A177717 A symmetrical triangle based on the Fibonacci Polynomials: p(x,n)=f(n,x)+x^(n-1)*f(n,1/x).

2, 1, 1, 2, 0, 2, 1, 2, 2, 1, 2, 0, 6, 0, 2, 1, 3, 4, 4, 3, 1, 2, 0, 11, 0, 11, 0, 2, 1, 4, 6, 10, 10, 6, 4, 1, 2, 0, 17, 0, 30, 0, 17, 0, 2, 1, 5, 8, 20, 21, 21, 20, 8, 5, 1

Offset: 0

Author

Roger L. Bagula, May 12 2010

Keywords

• nonn
• tabl
• uned

Comments

Row sums are:

{2, 2, 4, 6, 10, 16, 26, 42, 68, 110,...}

Examples

			{2},
{1, 1},
{2, 0, 2},
{1, 2, 2, 1},
{2, 0, 6, 0, 2},
{1, 3, 4, 4, 3, 1},
{2, 0, 11, 0, 11, 0, 2},
{1, 4, 6, 10, 10, 6, 4, 1},
{2, 0, 17, 0, 30, 0, 17, 0, 2},
{1, 5, 8, 20, 21, 21, 20, 8, 5, 1}
		

References

• Function form used from:http://functions.wolfram.com/HypergeometricFunctions/Fibonacci2General/26/01/02/0001/

Crossrefs

Cf. A168561

Programs

Mathematica

f[n_, z_] := (1/(2 Sqrt[4 + z^2])) (-HypergeometricPFQ[{}, {}, n ((-I) Pi - \ Log[(1/2) (z + Sqrt[4 + z^2])])] - HypergeometricPFQ[{}, {}, n (I Pi - Log[( 1/2) (z + Sqrt[4 + z^2])])] + 2 HypergeometricPFQ[{}, {}, n Log[(1/ 2) (z + Sqrt[4 + z^2])]]);
Table[CoefficientList[FullSimplify[ExpandAll[f[n, x] + x^(n - 1)*f[n, 1/x]]], x], {n, 1, 10}];
Flatten[%]

Formula

f(x,n)=(1/(2 Sqrt[4 + z^2])) (-HypergeometricPFQ[{}, {}, n ((-I) Pi - Log[(1/2) (z + \ Sqrt[4 + z^2])])] - HypergeometricPFQ[{}, {}, n (I Pi - Log[(1/2) (z + Sqrt[ 4 + z^2])])] + 2 HypergeometricPFQ[{}, {}, n Log[(1/2) (z + Sqrt[4 + z^2])]]);

p(x,n)=f(x,n)+x^(n-1)*f(1/x,n);

t(n,m)=coefficients(p(x,n))