A094438 Triangular array T(n,k) = Fibonacci(k+3)*C(n,k), k=0..n, n>=0.

2, 2, 3, 2, 6, 5, 2, 9, 15, 8, 2, 12, 30, 32, 13, 2, 15, 50, 80, 65, 21, 2, 18, 75, 160, 195, 126, 34, 2, 21, 105, 280, 455, 441, 238, 55, 2, 24, 140, 448, 910, 1176, 952, 440, 89, 2, 27, 180, 672, 1638, 2646, 2856, 1980, 801, 144, 2, 30, 225, 960, 2730, 5292, 7140, 6600, 4005, 1440, 233

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+3) and n-th alternating row sum is F(n-3).

Examples

			First few rows:
  2;
  2   3;
  2   6   5;
  2   9  15   8;
  2, 12, 30, 32, 13;
  2, 15, 50, 80, 65, 21;
		

Links

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

Crossrefs

Cf. A000045, A094435, A094436, A094437, A094439.

Cf. A094440, A094441, A094442, A094443, A094444.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Binomial(n,k)*Fibonacci(k+3) ))); # G. C. Greubel, Oct 30 2019

Magma

[Binomial(n,k)*Fibonacci(k+3): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 30 2019

Maple

with(combinat); seq(seq(fibonacci(k+3)*binomial(n,k), k=0..n), n=0..12); # G. C. Greubel, Oct 30 2019

Mathematica

Table[Fibonacci[k+3]Binomial[n,k],{n,0,12},{k,0,n}]//Flatten (* Harvey P. Dale, Dec 16 2017 *)

PARI

T(n,k) = binomial(n,k)*fibonacci(k+3);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

Sage

[[binomial(n,k)*fibonacci(k+3) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Oct 30 2019

Formula

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

T(n, k) = binomial(n,k)*Fibonacci(k+3).

Sum_{k=0..n} T(n,k) = Fibonacci(2*n+3).

Sum_{k=0..n} (-1)^k * T(n,k) = Fibonacci(n-3). (End)