Re: nice,erdos,can. math bull

In article <1164385109.354946.162770@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Chip Eastham <hardmath@xxxxxxxxx> wrote:

Ashay Burungale wrote:
any progress or idea?

You dropped off the original statement of this problem,
which was somewhat enigmatic. Let me try to restate
it for clarity.

For positive integers n,r: let C(n,r) denote "n choose r",
i.e. n!/(r!(n-r)!).

Show that if a is (strictly?) between n/2 and n, then
C(2a,a) does not divide C(2n,n).

[I suggest "strictly" not only because a = n always
divides, but since C(2,1) divides C(4,2).]

Perhaps consider prime p > 2 between a and 2a?
E.g. C(4,2) does not divide C(8,4) because there
are not enough factors of 3 "left over".

That won't work.
There are infinitely many cases where the denominator
of C(2n,n)/C(2a,a) is a power of 2 or 3, e.g. for n=a+1
where n is a power of 2 or 3, since
C(2a+2,a+1)/C(2a,a) = 2 (2a+1)/(a+1).

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.