Divisibility and Binomial Coefficients

sellersj at math.psu.edu sellersj at math.psu.edu
Sun May 4 19:06:07 CEST 2008


Following Franklin's idea, there are infinitely many such cases because

2*C(4n+1, 1) divides C(4n+1,2).

In fact, C(4n+1,2) / (2*C(4n+1,1)) = n.

James


> 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
>






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