A091144 a(n) = binomial(n^2, n)/(1+(n-1)*n).

1, 1, 2, 12, 140, 2530, 62832, 1997688, 77652024, 3573805950, 190223180840, 11502251937176, 779092434772236, 58448142042957576, 4811642166029230560, 431306008583779517040, 41820546066482630185200

Offset

0,3

Comments

Diagonal of array T(n,k) = binomial(kn,n)/(1+(k-1)n).

Number of paths up and left from (0,0) to (n^2-n,n) where x/y <= n-1 for all intermediate points. - Henry Bottomley, Dec 25 2003

Empirical: In the ring of symmetric functions over the fraction field Q(q, t), letting s(1^n) denote the Schur function indexed by (1^n), a(n) is equal to the coefficient of s(n) in nabla^(n)s(1^n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions, and s(n) denotes the Schur function indexed by the integer partition (n) of n. - John M. Campbell, Apr 06 2018

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.

Formula

From Henry Bottomley, Dec 25 2003: (Start)

a(n) = A014062(n)/A002061(n);

a(n) = A062993(n-2, n);

a(n) = A070914(n, n-1);

a(n) = A071201(n, n^2-n);

a(n) = A071201(n, n^2-n+1);

a(n) = A071202(n, n^2-n+1). (End)

Maple

A091144 := proc(n)
    binomial(n^2, n)/(1+n*(n-1)) ;
end proc: # R. J. Mathar, Feb 14 2015

Mathematica

Table[Binomial[n^2, n] / (n (n - 1) + 1), {n, 0, 20}] (* Vincenzo Librandi, Apr 07 2018 *)

Prog

(PARI) a(n) = binomial(n^2, n)/(n*(n-1)+1); \\ Altug Alkan, Apr 06 2018
(Magma) [Binomial(n^2, n)/(1+(n-1)*n): n in [0..20]]; // Vincenzo Librandi, Apr 07 2018
(GAP) List([0..20], n->Binomial(n^2, n)/(1+(n-1)*n)); # Muniru A Asiru, Apr 08 2018

Crossrefs

Cf. A002061, A014062, A062993, A070914, A071201, A071202.

Sequence in context: A093543 A367636 A287885 * A275829 A351731 A240387

Adjacent sequences: A091141 A091142 A091143 * A091145 A091146 A091147

Keyword

nonn,easy

Author

Paul Barry, Dec 22 2003

Status

approved