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 A197649 Sum(k*Fibonacci(2*k), k=0..n), Fibonacci(n)=A000045(n) 3
 0, 1, 7, 31, 115, 390, 1254, 3893, 11789, 35045, 102695, 297516, 853932, 2432041, 6881395, 19361995, 54214939, 151164018, 419910354, 1162585565, 3209268665, 8835468881, 24266461007, 66501634776, 181882282200, 496539007825, 1353272290399, 3682496714743 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS There are only a small number of Fibonacci identities that can be solved for n. Some of these are   1. n = (5*sum(F(2*k-1)^2,k=1..n)-F(4*n))/2 (Vajda #95).   2. n = (sum(k*F(k),k=0..n)+F(n+3)-2)/F(n+2). (A104286)   3. n = (a(n)+F(2*n))/F(2*n+1).   4. n = F(n+4)- 3 - sum(F(k+2)-1,k=0..1) (A001924) n can also be expressed in terms of phi=(1+sqrt(5))/2:   5. n = floor(n*phi^3)-floor(2*n*phi).   6. n = (floor(2*n*phi^2)-floor(2*n*phi))/2. LINKS E. Pérez Herrero, A small Fibonacci sum, Psychedelic Geometry Blogspot Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1). FORMULA a(n) = n*F(2*n+1)-F(2*n), For F(n)= Fibonacci(n) 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 G.f.  x*(1+x) / ( (x^2-3*x+1)^2 ). - R. J. Mathar, Oct 17 2011 a(n) = A001871(n-1)+A001871(n-2). - R. J. Mathar, Oct 17 2011 MAPLE a:=n->sum(k*fibonacci(2*k), n= 0..n):seq(a(n), n=0..25); MATHEMATICA Table[Sum[k*Fibonacci[2*k], {k, 0, n}], {n, 0, 50}] (* T. D. Noe, Oct 17 2011 *) CROSSREFS Cf. A023619 (inverse binomial transform) Sequence in context: A109756 A055580 A097786 * A006458 A091344 A032197 Adjacent sequences:  A197646 A197647 A197648 * A197650 A197651 A197652 KEYWORD nonn,easy AUTHOR Gary Detlefs, Oct 16 2011 EXTENSIONS Identity 4 added Dec 22 2012 by Gary Detlefs STATUS approved

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Last modified September 19 03:05 EDT 2018. Contains 315155 sequences. (Running on oeis4.)