A094435 Triangular array read by rows: T(n,k) = Fibonacci(k)*C(n,k), k = 1...n; n>=1.

1, 2, 1, 3, 3, 2, 4, 6, 8, 3, 5, 10, 20, 15, 5, 6, 15, 40, 45, 30, 8, 7, 21, 70, 105, 105, 56, 13, 8, 28, 112, 210, 280, 224, 104, 21, 9, 36, 168, 378, 630, 672, 468, 189, 34, 10, 45, 240, 630, 1260, 1680, 1560, 945, 340, 55, 11, 55, 330, 990, 2310, 3696, 4290, 3465, 1870, 605, 89

Offset: 1

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

Examples

			First few rows:
  1;
  2   1;
  3   3   2;
  4   6   8   3;
  5, 10, 20, 15,  5;
  6, 15, 40, 45, 30, 8;
		

Links

• G. C. Greubel, Rows n = 1..100 of triangle, flattened

Crossrefs

Cf. A000045.

Cf. A094436, A094437, A094438, A094439, A094440, A094441, A094442, A094443, A094444.

Programs

GAP

Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k)*Fibonacci(k) ))); # G. C. Greubel, Oct 30 2019

Magma

[Binomial(n,k)*Fibonacci(k): k in [1..n], n in [1..12]]; // G. C. Greubel, Oct 30 2019

Maple

with(combinat); seq(seq(binomial(n,k)*fibonacci(k), k=1..n), n=1..12); # G. C. Greubel, Oct 30 2019

Mathematica

Table[Fibonacci[k]*Binomial[n, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Oct 30 2019 *)

PARI

T(n,k) = binomial(n,k)*fibonacci(k);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

Sage

[[binomial(n,k)*fibonacci(k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019

Formula

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

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

Sum_{k=1..n} binomial(n,k)*Fibonacci(k) = Fibonacci(2*n).

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