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 = (-F(4*n) + 5*Sum_{k=1..n} F(2*k-1)^2)/2 (Vajda #95).
2. n = (F(n+3) - 2 + Sum_{k=0..n} k*F(k))/F(n+2). (A104286)
3. n = (a(n) + F(2*n))/F(2*n+1).
4. n = F(n+4) - 3 - Sum_{k=0..1} (F(k+2) - 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
Michael De Vlieger, Table of n, a(n) for n = 0..2384
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Honeycombs in the Pascal triangle and beyond, arXiv:2203.13205 [math.HO], 2022. See p. 5.
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), where 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) is the n-th harmonic number.
From R. J. Mathar, Oct 17 2011: (Start)
G.f.: x*(1+x) / ( (x^2-3*x+1)^2 ).
a(n) ~ c*n*(3 + sqrt(5))^n*2^(-n), where c = (5 + sqrt(5))/10. - Stefano Spezia, Mar 29 2022
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
KEYWORD
nonn,easy
AUTHOR
Gary Detlefs, Oct 16 2011
EXTENSIONS
Identity 4 added by Gary Detlefs, Dec 22 2012
STATUS
approved