A208514 Triangle of coefficients of polynomials u(n,x) jointly generated with A208515; see the Formula section.

1, 1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 4, 6, 7, 5, 1, 5, 8, 12, 13, 8, 1, 6, 10, 18, 24, 23, 13, 1, 7, 12, 25, 38, 46, 41, 21, 1, 8, 14, 33, 55, 78, 88, 72, 34, 1, 9, 16, 42, 75, 120, 158, 165, 126, 55, 1, 10, 18, 52, 98, 173, 255, 313, 307, 219, 89, 1, 11, 20, 63, 124, 238

Offset

1,5

Comments

u(n,n) = Fibonacci(n), A000045

u(n+1,n) = A208354(n)

col 1: A000012

col 2: A000027

col 3: A005843

col 4: A055998

col 5: A140090

Links

Table of n, a(n) for n=1..72.

Formula

u(n,x)=u(n-1,x)+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...2...2
1...3...4...3
1...4...6...7...5
First five polynomials u(n,x):
1
1 + x
1 + 2x + 2x^2
1 + 3x + 4x^2 + 3x^3
1 + 4x + 6x^2 + 7x^3 + 5x^4

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + 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[%]    (* A208514 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208515 *)

Crossrefs

Cf. A208515.

Sequence in context: A323899 A182630 A208805 * A179901 A209561 A283822

Adjacent sequences: A208511 A208512 A208513 * A208515 A208516 A208517

Keyword

nonn,tabl

Author

Clark Kimberling, Feb 28 2012

Status

approved