# [seqfan] Re: when does n+1 fail to divide binomial(n^2,n+1) ?

Robert Gerbicz robert.gerbicz at gmail.com
Sat May 8 22:04:51 CEST 2010

```2010/5/8 N. J. A. Sloane <njas at research.att.com>

>
> Dear Seqfans,
> Michel Lagneau has contributed the following sequence.
> After some editing, it reads as follows:
>
> %I A177234
> %S A177234
> 0,1,2,21,364,8855,278256,10737573,491796152,26088783435,1573664496040,
> %T A177234 106395830418878,7970714909592876,655454164338881388,
> %U A177234
> 58702034425556612832,5687847988198592380965,592867741295430227919600
> %S A177234
> 0,-1,2,21,364,8855,278256,10737573,491796152,26088783435,1573664496040,
> %T A177234 106395830418878,7970714909592876,655454164338881388,
> %U A177234
> 58702034425556612832,5687847988198592380965,592867741295430227919600
> %N A177234 a(n) = binomial(n^2, n)/(n+1) if n+1 divides binomial(n^2, n),
> otherwise a(n) = -1.
> %C A177234 Binomial(n^2, n)/(n+1) is an integer for 2 <= n <= 4000. When is
> the next time this divis\
> ion fails (if any)?
> %D A177234 H. Gupta and S. P. Khare, On C(k^2,k) and the product of the
> first k primes, Publ. Fac. E\
> lectrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598.
> %e A177234 a(3) = 21 because binomial(9,3)/(3+1) = 84/4 = 21.
> %p A177234 with(numtheory):n0:=25:T:=array(1..n0-1):for n from 2 to n0 do:
> T[n-1]:= binomial(n*n,n)/\
> (n+1):od:print(T):
> %Y A177234 Cf. A014062 A123312
> %K A177234 sign,new
> %O A177234 0,3
> %A A177234 Michel Lagneau (mn.lagneau2(AT)orange.fr), May 05 2010, May 08
> 2010
>
> Can someone find further terms (beyond n=1) where
> the division fails?
>
> Neil
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

Hi!

Since: binomial(n^2,n)=binomial(n^2,n+1)*(n+1)/(n^2-n) and
gcd(n+1,n^2-n)=gcd(n+1,(n+1)*(n-2)+2)=gcd(n+1,2)|2 it proves that
binomial(n^2,n)/(n+1) is an integer or half of an integer for every n. And
proves that it is an integer if n is even.

```