A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

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

Offset: 1

Author

Clark Kimberling, May 03 2004

Keywords

• nonn
• tabl

Comments

Row sums yield the even-subscripted Fibonacci numbers (A001906).

Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - Clark Kimberling, Oct 22 2019

Examples

			Triangle starts:
   1;
   1,  2;
   2,  3,   3;
   3,  8,   6,   4;
   5, 15,  20,  10,  5;
   8, 30,  45,  40, 15,  6;
  13, 56, 105, 105, 70, 21, 7;
  ...
T(4,3) = F(2)*C(4,2) = 1*6 = 6.
		

Links

• G. C. Greubel, Rows n = 1..100 of triangle, flattened
• M. Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Crossrefs

Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.

Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.

Cf. A367208, A367209, A367210, A367211.

Programs

GAP

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

Magma

/* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 15 2017

Maple

with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch

Mathematica

Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 14 2016 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)

PARI

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

Sage

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

Formula

From Peter Bala, Aug 17 2007: (Start)

With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.

Other identities and formulas include:

F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;

Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);

F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;

F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.

The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.

F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.

The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.

G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)

From Peter Bala, Jun 29 2016: (Start)

Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.

d/dx(F(n,x)) = n*F(n-1,x).

F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.

F(n,x - 1) = (-1)^(n-1)*F(n,-x).

F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)

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

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

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

From Clark Kimberling, Oct 30 2019: (Start)

F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,

gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1. Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)

Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - Clark Kimberling, Nov 11 2023

Extensions

Error in expansion of generating function corrected by Peter Bala, Sep 24 2008