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A163199
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Sum of the cubes of the first n even-indexed Fibonacci numbers, minus 1.
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6
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-1, 0, 27, 539, 9800, 176175, 3162159, 56744792, 1018249595, 18271762299, 327873509424, 5883451505855, 105574253853887, 1894453118539344, 33994581881622075, 610008020755286075, 10946149791725643704, 196420688230338021807, 3524626238354441796015, 63246851602149831726824
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OFFSET
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0,3
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COMMENTS
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Natural bilateral extension (brackets mark index 0): ..., 9800, 539, 27, 0, -1, [-1], 0, 27, 539, 9800, 176175, ... This is A163199-reversed followed by A163199. That is, A163199(-n) = A163199(n-1).
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LINKS
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FORMULA
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Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = Sum_{k=1..n} F(2k)^3 - 1.
a(n) = Sum_{k=2..n} F(2k)^3 for n > 0.
a(n) = (1/20)*(F(6n+3) - 12*F(2n+1) - 10).
a(n) = (1/4)*(F(2n+1)^3 - 3*F(2n+1) - 2).
a(n) = (1/4)*L(n)^2*F(n+1)^2*L(n-1)*F(n+2) = A163197(n) if n is even.
a(n) = (1/4)*F(n)^2*L(n+1)^2*F(n-1)*L(n+2) = A163195(n) if n is odd.
a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = -8.
a(n) - 22 a(n-1) + 77 a(n-2) - 77 a(n-3) + 22 a(n-4) - a(n-5) = 0.
G.f.: (-1 + 22*x - 50*x^2 + 22*x^3 - x^4)/(1 - 22*x + 77*x^2 - 77*x^3 + 22*x^4 - x^5) = -(1 - 22*x + 50*x^2 - 22*x^3 + x^4)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)).
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MATHEMATICA
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a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k]^3, {k, 1, n} ] - 1, -Sum[ Fibonacci[-2k]^3, {k, 1, -n - 1} ] - 1 ]
Accumulate[Fibonacci[Range[0, 40, 2]]^3]-1 (* Harvey P. Dale, Jan 03 2016 *)
LinearRecurrence[{22, -77, 77, -22, 1}, {-1, 0, 27, 539, 9800}, 50] (* or *) Table[(1/20)*(F(6n+3) - 12*F(2n+1) - 10), {n, 0, 25}] (* G. C. Greubel, Dec 09 2016 *)
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PROG
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(PARI) Vec(-(1 - 22*x + 50*x^2 - 22*x^3 + x^4)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)) + O(x^50)) \\ G. C. Greubel, Dec 09 2016
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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