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A093145
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Third binomial transform of Fibonacci(3n)/Fibonacci(3).
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6
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0, 1, 10, 80, 600, 4400, 32000, 232000, 1680000, 12160000, 88000000, 636800000, 4608000000, 33344000000, 241280000000, 1745920000000, 12633600000000, 91417600000000, 661504000000000, 4786688000000000, 34636800000000000
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OFFSET
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0,3
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COMMENTS
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Fifth binomial transform of 1,5,5,25,25,125. - Al Hakanson (hawkuu(AT)gmail.com), Jul 13 2009
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LINKS
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FORMULA
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G.f.: x/(1 - 10*x + 20*x^2).
a(n) = ((5+sqrt(5))^n - (5-sqrt(5))^n)/(2*sqrt(5)).
a(n) = Sum_{k=0..n} binomial(n, 2*k+ 1)*5^(n-k-1).
a(n) = 10*a(n-1) - 20*a(n-2), n > 1; a(0)=0, a(1)=1. - Zerinvary Lajos, Apr 26 2009
G.f.: A(x) = x*G(0)/(1-5*x) where G(k) = 1 + 5*x/(1-5*x - x*(1-5*x)/(x + (1-5*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 30 2012
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MATHEMATICA
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f[n_] := Block[{s = Sqrt@ 5}, Simplify[((1 + s)(5 + s)^n + (1 - s)(5 - s)^n)/2]]; Array[f, 21, 0] (* Or *)
a[n_] := 10 a[n - 1] - 20 a[n - 2]; a[0] = 0; a[1] = 1; Array[a, 22, 0] (* Or *)
CoefficientList[Series[x/(1 - 10 x + 20 x^2), {x, 0, 21}], x] (* Robert G. Wilson v, Mar 07 2011 *)
LinearRecurrence[{10, -20}, {0, 1}, 30] (* Harvey P. Dale, Jan 23 2019 *)
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PROG
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(Sage) [lucas_number1(n, 10, 20) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009
(Magma) [n le 2 select n - 1 else 10*Self(n-1)-20*Self(n-2): n in [1..25]]; // Vincenzo Librandi, Dec 30 2012
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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