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A symmetrical triangle based on the Fibonacci Polynomials: p(x,n)=f(n,x)+x^(n-1)*f(n,1/x)
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%I #2 Mar 30 2012 17:34:41

%S 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,

%T 10,6,4,1,2,0,17,0,30,0,17,0,2,1,5,8,20,21,21,20,8,5,1

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

%C Row sums are:

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

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

%F 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])]]);

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

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

%e {2},

%e {1, 1},

%e {2, 0, 2},

%e {1, 2, 2, 1},

%e {2, 0, 6, 0, 2},

%e {1, 3, 4, 4, 3, 1},

%e {2, 0, 11, 0, 11, 0, 2},

%e {1, 4, 6, 10, 10, 6, 4, 1},

%e {2, 0, 17, 0, 30, 0, 17, 0, 2},

%e {1, 5, 8, 20, 21, 21, 20, 8, 5, 1}

%t 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])]]);

%t Table[CoefficientList[FullSimplify[ExpandAll[f[n, x] + x^(n - 1)*f[n, 1/x]]], x], {n, 1, 10}];

%t Flatten[%]

%Y Cf. A168561

%K nonn,tabl,uned

%O 0,1

%A _Roger L. Bagula_, May 12 2010