A190959 - OEIS

%I #38 Dec 24 2023 01:22:24

%S 0,1,3,4,-3,-29,-72,-71,147,796,1653,979,-5328,-20879,-35997,-3596,

%T 169197,525571,730728,-435671,-4960653,-12703604,-13307547,23595379,

%U 137323872,293994721,195364803,-883879196,-3628461603,-6465988829,-1255658472,28562968729

%N a(n) = 3*a(n-1) - 5*a(n-2), with a(0)=0, a(1)=1.

%C This is the Lucas U(P=3, Q=5) sequence. - _R. J. Mathar_, Oct 24 2012

%C a(n+2)/a(n+1) equals the continued fraction 3 - 5/(3 - 5/(3 - 5/(3 - ... - 5/3))) with n 5's. - _Greg Dresden_, Oct 06 2019

%H G. C. Greubel, <a href="/A190959/b190959.txt">Table of n, a(n) for n = 0..1000</a>

%H Wikipedia, <a href="http://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 (3,-5).

%F G.f.: x/(1 - 3*x + 5*x^2). - _Philippe Deléham_, Oct 11 2011

%F E.g.f.: 2*exp(3*x/2)*sin(sqrt(11)*x/2)/sqrt(11). - _Stefano Spezia_, Oct 06 2019

%t LinearRecurrence[{3,-5}, {0,1}, 50]

%o (PARI) x='x+O('x^30); concat([0], Vec(x/(1-3x+5*x^2))) \\ _G. C. Greubel_, Jan 25 2018

%o (Magma) I:=[0,1]; [n le 2 select I[n] else 3*Self(n-1) - 5*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jan 25 2018

%Y Cf. A190958 (index to generalized Fibonacci sequences), A190970 (binomial transf.), A106852 (inv. bin. transf., shifted).

%K sign,easy

%O 0,3

%A _Vladimir Joseph Stephan Orlovsky_, May 24 2011