Divisibility and Binomial Coefficients

[franktaw at netscape.net]

You are both assuming that a < b, which was not in the original 
statement
of the problem.  What about cases where a = b?  Even if these were
intended to be excluded, enumerating them might help shed some light
on the subject.

We have at least the case:

C(5,1)+C(5,1) | C(5,2).

Franklin T. Adams-Watters

-----Original Message-----
From: drew at math.mit.edu

Nice question. I have no good reason for saying so, but I would guess 
there
are infinitely many solutions.

Here is a complete list for n <= 5000:

 C(19,3)+C(19,5) | C(19,8)
 C(34,6)+C(34,7) | C(34,13)
 C(41,5)+C(41,7) | C(41,12)
 C(89,7)+C(89,8) | C(89,15)
 C(104,3)+C(104,4) | C(104,7)
 C(359,5)+C(359,6) | C(359,11)
 C(398,20)+C(398,21) | C(398,41)
 C(495,12)+C(495,14) | C(495,26)
 C(527,7)+C(527,9) | C(527,16)
 C(1845,15)+C(1845,17) | C(1845,32)
 C(2309,5)+C(2309,6) | C(2309,11)
 C(2729,19)+C(2729,20) | C(2729,39)
 C(3539,35)+C(3539,36) | C(3539,71)
 C(4619,11)+C(4619,12) | C(4619,23)

It might be worth trying to prove that a and b cannot differ by more 
than 2.

Drew

On May 4 2008, Stefan Steinerberger wrote:

>Dear seqfans,
>
>JM Bergot has asked me whether there are a < m,n
>such that C(m, a) + C(n, a) divides C(m+n,a), where
>C(m,n) is the binomial coefficient. As the nature of the
>problem suggests, there is a vast number of solutions
>(for example C(4,3) + C(5,3) = 14, C(9,3) = 6*14).
>
>I find the "dual" question more interesting. Are there
>distinct a,b <= n such that C(n,a) + C(n,b) divides C(n,a+b)?
>I only found the following seven solutions
>
>C(19,3)+C(19,5) | C(19,8)
>C(34,6)+C(34,7) | C(34,13)
>C(41,5)+C(41,7) | C(41,12)
>C(89,7)+C(89,8) | C(89,15)
>C(104,3)+C(104,4) | C(104,7)
>C(359,5)+C(359,6) | C(359,11)
>C(398,20)+C(398,21) | C(398,41)
>
>Is there any reason to assume that the number of solutions is
>finite/infinite?
>
>Best wishes,
>Stefan