1,3

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

Comments from Peter Bala (pbala(AT)toucansurf.com), Aug 17 2007: (Start)

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

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

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 A081574.

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

Triangle starts:

1;

1, 2;

2, 3, 3;

3, 8, 6, 4;

T(4,3)=F(2)*C(4,2)=1*6=6

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 (Deutsch)

Cf. A094435, A000045.

Cf. A001906.

Cf. A132148.

Sequence in context: A025496 A099959 A099964 this_sequence A093736 A076938 A014589

Adjacent sequences: A094437 A094438 A094439 this_sequence A094441 A094442 A094443

nonn,tabl

Clark Kimberling (ck6(AT)evansville.edu), May 03 2004

Corrected error in expansion of generating function. - Peter Bala (pbala(AT)toucansurf.com), Sep 24 2008