%I #14 Sep 16 2020 19:53:50
%S 0,1,1,2,4,7,12,19,31,50,82,133,216,349,565,914,1480,2395,3876,6271,
%T 10147,16418,26566,42985,69552,112537,182089,294626,476716,771343,
%U 1248060,2019403,3267463,5286866,8554330,13841197,22395528,36236725,58632253,94868978,153501232,248370211,401871444,650241655,1052113099,1702354754
%N Expansion of x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).
%C The integer part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2).
%C The o.g.f. for the numerators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is 6*x/((1 + x - x^2)*(1 - 4*x - x^2)).
%C The o.g.f. for the denominators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is (1 + 3*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
%C Convolution of Fibonacci numbers and periodic sequence [1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, ...].
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicMean.html">Harmonic Mean</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,2,0,-1).
%F G.f.: x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).
%F a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - a(n-6).
%F a(n) = (9*sqrt(5)*(((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n) + 5*((-1)^n + 2*cos(Pi*n/3) - 3))/30.
%F a(n) = floor(3*F(n)*F(n+1)*F(n+2)/(2*F(n+1)*F(n+2)-(-1)^n)), where F(n) is the n-th Fibonacci number (A000045).
%F a(n) = floor(3*A065563(n)/A236428(n+1)).
%F a(n) = 3*A000045(n)/2 + ((-1)^n + 2*cos(Pi*n/3) - 3)/6.
%F a(n) ~ 3*phi^n/(2*sqrt(5)), where phi is the golden ratio (A001622).
%F Lim_{n->infinity} a(n+1)/a(n) = phi.
%e a(1) = floor(3/(1/F(1)+1/F(2)+1/F(3))) = floor(3/(1/1+1/1+1/2)) = 1;
%e a(2) = floor(3/(1/F(2)+1/F(3)+1/F(4))) = floor(3/(1/1+1/2+1/3)) = 1;
%e a(3) = floor(3/(1/F(3)+1/F(4)+1/F(5))) = floor(3/(1/2+1/3+1/5)) = 2, etc.
%t LinearRecurrence[{2, 0, -2, 2, 0, -1}, {0, 1, 1, 2, 4, 7}, 46]
%t Table[Floor[3 Fibonacci[n] Fibonacci[n + 1] Fibonacci[n + 2]/(2 Fibonacci[n + 1] Fibonacci[n + 2] - (-1)^n)], {n, 0, 45}]
%o (PARI) concat(0, Vec((x*(1-x+2*x^3-x^4)/((1-x)*(1+x)*(1-x+x^2))) + O(x^40))) \\ _Felix Fröhlich_, Dec 22 2016
%Y Cf. A000045, A001622, A004695, A065563, A236428.
%Y Cf. A062114 (the integer part of the harmonic mean of Fibonacci(n+1) and Fibonacci(n+2) for n>0).
%Y Cf. A074331 (the integer part of the geometric mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2)).
%K nonn,easy
%O 0,4
%A _Ilya Gutkovskiy_, Dec 22 2016