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a(n) = Sum_{k=1..n} k^3 * 4^(n-k).
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%I #17 Dec 29 2023 08:05:19

%S 0,1,12,75,364,1581,6540,26503,106524,426825,1708300,6834531,27339852,

%T 109361605,437449164,1749800031,6999204220,27996821793,111987293004,

%U 447949178875,1791796723500,7167186903261,28668747623692,114674990506935,458699962041564

%N a(n) = Sum_{k=1..n} k^3 * 4^(n-k).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (8,-22,28,-17,4).

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

%F a(n) = 8*a(n-1) - 22*a(n-2) + 28*a(n-3) - 17*a(n-4) + 4*a(n-5).

%F a(n) = (11*4^(n+1) - (9*n^3 + 36*n^2 + 60*n + 44))/27.

%F a(0) = 0; a(n) = 4*a(n-1) + n^3.

%o (PARI) a(n) = sum(k=1, n, k^3*4^(n-k));

%Y Cf. A000578, A000537, A213575, A066999.

%Y Cf. A097788, A368525.

%Y Cf. A014825, A368529.

%K nonn,easy

%O 0,3

%A _Seiichi Manyama_, Dec 29 2023