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a(n+2) = 5a(n+1) - 3a(n) (n >= 1); a(0) = 0, a(1) = 1, a(2) = 4.
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%I #16 Dec 27 2023 08:37:46

%S 0,1,4,17,73,314,1351,5813,25012,107621,463069,1992482,8573203,

%T 36888569,158723236,682950473,2938582657,12644061866,54404561359,

%U 234090621197,1007239421908,4333925245949,18647907964021,80237764082258

%N a(n+2) = 5a(n+1) - 3a(n) (n >= 1); a(0) = 0, a(1) = 1, a(2) = 4.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5, -3).

%F G.f.: (x-x^2)/(3*x^2-5*x+1) [From Harvey P. Dale, Aug 20 2011]

%t Join[{0},LinearRecurrence[{5,-3},{1,4},30]] (* _Harvey P. Dale_, Aug 20 2011 *)

%Y Cf. A018902; equals A095934 - A095939.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jul 13 2004

%E Extended by _Ray Chandler_, Jul 16 2004