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A219672
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a(n) = Sum_{k=0..n} binomial(n,k)^2*Fibonacci(k).
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3
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0, 1, 5, 20, 87, 405, 1924, 9225, 44625, 217528, 1066725, 5256087, 26001000, 129053365, 642376709, 3205403100, 16029187391, 80309053285, 403040543420, 2025751379997, 10195547237235, 51376594943136, 259180112907875, 1308811957775785, 6615383878581072
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OFFSET
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0,3
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LINKS
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FORMULA
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G.f.: (1/sqrt(1 - (3 + sqrt(5))*x + (3 - sqrt(5))/2*x^2) - 1/sqrt(1 - (3 - sqrt(5))*x + (3 + sqrt(5))/2*x^2))/sqrt(5)
a(n) ~ (1+sqrt(5))/4*sqrt((6-2*sqrt(5)+sqrt(2*sqrt(5)-2))/(10*Pi*n)) * ((3+sqrt(5))/2+sqrt(2+2*sqrt(5)))^n
D-finite Recurrence: (n-1)*n*(13*n^2 - 52*n + 49)*a(n) = 3*(n-1)*(2*n-5)*(13*n^2 - 26*n + 10)*a(n-1) - (7*n^2-14*n+6)*(13*n^2 - 52*n + 49)*a(n-2) + (n-2)*(182*n^3 - 819*n^2 + 1050*n - 351)*a(n-3) - (n-3)*(n-2)*(13*n^2 - 26*n + 10)*a(n-4)
a(n) = (hypergeom([-n,-n], [1], phi) - hypergeom([-n,-n], [1], 1-phi))/sqrt(5) = ((1-phi)^n * P_n(-sqrt(5)-2) - phi^n * P_n(sqrt(5)-2))/sqrt(5), where phi = (1+sqrt(5))/2, P_n(x) is the Legendre polynomial. - Vladimir Reshetnikov, Sep 28 2016
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MATHEMATICA
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Table[Sum[Binomial[n, k]^2*Fibonacci[k], {k, 0, n}], {n, 0, 20}]
FullSimplify@Table[((1 - GoldenRatio)^n LegendreP[n, -Sqrt[5] - 2] - GoldenRatio^n LegendreP[n, Sqrt[5] - 2])/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 28 2016 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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