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

N. J. A. Sloane njas at research.att.com
Sat May 8 21:45:46 CEST 2010

```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

```