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%I #12 Sep 08 2022 08:45:15
%S 0,1,6,31,144,637,2730,11467,47508,194953,794574,3222583,13023192,
%T 52491349,211161138,848231779,3403688796,13647040225,54685016022,
%U 219030629455,876994213920,3510591943981,14050213040826,56224387958011
%N a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k+1) * 3^(n-k-1)*(4/3)^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="/A099621/b099621.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 (6,-5,-12).
%F G.f.: x^2/((1-3*x)*(1-3*x-4*x^2)).
%F a(n) = 6*a(n-1) - 5*a(n-2) - 12*a(n-3).
%F From _G. C. Greubel_, Jun 06 2019: (Start)
%F a(n) = (4^(n+2) - 5*3^(n+1) - (-1)^n)/20.
%F E.g.f.: (-exp(-x) - 15*exp(3*x) + 16*exp(4*x))/20. (End)
%t Table[Sum[Binomial[n-k,k+1]3^(n-k-1) (4/3)^k,{k,0,Floor[n/2]}],{n,0,25}] (* or *) LinearRecurrence[{6,-5,-12},{0,1,6},30] (* _Harvey P. Dale_, Dec 13 2012 *)
%t Table[(4^(n+2)-5*3^(n+1)-(-1)^n)/20, {n,0,30}] (* _G. C. Greubel_, Jun 06 2019 *)
%o (PARI) vector(30, n, n--; (4^(n+2)-5*3^(n+1)-(-1)^n)/20) \\ _G. C. Greubel_, Jun 06 2019
%o (Magma) [(4^(n+2)-5*3^(n+1)-(-1)^n)/20: n in [0..30]]; // _G. C. Greubel_, Jun 06 2019
%o (Sage) [(4^(n+2)-5*3^(n+1)-(-1)^n)/20 for n in (0..30)] # _G. C. Greubel_, Jun 06 2019
%o (GAP) List([0..30], n-> (4^(n+2)-5*3^(n+1)-(-1)^n)/20) # _G. C. Greubel_, Jun 06 2019
%Y Cf. A094705, A099622.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Oct 25 2004