OFFSET
1,2
COMMENTS
NI(F(n+1)/a(n)) = (n,n,n,n,n,...), where NI(x) denotes the r-nested-interval sequence of x, and r = (1/1, 1/2, 1/3, 1/5, 1/8, ...), the reciprocals of Fibonacci numbers. Definitions follow. Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) , x <= r(n), and let L(1) = r(n(1))-r(n(1)+1). Let n(2) be the index n such that r(n(1)+1) < x <= r(n(1)+1) + L(1)r(n), and let L(2) = (r(n(2))-r(r(n)+1)L(1). Continue inductively to obtain the sequence (n(1), n(2), n(3), ... ), the r-nested interval sequence of x.
Conversely, given a sequence s= (n(1),n(2),n(3),...) of positive integers, the number x having satisfying NI(x) = s, is the sum of left-endpoints of nested intervals (r(n(k)+1), r(n(k))]; i.e., x = sum{L(k)r(n(k+1)+1), k >=1}, where L(0) = 1. Thus, for r = (1/F(n+1)), the number F(n+1)F(n+2)/a(n) is the only x for which NI(x) = (n,n,n,...).
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).
FORMULA
a(n) = F(n+1)F(n+2)-F(n), F = A000045 (Fibonacci numbers).
a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+2*x-3*x^2-2*x^3+x^4) / ((1+x)*(1-3*x+x^2)*(1-x-x^2)). - Colin Barker, Mar 06 2016
MATHEMATICA
f[n_] := Fibonacci[n]; u = Table[f[n + 1] f[n + 2] - f[n], {n, 1, 40}]
LinearRecurrence[{3, 1, -5, -1, 1}, {1, 5, 13, 37, 99}, 40] (* Harvey P. Dale, Jul 27 2021 *)
PROG
(PARI) a(n) = fibonacci(n+1)*fibonacci(n+2) - fibonacci(n); \\ Altug Alkan, Mar 06 2016
(Magma) [Fibonacci(n+1)*Fibonacci(n+2) - Fibonacci(n): n in [1..30]]; // Vincenzo Librandi, Mar 06 2016
(PARI) Vec(x*(1+2*x-3*x^2-2*x^3+x^4)/((1+x)*(1-3*x+x^2)*(1-x-x^2)) + O(x^50)) \\ Colin Barker, Mar 06 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 05 2016
STATUS
approved