A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.
1, 0, 3, 5, 16, 39, 105, 272, 715, 1869, 4896, 12815, 33553, 87840, 229971, 602069, 1576240, 4126647, 10803705, 28284464, 74049691, 193864605, 507544128, 1328767775, 3478759201, 9107509824, 23843770275, 62423800997, 163427632720, 427859097159, 1120149658761
Offset: 1
Examples
G.f. = x + 3*x^3 + 5*x^4 + 16*x^5 + 39*x^6 + 105*x^7 + 272*x^8 + 715*x^9 + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- John P. Bonomo and Montana Ferita, A Small Fib, College Math. J., 2023.
- Nurettin Irmak, Product of arbitrary Fibonacci numbers with distance 1 to Fibonomial coefficient, Turk J Math, (2017) 41: 825-828. See p. 828.
- C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Série 4, Vol. 19, 391-397.
- Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
Crossrefs
Programs
-
Magma
[Fibonacci(n)^2-Fibonacci(n-1)^2: n in [1..40]]; // Vincenzo Librandi, Jun 18 2014
-
Maple
a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,0,3>>)[1, 1]: seq(a(n), n=0..30); # Alois P. Heinz, Sep 28 2016
-
Mathematica
a[ n_] := Fibonacci[n + 1] Fibonacci[n - 2]; (* Michael Somos, Jun 17 2014 *) CoefficientList[Series[(1 - x)^2/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)
-
PARI
{a(n) = fibonacci( n + 1) * fibonacci( n - 2)}; -
PARI
a(n) = round(2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ Colin Barker, Sep 28 2016
-
PARI
lista(nn) = {my(p = (3*x-1)/(x^3-2*x^2-2*x+1)); for (n=1, nn, p = deriv(p, x); print1(subst(p, x, 0)/n!, ", "); ); } \\ Michel Marcus, May 22 2018
Formula
G.f.: x * (1 - x)^2 / ((1 + x) * (1 -3*x + x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = - A121646(n).
a(n) = -a(1-n) for all n in Z.
a(n) = A121801(n+1) / 2. - Michael Somos, Jun 17 2014
a(n) = a(n-1) + A000045(n-1)^2 - 2*(-1)^n, for n>1. - Alexander Samokrutov, Sep 07 2015
a(n) = F(n-1)*F(n) - (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = 2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5. - Colin Barker, Sep 28 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=3} 1/a(n) = (1/2) * A290565 - 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = (3/2) * (1/phi - 1/2), where phi is the golden ratio (A001622). (End)
Comments