A094437 Triangular array T(n,k) = Fibonacci(k+2)*C(n,k), k=0..n, n>=0.

1, 1, 2, 1, 4, 3, 1, 6, 9, 5, 1, 8, 18, 20, 8, 1, 10, 30, 50, 40, 13, 1, 12, 45, 100, 120, 78, 21, 1, 14, 63, 175, 280, 273, 147, 34, 1, 16, 84, 280, 560, 728, 588, 272, 55, 1, 18, 108, 420, 1008, 1638, 1764, 1224, 495, 89, 1, 20, 135, 600, 1680, 3276, 4410, 4080, 2475, 890

Offset: 0

Author

Clark Kimberling, May 03 2004

Keywords

• nonn
• easy
• tabl

Comments

Let F(n) denote the n-th Fibonacci number (A000045). Then n-th row sum of T is F(2n+2) and n-th alternating row sum is -F(n-2).

A094437 is jointly generated with A094436 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. [Clark Kimberling, Feb 26 2012]

Subtriangle of the triangle given by (1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 28 2012

Examples

			First four rows:
  1;
  1 2;
  1 4 3;
  1 6 9 5;
sum = 1+6+9+5=21=F(8); alt.sum = 1-6+9-5=-1=-F(1).
T(3,2)=F(4)*C(3,2)=3*3=9.
From _Philippe Deléham_, Apr 28 2012: (Start)
(1, 0, 0, 1, 0, 0, ...) DELTA (0, 2, -1/2, -1/2, 0, 0, ...) begins :
  1;
  1, 0;
  1, 2,  0;
  1, 4,  3,  0;
  1, 6,  9,  5, 0;
  1, 8, 18, 20, 8, 0; . (End)
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened

Crossrefs

Cf. A000045.

Cf. A094435, A094436, A094438, A094439, A094440, A094441, A094442, A094443, A094444.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Binomial(n,k)*Fibonacci(k+2) ))); # G. C. Greubel, Oct 30 2019

Magma

[Binomial(n,k)*Fibonacci(k+2): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 30 2019

Maple

with(combinat); seq(seq(fibonacci(k+2)*binomial(n,k), k=0..n), n=0..12); # G. C. Greubel, Oct 30 2019

Mathematica

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
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[%]  (* A094436 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A094437 *)
(* Second program *)
Table[Fibonacci[k+2]*Binomial[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 30 2019 *)

PARI

T(n,k) = binomial(n,k)*fibonacci(k+2);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

Sage

[[binomial(n,k)*fibonacci(k+2) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Oct 30 2019

Formula

From Philippe Deléham, Apr 28 2012: (Start)

As DELTA-triangle T(n,k):

G.f.: (1-x-y*x+2*y*x^2-y^2*x^2)/(1-2*x-y*x+x^2+y*x^2-y^2*x^2).

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = 1, T(2,1) = 2, T(1,1) = T(2,2) = 0 and T(n,k) = 0 if k<0 or if k>n. (End)

From G. C. Greubel, Oct 30 2019: (Start)

T(n, k) = binomial(n, k)*Fibonacci(k+2).

Sum_{k=0..n} T(n,k) = Fibonacci(2*n+2).

Sum_{k=0..n} (-1)^(k+1) * T(n,k) = Fibonacci(n-2). (End)