A094439 Triangular array T(n,k) = Fibonacci(k+4)*C(n,k), k=0..n, n>=0.

3, 3, 5, 3, 10, 8, 3, 15, 24, 13, 3, 20, 48, 52, 21, 3, 25, 80, 130, 105, 34, 3, 30, 120, 260, 315, 204, 55, 3, 35, 168, 455, 735, 714, 385, 89, 3, 40, 224, 728, 1470, 1904, 1540, 712, 144, 3, 45, 288, 1092, 2646, 4284, 4620, 3204, 1296, 233, 3, 50, 360, 1560, 4410, 8568, 11550, 10680, 6480, 2330, 377

Offset: 0

Author

Clark Kimberling, May 03 2004

Keywords

• nonn
• tabl

Comments

Let F(n) denote the n-th Fibonacci number (A000045). Then n-th row sum of T is F(2n+4) and n-th alternating row sum is -F(n-4).

Examples

			First few rows:
  3;
  3,  5;
  3, 10,  8;
  3, 15, 24,  13;
  3, 20, 48,  52,  21;
  3, 25, 80, 130, 105, 34;
		

Links

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

Crossrefs

Cf. A000045, A007318, A094435, A094436, A094437, A094438, A094440, A094441, A094442, A094443, A094444.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Binomial(n,k)* Fibonacci(k+4) ))); # G. C. Greubel, Oct 30 2019

Magma

[Binomial(n,k)*Fibonacci(k+4): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 30 2019

Maple

with(combinat); seq(seq(fibonacci(k+4)*binomial(n,k), k=0..n), n=0..12); # G. C. Greubel, Oct 30 2019

Mathematica

Table[Fibonacci[k+4]*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+4);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

Sage

[[binomial(n,k)*fibonacci(k+4) 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+4).

Sum_{k=0..n} T(n,k) = Fibonacci(2*n+4).

Sum_{k=0..n} (-1)^(k+1) * T(n,k) = Fibonacci(n-4). (End)