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a(n)= 5*a(n-1) -a(n-2) -a(n-3).
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%I #10 Jan 21 2023 14:43:33

%S 1,1,1,3,13,61,289,1371,6505,30865,146449,694875,3297061,15643981,

%T 74227969,352198803,1671122065,7929183553,37622596897,178512678867,

%U 847011613885,4018922793661,19069089675553,90479513970219

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

%D Roger Bagula, Factoring Double Fibonacci Sequences, 2000

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (5,-1,-1)

%F G.f.: (1-4*x-3*x^2)/(1-5*x+x^2+x^3) [Sep 28 2009]

%t F[1] = 1; F[2] = 1; F[3] = 1; F[n__] := F[n] = 5*F[n - 1] - F[n - 2] - F[n - 3] a = Table[Abs[F[n]], {n, 1, 50}]

%t LinearRecurrence[{5,-1,-1},{1,1,1},30] (* _Harvey P. Dale_, Jan 21 2023 *)

%Y Cf. A056015, A056016.

%K nonn,easy

%O 0,4

%A _Roger L. Bagula_, Jun 05 2005

%E Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009