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%I #18 Sep 08 2022 08:44:58
%S 1,22,399,7164,128557,2306866,41395035,742803768,13329072793,
%T 239180506510,4291920044391,77015380292532,1381984925221189,
%U 24798713273688874,444994854001178547,7985108658747524976,143286961003454271025,2571180189403429353478
%N a(n) = (F(6*n+5) - 1)/4, where F=A000045 (the Fibonacci sequence).
%H Colin Barker, <a href="/A049663/b049663.txt">Table of n, a(n) for n = 0..700</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (19,-19,1).
%F G.f.: (1+3*x)/( (1-x)*(x^2-18*x+1) ). - _R. J. Mathar_, Oct 26 2015
%F a(n) = A049664(n+1) + 3*A049664(n). - _R. J. Mathar_, Oct 26 2015
%F From _Colin Barker_, Mar 04 2016: (Start)
%F a(n) = (-1/4+1/40*(9+4*sqrt(5))^(-n)*(25-11*sqrt(5)+(9+4*sqrt(5))^(2*n)*(25+11*sqrt(5)))).
%F a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3) for n>2. (End)
%t (Fibonacci[6*Range[0,20]+5]-1)/4 (* or *) LinearRecurrence[{19,-19,1},{1,22,399},20] (* _Harvey P. Dale_, Sep 22 2016 *)
%o (PARI) Vec((1+3*x)/((1-x)*(1-18*x+x^2)) + O(x^25)) \\ _Colin Barker_, Mar 04 2016
%o (PARI) for(n=0,30, print1((fibonacci(6*n+5) - 1)/4, ", ")) \\ _G. C. Greubel_, Dec 02 2017
%o (Magma) [(Fibonacci(6*n+5) - 1)/4: n in [0..30]]; // _G. C. Greubel_, Dec 02 2017
%K nonn,easy
%O 0,2
%A _Clark Kimberling_
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