A142470 Triangle T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k), read by rows.

1, 1, 1, 1, 8, 1, 1, 30, 30, 1, 1, 80, 300, 80, 1, 1, 175, 1750, 1750, 175, 1, 1, 336, 7350, 19600, 7350, 336, 1, 1, 588, 24696, 144060, 144060, 24696, 588, 1, 1, 960, 70560, 790272, 1728720, 790272, 70560, 960, 1, 1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1

Offset: 0

Author

Roger L. Bagula, Sep 20 2008

Keywords

• nonn
• tabl

Comments

Row sums are 1, 2, 10, 62, 462, 3852, 34974, 338690, 3452306, 36683660, 403472368, ...

From Peter Bala, May 08 2012: (Start)

Define the action of the operator L on a sequence { a(i) }{0<=i<=n} by L{ a(i) }{0<=i<=n} = { a(i)^2 - a(i-1)*a(i+1) }_{0<=i<=n} with the conventions a(-1) = a(n+1) = 0. Extend the action of L to a lower triangular array T by letting L act on the rows of T. Then L acting on Pascal's triangle A007318 produces the triangle of Narayana numbers A001263 and L applied to A001263 produces the present triangle.

Since the Narayana polynomials are real-rooted it follows by a theorem of Branden that the row polynomials of this array are also real-rooted.

(End)

Examples

			The triangle begins as:
  1;
  1,    1;
  1,    8,      1;
  1,   30,     30,       1;
  1,   80,    300,      80,        1;
  1,  175,   1750,    1750,      175,        1;
  1,  336,   7350,   19600,     7350,      336,       1;
  1,  588,  24696,  144060,   144060,    24696,     588,      1;
  1,  960,  70560,  790272,  1728720,   790272,   70560,    960,    1;
  1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened
• Petter Brändén, Iterated sequences and the geometry of zeros, arXiv:0909.1927 [math.CO], 2009-2010; J. Reine Angew. Math. 658 (2011), 115-131.

Crossrefs

Cf. A001263, A008459, A056939.

Programs

Magma

A142470:= func< n,k | ( (k+2)/(2*Binomial(k+2, 2)^2) )*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k) >;
[A142470(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 03 2021

Mathematica

f[n_, k_]:= f[n, k]= Binomial[n, k]*Product[j!*(n+j)!/((k+j)!*(n-k+j)!), {j,1,2}];
T[n_, k_]:= Binomial[n, k]*f[n, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 03 2021 *)

Sage

def A142470(n, k): return (2/((k+1)^2*(k+2)))*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k)
flatten([[A142470(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 03 2021

Formula

Let f(n, k) = binomial(n, k)*Product_{j=1.2} ( j!*(n+j)!/((k+j)!*(n-k+j)!) ), then T(n, k) = 2^(k-n)*f(n, k)*Sum_{j=k..n} binomial(n, j)*binomial(j, k) = binomial(n, k)*f(n, k).

From Peter Bala, May 08 2012: (Start)

T(n, k) = C(n, k)^2 * Product {i=1..2} i!*(n+i)!/((k+i)!*(n-k+i)!) = C(n, k)*C(n+2, k)*C(n+2, k+1)*C(n+2, k+2)/(C(n+2, 1)*C(n+2, 2)).

T(n, k) = 2/((n+1)*(n+2)*(n+3))*C(n, k)*C(n+1, k)*C(n+2, k+2)*C(n+3, k+1) = C(n, k)*A056939(n, k).

(End)

T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k). - G. C. Greubel, Apr 03 2021

Extensions

Edited by G. C. Greubel, Apr 03 2021