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a(n) = Sum_{k=0..floor(n/2)} C(n-k,k+2)*3^(n-k-2)*(4/3)^k.
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%I #9 Dec 16 2024 05:44:34

%S 0,0,1,9,58,318,1591,7503,33976,149436,643261,2724357,11395654,

%T 47210154,194121811,793526571,3228811492,13090123272,52917410041,

%U 213437246145,859342367890,3455021317590,13875655896751,55677180731079

%N a(n) = Sum_{k=0..floor(n/2)} C(n-k,k+2)*3^(n-k-2)*(4/3)^k.

%C In general a(n) = Sum_{k=0..floor(n/2)} C(n-k,k+2)*u^(n-k-2)*(v/u)^k has g.f. x^2/((1-u*x)^2(1-u*x-v*x^2)) and satisfies the recurrence a(n) = 3u*a(n-1)-(3u^2-v)*a(n-2)+(u^3-2uv)*a(n-3)+u^2^v*a(n-4).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (9,-23,3,36).

%F G.f.: x^2/((1-3*x)^2*(1-3*x-4*x^2)).

%F a(n) = 9*a(n-1)-23*a(n-2)+3*a(n-3)+36*a(n-4).

%F a(n) = -(n/4+13/16)*3^n +(-1)^n/80 +4^(n+1)/5 . - _R. J. Mathar_, Dec 16 2024

%Y Cf. A099623.

%K easy,nonn

%O 0,4

%A _Paul Barry_, Oct 25 2004