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A209563
Triangle of coefficients of polynomials u(n,x) jointly generated with A209564; see the Formula section.
3
1, 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 3, 8, 10, 1, 1, 3, 8, 19, 15, 1, 1, 3, 8, 21, 40, 21, 1, 1, 3, 8, 21, 53, 76, 28, 1, 1, 3, 8, 21, 55, 125, 133, 36, 1, 1, 3, 8, 21, 55, 142, 273, 218, 45, 1, 1, 3, 8, 21, 55, 144, 354, 554, 339, 55, 1, 1, 3, 8, 21, 55, 144, 375, 839, 1053
OFFSET
1,5
COMMENTS
A209563: first k terms of row n are F(2),...,F(2k), where F = A000045 (Fibonacci numbers) and k=floor ((n+1)/2).
A209564: first k terms of row n are F(1), ..., F(2k-1), where k=floor ((n+2)/2).
For a discussion and guide to related arrays, see A208510.
FORMULA
u(n,x)=x*u(n-1,x)+v(n-1,x),
v(n,x)=x*u(n-1,x)+x*v(n-1,x) +1,
where u(1,x)=1, v(1,x)=1.
EXAMPLE
First five rows:
1
1...1
1...3...1
1...3...6...1
1...3...8...10...1
First three polynomials v(n,x): 1, 1 + x, 1 + 3x + x^2.
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
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[%] (* A209563 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A209564 *)
CROSSREFS
Sequence in context: A094644 A113046 A245541 * A308624 A133825 A365968
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Mar 10 2012
STATUS
approved