A188599 - OEIS

%I #59 Jul 03 2021 19:25:14

%S 0,1,6,11,-84,-779,-2574,4031,88536,430441,369246,-8545549,-60504444,

%T -149387939,616283466,7432399271,29187308976,-10686127919,

%U -793799491914,-4495643753509,-7128875223204,69617842498501,595928935571106,1835127550964111,-3887458083492984

%N Expansion of x/(1-6*x+25*x^2).

%C Original name: G.f.: 1/(1-6*x+25*x^2).

%C Suggested by _Philippe Flajolet_ as an example of a simple formula for which the general term is hard to guess because 1-6*x+25*x^2 has 2 complex roots of equal size and modulus 1.

%C The Lucas sequence U_n(6,25). - _Peter Bala_, Feb 02 2017

%D Discussion in 1993 at the FPSAC 1993 in Florence.

%H Alois P. Heinz, <a href="/A188599/b188599.txt">Table of n, a(n) for n = 0..1000</a> (first 110 terms from Vincenzo Librandi)

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-25).

%F a(n) = ((3+4*i)^n-(3-4*i)^n)/8/i, where i=sqrt(-1). - Denis Excoffier, Jan 19 2013

%F From _Peter Bala_, Feb 02 2017: (Start)

%F a(n) = (1/4)*( Re((2 - i)^n)*Im((2 + i)^n) - Re((2 + i)^n)*Im((2 - i)^n) ).

%F a(n) = (1/2) * the directed or signed area of the triangle in the complex plane with vertices at the points 0, (2 - i)^n and (2 + i)^n. (End)

%F a(n) = 5^n*sin(n*arctan(1/2))*cos(n*arctan(1/2))/2. - _Peter Luschny_, Feb 02 2017

%F E.g.f.: (1/4)*exp(3*x)*sin(4*x). - _Stefano Spezia_, Feb 01 2020

%p x/(1-6*x+25*x^2):series(%,x,44):seriestolist(%);

%t Table[Im[(3 + 4*I)^n]/4, {n, 0, 22}] (* _Jean-François Alcover_, Jun 14 2011 *)

%t CoefficientList[Series[x/(1-6*x+25*x^2),{x,0,30}],x] (* _Harvey P. Dale_, Dec 01 2018 *)

%t LinearRecurrence[{6,-25},{0,1},30] (* _Harvey P. Dale_, Jul 03 2021 *)

%o (PARI) Vec(x/(1-6*x+25*x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Jun 14 2011

%K sign,easy

%O 0,3

%A _Simon Plouffe_, Apr 06 2011

%E Minor edits by _N. J. A. Sloane_, Apr 06 2011

%E Minor modification to Name by _Peter Bala_, Feb 02 2017