
simply because each of a, b, c, d only decreases at
each step.
But it seems (see Ref. above)
that it's proven (or only suggested) to be true
also for signed differences,
but I leave this more difficult case to SEQ gurus.
In some particular cases it's possible to
calculate exactly all steps
(I consider only unsigned differences) .
For example,
let a=n; b=n^2; c=n^3; d=n^4;
then we have subsequently (or successively?):
k=0: a=n, b=n^2, c=n^3, d=n^4;
k=1: a=n( n-1), b=(n-1)*n^2, c=(n-1)*n^3,
d=n(n^3-1);
k=2: a=n(n-1)^2, b=(n-1)^2*n^2, c=n(n^2-1),
d=n^2*(n^2-1);
k=3: a=n(n-1)^3, b= n(1+n -3n^2+n^3),
c=n(n+1)(n-1)^2, d=n(n-1)(1+n^2 );
k=4: a=2n(n-1), b=2(n-1)n^2, c=a, d=b;
k=5: a=b=c=d=2n(n-1)^2;
k=6: a=b=c=d=0.
We say that in k=6 steps we come to finish, at any n.
I submit the SEQ giving the number of k(n)
for the case
k=0: a=Prime[n], b=Prime[n+1], c=Prime[n+2],
d=Prime[n+3]}.
Maximal found k is 12 at n=22059,
three cases with k=11 are at
n=18024, 41761, 84938 (more entries need to submit to
OEIS!),
and there are many cases with k=10 (and less, of
course).
Thanks for your attention, Zak
PS It's evident expansion for polygons,
some1 may wish to consider it.
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