A269803 a(n) = F(n+1)*F(n+2) - F(n), where F = A000045 (Fibonacci numbers).

1, 5, 13, 37, 99, 265, 701, 1849, 4861, 12761, 33463, 87697, 229737, 601693, 1575629, 4125661, 10802107, 28281881, 74045509, 193857841, 507533181, 1328750065, 3478730543, 9107463457, 23843695249, 62423679605, 163427436301, 427858779349, 1120149144531

Offset: 1

Author

Clark Kimberling, Mar 05 2016

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A000045, A269802.

Programs

Magma

[Fibonacci(n+1)*Fibonacci(n+2) - Fibonacci(n): n in [1..30]]; // Vincenzo Librandi, 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 *)

PARI

a(n) = fibonacci(n+1)*fibonacci(n+2) - fibonacci(n); \\ Altug Alkan, 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

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