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a(n) = Sum_{k=0..n} 3^(n-k) * floor(k/3).
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%I #22 Dec 22 2023 10:34:09

%S 0,0,0,1,4,13,41,125,377,1134,3405,10218,30658,91978,275938,827819,

%T 2483462,7450391,22351179,67053543,201160635,603481912,1810445743,

%U 5431337236,16294011716,48882035156,146646105476,439938316437,1319814949320,3959444847969,11878334543917

%N a(n) = Sum_{k=0..n} 3^(n-k) * floor(k/3).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3,1,-4,3).

%F a(n) = a(n-3) + (3^(n-2) - 1)/2.

%F a(n) = 1/2 * Sum_{k=0..n} floor(3^k/13) = Sum_{k=0..n} floor(3^k/26).

%F a(n) = 4*a(n-1) - 3*a(n-2) + a(n-3) - 4*a(n-4) + 3*a(n-5).

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

%F a(n) = (floor(3^(n+1)/26) - floor((n+1)/3))/2.

%o (PARI) a(n, m=3, k=3) = (k^(n+1)\(k^m-1)-(n+1)\m)/(k-1);

%Y Partial sums of A033139.

%Y Column k=3 of A368343.

%Y Cf. A097137.

%K nonn,easy

%O 0,5

%A _Seiichi Manyama_, Dec 22 2023