A177784 a(n) = binomial(n^2, n) / ( n*(n+1) ).

1, 7, 91, 1771, 46376, 1533939, 61474519, 2898753715, 157366449604, 9672348219898, 664226242466073, 50419551102990876, 4193002458968329488, 379189865879906158731, 37054233830964389244975

Offset: 2

Author

Michel Lagneau, May 13 2010

Keywords

• nonn

Comments

All terms are integer because n and n+1 divide the binomial (cf. A060545, A177234).

Empirical: In the ring of symmetric functions over the fraction field Q(q, t), letting s(n) denote the Schur function indexed by n, a(n)*(-1)^(n+1) is equal to the coefficient of s(n) in nabla^(n)s(n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions. - John M. Campbell, Nov 18 2017

Examples

			For n = 3, binomial(9,3)/(3*4) =84/12 = 7.
For example, the coefficient of s(3) in nabla(nabla(nabla(s(3)))) is equal to q^6*t^2+q^5*t^3+q^4*t^4+q^3*t^5+q^2*t^6+q^4*t^3+q^3*t^4, and if we let q and t be equal to 1, this coefficient reduces to 7 = a(3). - _John M. Campbell_, Nov 18 2017
		

Links

• G. C. Greubel, Table of n, a(n) for n = 2..335

Crossrefs

Cf. A060545, A177234.

Programs

Magma

[Binomial(n^2,n)/(2*Binomial(n+1,2)): n in [2..30]]; // G. C. Greubel, Jul 18 2024

Maple

A177784 := proc(n)
        binomial(n^2,n)/(n^2+n) ;
end proc:
seq(A177784(n),n=2..20) ; # R. J. Mathar, Nov 07 2011

Mathematica

Table[Binomial[n^2,n]/(2*Binomial[n+1,2]), {n,2,30}] (* G. C. Greubel, Jul 18 2024 *)

SageMath

[binomial(n^2,n)//(n*(n+1)) for n in range(2,31)] # G. C. Greubel, Jul 18 2024