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A094440 Triangular array T(n,k) = F(n+1-k)*C(n,k-1), k = 1,2,3,...,n; n >= 1. 6

%I

%S 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,

%T 7,21,104,224,280,210,112,28,8,34,189,468,672,630,378,168,36,9,55,340,

%U 945,1560,1680,1260,630,240,45,10,89,605,1870,3465,4290,3696,2310,990,330

%N Triangular array T(n,k) = F(n+1-k)*C(n,k-1), k = 1,2,3,...,n; n >= 1.

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

%F From _Peter Bala_, Aug 17 2007: (Start)

%F 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.

%F Other identities and formulas include:

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

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

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

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

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

%F 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.

%F 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)

%F From _Peter Bala_, Jun 29 2016: (Start)

%F 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.

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

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

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

%F 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)

%e Triangle starts:

%e 1;

%e 1, 2;

%e 2, 3, 3;

%e 3, 8, 6, 4;

%e ...

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

%p 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_

%t Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* _Harvey P. Dale_, Sep 14 2016 *)

%o (MAGMA) /* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Aug 15 2017

%Y Cf. A000045, A001906, A030191, A081574, A094435, A132148.

%K nonn,tabl

%O 1,3

%A _Clark Kimberling_, May 03 2004

%E Error in expansion of generating function corrected by _Peter Bala_, Sep 24 2008

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Last modified July 17 14:44 EDT 2019. Contains 325106 sequences. (Running on oeis4.)