|
%I #18 Sep 08 2022 08:46:16
%S 1,5,13,37,99,265,701,1849,4861,12761,33463,87697,229737,601693,
%T 1575629,4125661,10802107,28281881,74045509,193857841,507533181,
%U 1328750065,3478730543,9107463457,23843695249,62423679605,163427436301,427858779349,1120149144531
%N a(n) = F(n+1)*F(n+2) - F(n), where F = A000045 (Fibonacci numbers).
%C 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.
%C 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,...).
%H Clark Kimberling, <a href="/A269803/b269803.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-5,-1,1).
%F a(n) = F(n+1)F(n+2)-F(n), F = A000045 (Fibonacci numbers).
%F a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) + a(n-5).
%F 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
%t f[n_] := Fibonacci[n]; u = Table[f[n + 1] f[n + 2] - f[n], {n, 1, 40}]
%t LinearRecurrence[{3,1,-5,-1,1},{1,5,13,37,99},40] (* _Harvey P. Dale_, Jul 27 2021 *)
%o (PARI) a(n) = fibonacci(n+1)*fibonacci(n+2) - fibonacci(n); \\ _Altug Alkan_, Mar 06 2016
%o (Magma) [Fibonacci(n+1)*Fibonacci(n+2) - Fibonacci(n): n in [1..30]]; // _Vincenzo Librandi_, Mar 06 2016
%o (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
%Y Cf. A000045, A269802.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Mar 05 2016
|
