A177234 a(n) = binomial(n^2, n)/(n+1).

2, 21, 364, 8855, 278256, 10737573, 491796152, 26088783435, 1573664496040, 106395830418878, 7970714909592876, 655454164338881388, 58702034425556612832, 5687847988198592380965, 592867741295430227919600

Offset: 2

Author

Michel Lagneau, May 05 2010, May 08 2010

Keywords

• nonn

Comments

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

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

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

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

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

Proof 2 from Max Alekseyev, May 08 2010:

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

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.

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.

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

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

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.

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

Examples

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

References

• 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.

Links

• G. C. Greubel, Table of n, a(n) for n = 2..335
• 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. [PDF] [From _R. J. Mathar_, May 09 2010]

Crossrefs

Cf. A014062 A123312, A177456, A177784, A177788.

Programs

Magma

[Binomial(n^2,n)/(n+1): n in [2..30]]; // G. C. Greubel, Apr 27 2024

Maple

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):

Mathematica

Table[Binomial[n^2,n]/(n+1), {n,2,30}] (* G. C. Greubel, Apr 27 2024 *)

SageMath

[binomial(n^2,n)/(n+1) for n in range(2,31)] # G. C. Greubel, Apr 27 2024

Formula

From G. C. Greubel, Apr 27 2024: (Start)

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

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

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

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