A197649 a(n) = Sum_{k=0..n} k*Fibonacci(2*k).

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

Offset: 0

Author

Gary Detlefs, Oct 16 2011

Keywords

• nonn
• easy

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.
• Rigoberto Flórez, Leandro Junes, Luisa M. Montoya, and José L. Ramírez, Counting Subwords in Non-Decreasing Dyck Paths, J. Int. Seq. (2025) Vol. 28, Art. No. 25.1.6. See pp. 15, 17, 19.
• E. Pérez Herrero, A small Fibonacci sum, Psychedelic Geometry Blogspot
• Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Crossrefs

Cf. A023619 (inverse binomial transform).

Cf. A001871, A001924, A104286.

Programs

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 *)

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) = A001871(n-1) + A001871(n-2). (End)

a(n) ~ c*n*(3 + sqrt(5))^n*2^(-n), where c = (5 + sqrt(5))/10. - Stefano Spezia, Mar 29 2022

E.g.f.: 2*exp(3*x/2)*(5*x*cosh(sqrt(5)*x/2) + sqrt(5)*(2*x - 1)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Mar 04 2025

Extensions

Identity 4 added by Gary Detlefs, Dec 22 2012