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%I #11 Jul 22 2022 19:00:02
%S 0,1,8,53,316,1785,9744,51997,273092,1417889,7299160,37334661,
%T 190028748,963565513,4871514656,24572321645,123720601684,622038982257,
%U 3123938806632,15674669614549,78593250398300,393845861293721
%N a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1)*(5/4)^k.
%C In general a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1) * u^(n-k-1) * (v/u)^(k-1) has g.f. x^2/((1-u*x) * (1-u*x-v*x^2)) and satisfies the recurrence a(n) = 2*u*a(n-1) - (u^2 - v)*a(n-2) - u*v*a(n-3).
%H G. C. Greubel, <a href="/A099622/b099622.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-11,-20).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1)*(5/4)^k.
%F a(n) = 8*a(n-1) - 11*a(n-2) - 20*a(n-3).
%F G.f.: x^2/((1-4*x)*(1-4*x-5*x^2)) = x^2/((1+x)*(1-4*x)*(1-5*x)).
%F From _G. C. Greubel_, Jul 22 2022: (Start)
%F a(n) = (1/30)*(5^(n+2) - 6*4^(n+1) - (-1)^n).
%F E.g.f.: (1/30)*(25*exp(5*x) - 24*exp(4*x) - exp(-x)). (End)
%t LinearRecurrence[{8,-11,-20},{0,1,8},30] (* _Harvey P. Dale_, Nov 05 2017 *)
%o (Magma) [(5^(n+2) -6*4^(n+1) -(-1)^n)/30: n in [0..40]]; // _G. C. Greubel_, Jul 22 2022
%o (SageMath) [(5^(n+2) -6*4^(n+1) -(-1)^n)/30 for n in (0..40)] # _G. C. Greubel_, Jul 22 2022
%Y Cf. A094705, A099582, A099621.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Oct 25 2004