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A209564
Triangle of coefficients of polynomials v(n,x) jointly generated with A209559; see the Formula section.
3
1, 1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 5, 11, 5, 1, 2, 5, 13, 21, 6, 1, 2, 5, 13, 32, 36, 7, 1, 2, 5, 13, 34, 72, 57, 8, 1, 2, 5, 13, 34, 87, 148, 85, 9, 1, 2, 5, 13, 34, 89, 212, 281, 121, 10, 1, 2, 5, 13, 34, 89, 231, 485, 499, 166, 11, 1, 2, 5, 13, 34, 89, 233, 585, 1039
OFFSET
1,3
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...2
1...2...3
1...2...5...4
1...2...5...11...1
First three polynomials v(n,x): 1, 1+2x , 1+2x+3x^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: A065158 A364842 A181842 * A029653 A067763 A343863
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Mar 10 2012
STATUS
approved