login
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n)= (a(n-1)+ c)/(a(n-1)+1); with c=5 and a(1)=0. Also product of the powers of two and five times the Fibonacci numbers.
2

%I #15 Mar 08 2021 11:44:26

%S 0,5,10,40,120,400,1280,4160,13440,43520,140800,455680,1474560,

%T 4771840,15441920,49971200,161710080,523304960,1693450240,5480120320,

%U 17734041600,57388564480,185713295360,600980848640,1944814878720

%N Numerators of fractions in the approximation of the square root of 5 satisfying: a(n)= (a(n-1)+ c)/(a(n-1)+1); with c=5 and a(1)=0. Also product of the powers of two and five times the Fibonacci numbers.

%C For denominators see: A084057 (= product of Lucas numbers (excluding first number (2)), and powers of 2).

%H G. C. Greubel, <a href="/A163305/b163305.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,4).

%F From _Colin Barker_, Jun 20 2012: (Start)

%F a(n) = 2*a(n-1) + 4*a(n-2).

%F G.f.: 5*x^2/(1-2*x-4*x^2). (End)

%F a(n) = 5*A063727(n-1). - _R. J. Mathar_, Mar 08 2021

%t LinearRecurrence[{2,4},{0,5},30] (* _Harvey P. Dale_, Mar 01 2016 *)

%o (PARI) a(n)=5*fibonacci(n-1)*2^(n-2) \\ _Franklin T. Adams-Watters_, Aug 06 2009

%Y Cf. A000032, A000079, A084057, A000045.

%K nonn,easy

%O 1,2

%A _Mark Dols_, Jul 24 2009

%E More terms from _Franklin T. Adams-Watters_, Aug 06 2009