OFFSET
0,3
COMMENTS
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).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. M. Bătineţu-Giurgiu and N. Stanciu, Problem B-1179, The Fibonacci Quarterly, Volume 53, Number 4 (November 2015), p. 366.
Index entries for linear recurrences with constant coefficients, signature (6,-9,-6,20,-6,-9,6,-1).
FORMULA
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).
MAPLE
with(combinat, fibonacci): A292465:=seq(n*fibonacci(n)*fibonacci(n+1), n=0..10^2); # Muniru A Asiru, Sep 26 2017
MATHEMATICA
Table[n Fibonacci[n] Fibonacci[n+1], {n, 0, 30}]
PROG
(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)
A292465:=List([0..10^3], n->n*Fibonacci(n)*Fibonacci(n+1)); # Muniru A Asiru, Sep 26 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Sep 17 2017
STATUS
approved