OFFSET
1,2
COMMENTS
Each row begins with a power of 2 and ends with a Fibonacci number.
Alternating row sums: 1,0,0,0,0,0,0,0,0,0,0,...
For a discussion and guide to related arrays, see A208510.
As triangle T(n,k) with 0<=k<=n, it is (2, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 07 2012
FORMULA
u(n,x)=u(n-1,x)+(x+1)*v(n-1,x),
v(n,x)=(x+1)*u(n-1,x)+(x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
As triangle T(n,k), 0<=k<=n : T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 1, T(1,0) = T(1,1) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 07 2012
G.f.: (-1-x*y)*x*y/(-1+x*y+x^2*y^2+2*x+x^2*y). - R. J. Mathar, Aug 12 2015
EXAMPLE
First five rows:
1
2....2
4....7....3
8....20...17...5
16...52...65...37...8
First three polynomials v(n,x): 1, 2 + 2x, 4 + 7x + 3x^2.
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A209141 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A209142 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Mar 06 2012
STATUS
approved