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Nearest integer to absolute value of the function f(n) where f(n) is the derivative of F(n) = ((1/2+sqrt(5)/2)^n-(1/2-sqrt(5)/2)^n)/sqrt(5) with respect to n.
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%I #19 Apr 05 2016 22:32:22

%S 1,1,1,1,2,2,4,6,10,16,26,43,69,112,181,294,475,768,1243,2012,3255,

%T 5267,8523,13790,22313,36103,58416,94519,152934,247453,400387,647841,

%U 1048228,1696069,2744297,4440365,7184662,11625027,18809689,30434716,49244405,79679122,128923527,208602649,337526177

%N Nearest integer to absolute value of the function f(n) where f(n) is the derivative of F(n) = ((1/2+sqrt(5)/2)^n-(1/2-sqrt(5)/2)^n)/sqrt(5) with respect to n.

%C F(n) is the Fibonacci(n) for integer n.

%C Since F(n) is the sum of F(n-1) and F(n-2), the derivative of F(n) is simply the sum of the derivatives of F(n-1) and F(n-2). So sum of the two consecutive terms is generally equal to next term of this sequence.

%o (PARI) f(n) = -((sqrt(5)-1)^n*(log(-1)-log(2)+log(sqrt(5)-1))*(-1)^n+(1+sqrt(5))^n*(log(2)-log(sqrt(5)+1)))/(sqrt(5)*2^n);

%o a(n) = round(abs(f(n)));

%Y Cf. A000045.

%K nonn

%O 0,5

%A _Altug Alkan_, Apr 05 2016