Divisibility and Binomial Coefficients

[Max Alekseyev]

On Sun, May 4, 2008 at 7:23 AM, Max Alekseyev <maxale at gmail.com> wrote:
> On Sun, May 4, 2008 at 5:12 AM, Stefan Steinerberger
> <stefan.steinerberger at gmail.com> wrote:
>
>> 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)?
>
> It seems that there is also restriction a+b <= n. Otherwise, there
> would be a lot of solutions dividing C(n,a+b) = 0 when a+b>n.
>
>> 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?
>
> Hard to say. But most of your solutions are of the form:
>
> C(n,a)+C(n,a+1) | C(n,2a+1)
>
> which is simply C(n+1,a+1) | C(n,2a+1).
>
> The next solution of this form is
>
> n = 2309, a = 5

Further solutions:

2729, 19
3539, 35
4619, 11
8644, 18

Regards,
Max