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A197649 Sum(k*Fibonacci(2*k), k=0..n), Fibonacci(n)=A000045(n) 3

%I

%S 0,1,7,31,115,390,1254,3893,11789,35045,102695,297516,853932,2432041,

%T 6881395,19361995,54214939,151164018,419910354,1162585565,3209268665,

%U 8835468881,24266461007,66501634776,181882282200,496539007825,1353272290399,3682496714743

%N Sum(k*Fibonacci(2*k), k=0..n), Fibonacci(n)=A000045(n)

%C There are only a small number of Fibonacci identities that can be solved for n. Some of these are

%C 1. n = (5*sum(F(2*k-1)^2,k=1..n)-F(4*n))/2 (Vajda #95).

%C 2. n = (sum(k*F(k),k=0..n)+F(n+3)-2)/F(n+2). (A104286)

%C 3. n = (a(n)+F(2*n))/F(2*n+1).

%C 4. n = F(n+4)- 3 - sum(F(k+2)-1,k=0..1) (A001924)

%C n can also be expressed in terms of phi=(1+sqrt(5))/2:

%C 5. n = floor(n*phi^3)-floor(2*n*phi).

%C 6. n = (floor(2*n*phi^2)-floor(2*n*phi))/2.

%H E. Pérez Herrero, <a href="http://psychedelic-geometry.blogspot.com.es/2010/05/small-fibonacci-sum_13.html">A small Fibonacci sum</a>, Psychedelic Geometry Blogspot

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (6,-11,6,-1).

%F a(n) = n*F(2*n+1)-F(2*n), For F(n)= Fibonacci(n)

%F a(n) = ((F(2*n+1)*((n-1)*h(n-1)-(n-1)*h(n-2))-h(n)*F(2*n))/h(n), n>2. Where h(n)=n-th harmonic number

%F G.f. x*(1+x) / ( (x^2-3*x+1)^2 ). - _R. J. Mathar_, Oct 17 2011

%F a(n) = A001871(n-1)+A001871(n-2). - _R. J. Mathar_, Oct 17 2011

%p a:=n->sum(k*fibonacci(2*k),n= 0..n):seq(a(n), n=0..25);

%t Table[Sum[k*Fibonacci[2*k], {k, 0, n}], {n, 0, 50}] (* _T. D. Noe_, Oct 17 2011 *)

%Y Cf. A023619 (inverse binomial transform)

%K nonn,easy

%O 0,3

%A _Gary Detlefs_, Oct 16 2011

%E Identity 4 added Dec 22 2012 by Gary Detlefs

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Last modified February 22 19:36 EST 2018. Contains 299469 sequences. (Running on oeis4.)