A209971 - OEIS

%I #20 Jan 22 2019 15:57:41

%S 0,2,4,8,16,34,76,176,416,994,2388,5752,13872,33474,80796,195040,

%T 470848,1136706,2744228,6625128,15994448,38613986,93222380,225058704,

%U 543339744,1311738146,3166815988,7645370072,18457556080,44560482178,107578520380,259717522880

%N a(n) = A000129(n) + n.

%D Paul Brickman, Problem in August 2011 issue of Fibonacci Quarterly. [Brickman has several problems in this issue, and I am not sure now which one I was referring to. - _N. J. A. Sloane_, Jan 22 2019]

%H Colin Barker, <a href="/A209971/b209971.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,0,1).

%F G.f.: 2*x*(-1+2*x) / ( (x^2+2*x-1)*(x-1)^2 ). a(n) = 2*A100131(n-1). - _R. J. Mathar_, Mar 27 2012

%F From _Colin Barker_, Nov 06 2017: (Start)

%F a(n) = (-(1-sqrt(2))^n + (1+sqrt(2))^n) / (2*sqrt(2)) + n.

%F a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4) for n>3.

%F (End)

%o (PARI) concat(0, Vec( 2*x*(1 - 2*x) / ((1 - x)^2*(1 - 2*x - x^2)) + O(x^50))) \\ _Colin Barker_, Nov 06 2017

%Y Cf. A000129.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Mar 25 2012