A177234 - OEIS

%I #9 Apr 28 2024 11:38:02

%S 2,21,364,8855,278256,10737573,491796152,26088783435,1573664496040,

%T 106395830418878,7970714909592876,655454164338881388,

%U 58702034425556612832,5687847988198592380965,592867741295430227919600

%N a(n) = binomial(n^2, n)/(n+1).

%C Theorem: binomial(n^2, n)/(n+1) is an integer for n >= 2.

%C Proof 1 from William J. Keith, May 08 2010:

%C binomial(n^2, n) * 1/(n+1)

%C = (n^2)*(n^2-1)*(n^2-2)!/((n^2-n)!*n*(n-1)*(n-2)!) * 1/(n+1)

%C = n*(n^2-2)!/((n^2-n)!*(n-2)!) = n * binomial(n^2-2,n-2). QED

%C Proof 2 from Max Alekseyev, May 08 2010:

%C Recall that the valuation of m! w.r.t. prime p equals the sum floor(m/p^i) over i=1,2,3,...

%C Moreover, if m=a+b where a and b are nonnegative integers, then floor(m/p^i) - floor(a/p^i) - floor(b/p^i) >= 0.

%C Let n>1. To prove that binomial(n^2, n)/(n+1) is an integer, it is enough to show that its valuation w.r.t. any prime p is nonnegative.

%C It is clear that trouble may come only from primes dividing n+1.

%C Let valuation(n+1,p)=k > 0, i.e., n+1=p^k*m where prime p does not divide m.

%C Then n = p^k*m - 1, n^2 = p^(2k)*m^2 - 2*p^k*m + 1 and n^2 - n = p^(2k)*m^2 - 3*p^k*m + 2.

%C It is easy to check that floor(n^2/p^i) - floor(n/p^i) - floor((n^2-n)/p^i) = 1 for i=1,2,...,k if p>2 and for i=2,3,...,k+1 if p=2, implying that valuation(binomial(n^2, n)/(n+1),p) >= 0. QED

%D H. Gupta and S. P. Khare, On C(k^2,k) and the product of the first k primes, Publ. Fac. Electrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598.

%H G. C. Greubel, <a href="/A177234/b177234.txt">Table of n, a(n) for n = 2..335</a>

%H H. Gupta and S. P. Khare, <a href="http://www.ams.org/mathscinet-getitem?mr=453615">On C(k^2,k) and the product of the first k primes</a>, Publ. Fac. Electrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598. [<a href="http://pefmath2.etf.rs/files/101/580.pdf">PDF</a>] [From _R. J. Mathar_, May 09 2010]

%F From _G. C. Greubel_, Apr 27 2024: (Start)

%F a(n) = A014062(n)/(n+1).

%F a(n) = A177456(n)/(n-1).

%F a(n) = n*A177784(n).

%F a(n) = A177788(n)/n. (End)

%e a(3) = 21 because binomial(9,3)/(3+1) = 84/4 = 21.

%p 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):

%t Table[Binomial[n^2,n]/(n+1), {n,2,30}] (* _G. C. Greubel_, Apr 27 2024 *)

%o (Magma)

%o [Binomial(n^2,n)/(n+1): n in [2..30]]; // _G. C. Greubel_, Apr 27 2024

%o (SageMath)

%o [binomial(n^2,n)/(n+1) for n in range(2,31)] # _G. C. Greubel_, Apr 27 2024

%Y Cf. A014062 A123312, A177456, A177784, A177788.

%K nonn

%O 2,1

%A _Michel Lagneau_, May 05 2010, May 08 2010