# [seqfan] Re: Divisibility and Binomial Coefficients

Max Alekseyev maxale at gmail.com
Thu Jul 1 18:04:20 CEST 2010

```Here are some larger terms of A140601:

C(n,a) + C(n,a+1) divides C(n,2a+1) for

n=36569, a=47
n=255959, a=47
n=618799, a=25

C(n,a) + C(n,a+2) divides C(n,2a+2) for

n=36098, a=34

Would anybody like to cover the gap between current brute-force search
bound 25325 for A140601 and the terms 36098, 36569 to put them in the
sequence?
Unfortunately, I don't have resources for that.

Regards,
Max

On Fri, May 16, 2008 at 8:47 PM,  <drew at math.mit.edu> wrote:
> Stefan,
>
> I wanted to follow up with the results of a computation I left running as a
> background task this week. Below is a complete of solutions to
>
>  bin(n,a)+bin(n,b)=bin(n,a+b)
>
> with integers 0<=a<b<=n for n up to 18951 (this number just happens to be
> how far it got before I needed to reboot the machine for other reasons).
> There are 18 solutions, all of which have b-a<=2 (but the program checked
> all possible values of a and b, not just those differing by 1 or 2).
>
> bin(19,3)+bin(19,5) divides bin(19,8)
> bin(34,6)+bin(34,7) divides bin(34,13)
> bin(41,5)+bin(41,7) divides bin(41,12)
> bin(89,7)+bin(89,8) divides bin(89,15)
> bin(104,3)+bin(104,4) divides bin(104,7)
> bin(359,5)+bin(359,6) divides bin(359,11)
> bin(398,20)+bin(398,21) divides bin(398,41)
> bin(495,12)+bin(495,14) divides bin(495,26)
> bin(527,7)+bin(527,9) divides bin(527,16)
> bin(1845,15)+bin(1845,17) divides bin(1845,32)
> bin(2309,5)+bin(2309,6) divides bin(2309,11)
> bin(2729,19)+bin(2729,20) divides bin(2729,39)
> bin(3539,35)+bin(3539,36) divides bin(3539,71)
> bin(4619,11)+bin(4619,12) divides bin(4619,23)
> bin(8644,18)+bin(8644,19) divides bin(8644,37)
> bin(12923,34)+bin(12923,36) divides bin(12923,70)
> bin(14135,30)+bin(14135,31) divides bin(14135,61)
> bin(15774,24)+bin(15774,26) divides bin(15774,50)
>
> Cheers,
>
> Drew
>
> p.s. I notice the sequence of n for which such a solution exists is not in
> the OEIS. Have you, or do you plan to submit such a sequence?
>
> On May 5 2008, Stefan Steinerberger wrote:
>
>> First, thanks for all your interest and thoughts.
>> The computed numbers lead me to conjecture that
>> both problems
>>
>> C(n,a)+C(n,a+1) | C(n,2a+1)
>> C(n,a)+C(n,a+2) | C(n,2a+2)
>>
>> have infinitely many solutions and that Drew's observation
>> about the fact that the two variables never differ by more than
>> 2 is correct. Sadly, no idea for a proof (I don't suppose one
>> could prove the "differ by 2" statement by using Legendre's
>> theorem to count the multiplicity of prime factors of both terms?)
>>
>> In any case, thanks for all your interest/time/computing power,
>> Stefan
>>
>

```