A210220 T(n, k) = -binomial(2*n-k+2, k+1)*hypergeom([2*n-k+3, 1], [k+2], 2). Triangle read by rows, T(n, k) for 1 <= k <= n.
1, 2, 2, 3, 6, 3, 4, 12, 13, 4, 5, 20, 34, 24, 5, 6, 30, 70, 80, 40, 6, 7, 42, 125, 200, 166, 62, 7, 8, 56, 203, 420, 496, 314, 91, 8, 9, 72, 308, 784, 1211, 1106, 553, 128, 9, 10, 90, 444, 1344, 2576, 3108, 2269, 920, 174, 10, 11, 110, 615, 2160, 4956, 7476, 7274, 4352, 1461, 230, 11
Offset: 1
Examples
First five rows: 1 2...2 3...6....3 4...12...13...4 5...20...34...24...5 First three polynomials v(n,x): 1, 2 + 2x , 3 + 6x + 3x^2.
Programs
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Maple
T := (n,k) -> -binomial(2*n-k+2, k+1)*hypergeom([2*n-k+3, 1], [k+2], 2): seq(seq(simplify(T(n,k)), k=1..n), n=1..10); # Peter Luschny, Oct 31 2019
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Mathematica
u[1, x_] := 1; v[1, x_] := 1; z = 16; u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1; v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n, 1, z}]; TableForm[cu] Flatten[%] (* A210219 *) Table[Expand[v[n, x]], {n, 1, z}] cv = Table[CoefficientList[v[n, x], x], {n, 1, z}]; TableForm[cv] Flatten[%] (* A210220 *) (* alternate program *) T[n_,k_]:=Sum[Binomial[2*j+k-2,k-1],{j,1,n-k+1}];Flatten[Table[T[n,k],{n,1,11},{k,1,n}]] (* Detlef Meya, Dec 05 2023 *)
Formula
First and last term in row n: n.
Column 2: n*(n-1).
Column 3: A016061.
Column 4: A112742.
Row sums: -1+(even-indexed Fibonacci numbers).
Periodic alternating row sums: 1,0,0,1,0,0,1,0,0,...
u(n,x)=x*u(n-1,x)+v(n-1,x)+1,
v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
T(n,k) = Sum_{j=1..n-k+1} binomial(2*j+k-2,k-1). - Detlef Meya, Dec 05 2023
Extensions
New name from Peter Luschny, Oct 31 2019
Comments