Divisibility and Binomial Coefficients

[Maximilian Hasler]

On Sun, May 4, 2008 at 12:54 PM,  <franktaw at netscape.net> wrote:
> 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

There was written "distinct a,b".

I agree that the problem is nice.
I observed that for the given solutions with b=a+2, we have a+1 = b-1
= (a+b)/2 ; the following PARI script finds them again and 2 more:

gp > for(i=1,999,for(j=2*i,100000,binomial(j,i*2)%(binomial(j,i-1)+binomial(j,i+1))&next;print1([i,j]",");next(2)))
[4, 19],[6, 41],[8, 527],[13, 495],[16, 1845],[25, 15774],[35, 12923],

In fact I suspected that there be a subsequence with i=2^k, but I did
not find the j for i=32 : must be more than 10^5.

Maximilian



As I read Stefan's message, a and b are required to be distinct and <= n 
(and perhaps we should add that a, b, and n are nonnegative integers).

>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

Otherwise, C(n,1)+C(n,1) divides C(n,2) for all n > 1 congruent to 1 mod 4, 
C(n,2)+C(n,2) divides C(n,4) for all n > 24 congruent to 2, 3, or 11 mod 
24, and in general for any fixed k there are infintely many n for which 
C(n,k)+C(n,k) divides C(n,2k).

Regards,

Drew

On May 4 2008, franktaw at netscape.net wrote:

>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
>