A156085 One fourth of the sum of the squares of the first n Fibonacci numbers with index divisible by 3.

0, 1, 17, 306, 5490, 98515, 1767779, 31721508, 569219364, 10214227045, 183286867445, 3288949386966, 59017802097942, 1059031488375991, 19003548988669895, 341004850307682120, 6119083756549608264, 109802502767585266633, 1970325966059985191129

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -5490, -306, -17, -1, 0, [0], 1, 17, 306, 5490, 98515, ... This is (-A156085)-reversed followed by A156085. That is, A156085(-n) = -A156085(n-1).

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..798
• C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7
• Index entries for linear recurrences with constant coefficients, signature (17, 17, -1).

Crossrefs

Cf. A156084, A156090, A156091.

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (1/4) Fibonacci[3k]^2, {k, 1, n} ], -Sum[ (1/4) Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[3*Range[0,20]]^2]/4 (* or *) LinearRecurrence[{17,17,-1},{0,1,17},30] (* Harvey P. Dale, Aug 17 2014 *)

PARI

concat(0, Vec(x/((1+x)*(1-18*x+x^2)) + O(x^25))) \\ Colin Barker, Mar 04 2016

PARI

a(n) = sum(k=1, n, fibonacci(3*k)^2)/4; \\ Michel Marcus, Mar 04 2016

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

a(n) = (1/4) sum_{k=1..n} F(3k)^2.

Closed form: a(n) = L(6n+3)/80 - (-1)^n/20.

Factored closed form: a(n) = (1/16) F(n) F(n+1) (L(n) - 1)(L(n) + 1)(L(2n+2) - 1) if n is even; a(n) = (1/16) F(n) F(n+1) (L(n+1) - 1)(L(n+1) + 1)(L(2n) - 1) if n is odd.

Recurrence: a(n) - 17 a(n-1) - 17 a(n-2) + a(n-3) = 0.

G.f.: A(x) = x/(1 - 17 x - 17 x^2 + x^3) = x/((1 + x)(1 - 18 x + x^2)).

a(n) = ((9+4*sqrt(5))^(-n)*(2-sqrt(5)-4*(-9-4*sqrt(5))^n+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)))/80. - Colin Barker, Mar 04 2016