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A292465
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a(n) = n*F(n)*F(n+1), where F = A000045.
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1
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0, 1, 4, 18, 60, 200, 624, 1911, 5712, 16830, 48950, 140976, 402624, 1141933, 3219580, 9031050, 25219824, 70153016, 194466672, 537404835, 1480993800, 4071156726, 11165970794, 30561658848, 83490220800, 227687745625, 619938027124, 1685442626946, 4575973716132
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OFFSET
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0,3
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COMMENTS
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Inequality proposed by Bătineţu-Giurgiu and Stanciu (see References): Let {x(n)}_{n>=1} be a sequence of real numbers. Prove that 2*(Sum_{k=1..n} F(k)*sin(x(k)))*(Sum_{k=1..n} F(k)*cos(x(k))) <= n*F(n)*F(n+1).
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LINKS
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D. M. Bătineţu-Giurgiu and N. Stanciu, Problem B-1179, The Fibonacci Quarterly, Volume 53, Number 4 (November 2015), p. 366.
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FORMULA
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G.f.: x*(1 - 2*x + 3*x^2 - 6*x^3 + 6*x^4 - 2*x^5)/((1 - x)^2*(1 + x)^2*(1 - 3*x + x^2)^2).
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MAPLE
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with(combinat, fibonacci): A292465:=seq(n*fibonacci(n)*fibonacci(n+1), n=0..10^2); # Muniru A Asiru, Sep 26 2017
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MATHEMATICA
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Table[n Fibonacci[n] Fibonacci[n+1], {n, 0, 30}]
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PROG
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(Magma) [n*Fibonacci(n)*Fibonacci(n+1): n in [0..35]];
(PARI) a(n) = n*fibonacci(n)*fibonacci(n+1); \\ Altug Alkan, Sep 17 2017
(GAP)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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